Page 2 of Packet Answers
FORCES PACKET KEY
Problem Set 1: Drawing Free Body Diagrams.
#)
#(
Draw a free body diagram of each situation. Unless otherwise specified, consider frictional forces
as well. (yes, you will need to use a separate sheet of paper)
1. A package, dropped from a plane falling through the air - first FBD below
!$
2. The same package, once it has reached terminal velocity
!!"#
Page 2 of Packet Answers
!!"#
6. A car is coasting to the right and slowing down
N
!!
!$
!$
!$
3. A baseball just hit, flying over second base
7. A child sledding down a frictionless hill
$
!$
N
4. A book being pushed across a table at a constant velocity (considering
friction)
!$
!'
!%&&
!$
8. A child sledding down a hill that does have friction
% &' !'
$
!$
5. A picture hanging on a wall, suspended by two wires (see picture à)
9. A child sledding down a hill being pushed from behind by her sister
1
2
Page 3 of Packet Answers
C. Vector Nature of Forces
% &' !'
$
!$
Read Section 4.4 page 90 C&J Physics
!%&&
Forces can be added to form a resultant, and they can be resolved into x and y components.
10. Bernie is floating in an inflated inner tube along the Deschutes river. The river is flowing at a
rate of 2.0 m/s. Bernie uses his paddle to exert a 50 N force on the raft in a direction of 60o to
the current. How much of his force will actually help move him across the river?
The fraction of 50 N that helps
in the y-direction is the ycomponent of the force.
!+ = ! *,-+ = 50 sin 60 = 43N
F
600
!* = ! )&*+ = 50 cos 60 = 25 N
11. Marco is pulling his sister Estelle in a wagon. He exerts a force of 50N at an angle of 30o to
the horizontal. How much of his 50N effort is going in to pulling the wagon forward? If he
pulls with this force for a distance of 30m, what is his final velocity?
F
300
!* = ! )&*+ = 50 cos 30 = 43N
Is there enough info to calculate the
answer?
3
4
Page 4 of Packet Answers
Page 4 of Packet Answers
E. Using Newton’s Second Law in Problem Solving
14. Elspeth pushes a 20kg box along the floor with a force of 40 N. Her annoying brother is
pushing in the opposite direction with a force of 15N. The friction between the box and the
floor is 5 N.
a. Draw the FBD.
b. Write the net force equation in the two directions of interest.
$
c. Calculate the rate of acceleration of the box in the two directions of interest.
!, or N
To use Newton’s second law, remember these key points:
Þ The NET FORCE is the SUM OF ALL THE FORCES
Þ ««NET FORCE is NOT a type of force. It is a RESULTANT force««
Þ The Greek letter sigma, S, means “sum”. So, S F = FNET – a vector sum
Þ The sum of the forces acting on an object (Fnet or ΣF) determines its change in motion (change in
velocity) such that ΣF=ma
Þ Look for NET FORCE in TWO directions to understand the motion of an object
• Net force along the surface on which the body is moving
• Net force PERPENDICULAR to the surface on which the body is moving
Þ If the sum of the forces is zero, there is no change in motion
0
. !* = !-./&-01 − !' − !2#3
. !+ = $ − !$
!$
Or (weight)
∑ !* = 23* from Newton’s second Law
!-./&-01 − !' − !2#3 = 23*
Example: A block of mass m is being pulled along a horizontal frictionless surface by a force P.
4
(
12. What is the net force needed to accelerate a 1000 kg car at a rate of ) g? (Ans: 5000N)
a. Does this net force have to be one force or could be many forces? Could be many; it is the
vector sum of all forces
b. Give an example of a scenario of multiple forces on a car such that the net
force
is what you calculated in the problem. One example: engine thrust
N
(Fapp) forward 8000N, friction (Ff) opposing motion (backward) 2000N
!%&&
and air resistance (!!"# ) backward 1000N. Weight (Fg) and normal (N) ""
cancel (9800N each).
c. Draw a FBD to represent your scenario.
d. Write the net force equation in the two directions of interest.
S Fx = Fapp - Ff – !!"#
S Fy = N - Fg = 0
!!"#
. !+ = $ − !$
78595(9
!-./&-01
!$
3* = !"#$!%& 6 ' ()* =
= 1 2/* )
)8
If the problem also asks you to calculate Normal force, you would use the Y net force equation
. !+ = $ − !$ = 0 78)39*8 3+ = 0
$ = !$ = 196 N
15. An object’s weight can be determined using Newton’s second law. It is simply the Force (in
Newtons) that an object is experiencing due to the gravitational acceleration of the earth. (9.8
m/s2). What is the weight of a 65kg woman at the surface of the Earth? What would her
weight be if she were on Mercury where the acceleration of gravity is 3.7 m/s2?
!$
13. What net force is required to bring a 1500kg cart to rest from a speed of 100 km/hr within a
distance of 55 m? (Ans: -1.1 x104 N)
a. Give an example of a scenario of multiple forces on a car such that the net force is what you
calculated in the problem. Car is braking, moving right and applying a force left, opposite
motion. Friction also acts to the left, opposite motion.
b. Draw a FBD to represent your scenario.
c. Write the net force equation in the two directions of interest.
. !* = − !' − !!&&
54 54
!'
Givens:
2 = 20 ;<
!-./&-01 = 40 $
!2#3
!2#3 = 15 $
!' = 5 $
!$ = 2< = 20(9.8) = 196 $
3+ = 0 2/* )
3* = ?
m = 65 kg; On earth acceleration due to gravity = 9.8 m/s2
Weight = !$ = 2< = 65(9.8) = 637 $ on Earth
Weight = !$ = 2< = 65(3.7) = 240.5 $ on Mercury
N
""
!!$$
!$
5
6
Page 5 of Packet Answers
Page 5 of Packet Answers
F. Understanding the Normal Force
18. What is the normal force acting on a 8-kg mass which is being pushed at a constant
velocity of 5 m/s by a force of 46 Newtons along a rigid handle that makes an angle
of 31º to the horizontal?
Givens:
2 = 8 ;<
. !* = !* − %
∑ !* = 23* from Newton’s second Law !$ = 2< = 8(9.8) = 78.4 $
%
! = 46 $
$
!* − % = 23* = 0
+ = 318
%
H* = 5 2/*
!$
. !+ = !+ + $ − !$
3+ = 0 2/* )
!
∑ !+ = 23+ from Newton’s second Law 3* = 0 2/* )
!$
$ − !$ − !+ = 0
!+ = ! *,-+
Solving for N
$ = !$ + !+ = 78.4 + 46J,- 31 = 78.4 + 23.69 = 102.09N ~ 102N
Read Section 4.8 page 97 of C&J Physics
The normal force is the name given to a supporting force. For
example, you stand on the ground and you exert a force on the ground. But, according to
Newton’s third law, the ground is also exerting a force on YOU! (Otherwise it would be an
unbalanced situation and you would be falling toward the center of the Earth!). That supporting
force, provided by the ground is called the normal force. In this case it is equal in magnitude and
opposite in direction to your weight. However, sometimes the normal force is less than your
weight (think quicksand) or more than your weight (bouncing up on a trampoline). In these cases
you are no longer in equilibrium.
Directions: On a separate sheet of paper: Show your work
• Draw a FBD
• Write the Net Force Equations in TWO directions,
• Find normal force N = ?
$ − !$ = 0
If she is accelerating upward, SFy > 0 so N – Fg > 0 and N > Fg
A bathroom scale measures normal force, so in this scenario the reading would be greater than when
she is at rest and normal force equals her weight (force of gravity).
$
20. If she is accelerating upwards at a rate of 2.2m/s2, what is the reading on the
bathroom scale?
∑ !+ = $ − !$
and ∑ !+ = 23+
so set them equal and N – Fg = may
N – 637 = 65(2.2)
N = 780N
!$
17. What is the normal force acting on a 8-kg mass which is being pulled at a constant
velocity of 5 m/s by a rope having a tension of 46 Newtons at an angle of 31º to the
horizontal?
∑ !* = 23* from Newton’s second Law
!* − % = 23* = 0
. !+ = !+ + $ − !$
Givens:
2 = 8 ;<
!$ = 2< = 8(9.8) = 78.4 $
! = 46 $
+ = 318
H* = 5 2/*
3+ = 0 2/* )
3* = 0 2/* )
!+ = ! *,-+
$
!
%
!$
$
+ !
!$
∑ !+ = 23+ from Newton’s second Law
!+ + $ − !$ = 0
Solving for N
$ = !$ − !+ = 78.4 − 46J,- 31 = 78.4 – 23.69 = 54.7N ~ 55N
Givens:
2 = 65 ;<
!$ = 2< = 65(9.8) = 637 $
3+ = 2.2 2/* )
21. As she approaches the top floor the elevator begins to slow down. How does the reading on
the bathroom scale change?
• ∑ !+ = $ − !$
• ∑ !* = 0
!+
%
!$
. !* = 0
Givens:
2 = 8 ;<
!$ = 2< = 8(9.8) = 78.4 $
3+ = 0 2/* )
3* = 0 2/* )
$ = !$ = 78.4 N
. !* = !* − %
$
. !+ = $ − !$
∑ !* = 0 No forces in the x-direction
∑ !+ = 23+ from Newton’s second Law
!
$ =19.!$Consider
= 78.4 Na 65kg woman standing in an elevator that is accelerating upwards. If she
was standing on a bathroom scale, would the reading on the scale be greater or less
than when she is at rest?
16. What is the normal force acting on a 8-kg mass that is at rest on a horizontal surface?
. !+ = $ − !$
$
!*
$
!$
If she is accelerating downward (slowing down), SFy < 0 so N – Fg < 0 and N < Fg
A bathroom scale measures normal force, so in this scenario the reading would be less than when
she is at res.
22. If she decelerates at a rate of -2.2 m/s2, what will the reading on the bathroom scale be?
∑ !+ = $ − !$
Givens:
and ∑ !+ = 23+
2 = 65 ;<
so set them equal and N – Fg = may
!$ = 2< = 65(9.8) = 637 $
N – 637 = 65(-2.2)
3+ = −2.2 2/* )
N = 494N
$ = !$ = 78.4 N
7
8
Page 5 of Packet Answers
220 = 216ay ay = 1.02 m/s2
23. Jack and Jill are standing on friction free ice. Jack has a mass of 82 kg and Jill’s mass is 48kg.
Jill pushes on Jack with a force of 45 N due east. What is the acceleration (magnitude and
direction) of each skater? If Jill’s push is of constant magnitude and lasts for 2.3 seconds,
what is the final velocity of each of the skaters?
We have to break this into two force diagrams: one for Jack, and one for Jill. FJill represents the
force of Jill’s push on Jack, and vice versa.
Jill
25. After a lead-off single in the 8th inning, Earl makes an effort to steal second base. As he hits
the dirt on his head first dive, his 73.2 kg body encounters 249 N of friction force. Construct a
free body diagram depicting the types of forces acting upon Earl. Then determine the net force
and acceleration.
Jack
N
N
∑ !+ = $ − !$ = 0
!'
N
∑ !* = −!' = 23*
!<%=>
!<"..
!$
S Fx = -FJack = max
S Fy = N - Fg = 0
m = 48kg
Newton’s 3rd law tells us FJack = -FJill = 45N
-FJack = max
-45 = 48a
a= -0.94 m/s2 or 0.94 m/s2 west
t = 2.3s
a= -0.94 m/s2
v0 = 0
vf = ?
vf = v0 + at
vf =0 + -0.94(2.3) = -2.2 m/s west
Page 6 of Packet Answers
!$
Givens:
2 = 73.2 ;<
!$ = 2< = 73.2(9.8) = 7173.6 $
!' = 249 $
−!' = 23*
rewriting in terms of ax:
ax = -Ff / m = -249/ 73.2 = -3.40 m/s2 (decelerating)
!$
26. More Elevator Problems… (conceptual)
What would happen to the reading if… (remember the scale reading is normal force)
a) You pushed up on the ceiling? increase
b) You pushed down on a sink? decrease (the sink also pushes up on you…)
c) You pushed sideways on the wall? No change
d) You jumped? decrease
e) You were in an elevator not moving? no change
f) You were in an elevator accelerating upward? increase
g) You were in an elevator accelerating downward? decrease
i) You were in an elevator with constant speed? no change
S Fx = FJill = max
S Fy = N - Fg = 0
m = 82kg
FJill = 45N
FJill = max
45 = 82a
a = 0.55 m/s2 east
t = 2.3s
a= 0.55 m/s2
v0 = 0
vf = ?
vf = v0 + at
vf =0 + 0.55(2.3) = 1.3 m/s east
G. Tension Force
Read Section 4.10, C&J Physics
24. The shipment of the new physics supplies has arrived. They are placed on the freight elevator
and transported up to the third floor for delivery to the physics rooms. The free body diagram
at the right depicts the forces acting upon the freight elevator as it begins its ascent through the
elevator shaft. Use force values to determine the net force, the mass and the acceleration of the
elevator. The values of the individual forces are: !: = 2340 N; !$ = 2120 N; !;3#6( =
!;3#6) = 276 N
S Fx = Fnorm2- Fnorm1 = 0 since it is not moving in the horizontal direction
S Fy = FT – Fg = may
Tension usually arises in the use of ropes or cables to transmit a force. Consider a
block being pulled by a rope. The person doing the pulling at one end of the rope is
not in contact with the block, and cannot exert a direct force on the block. Rather a
force is exerted on the rope, which transmits that force to the block. The force
experienced by the block from the rope is called the tension force.
Almost all situations you will be presented with in classical mechanics deal with massless ropes or
cables.
Fg = mg
2120 = m(9.8) so m=216.32 = 216kg
now to solve for ay: S Fy = FT – Fg = may
2340 – 2120 = 216(ay)
9
10
S Fy = T – Fg = may
27. The man pictured to the right pulls a 5.0kg crate so that it accelerates at a rate of
2.0m/s2, what is the tension in the rope? Draw a FBD.
#
Fg = 29.4N
m = 3.0kg
ay= 0 m/s2
v = 0.50 m/s (irrelevant!)
T=?
S Fy = T – Fg = may
solving for Fg = mg = 5.0(9.8) = 49N
Page 6 of Packet Answers
solving for T : T – Fg = may
T = Fg + may = 49 + (5.0)(2.0) = 59N
!$ 28.
Consider a 3.0 kg ball hanging from a rope. If the ball and
the rope are at rest, what is the tension in the rope? Draw a FBD.
#
!$
T – Fg = may
T – Fg = may = 0 m/s2
T = Fg = 29.4N
S Fy = T – Fg = may = 0 (at rest, no acceleration)
solving for Fg = mg = 3.0(9.8) = 29.4N
solving for T : T – Fg = 0
Fg = T = 29.4N
#
!$
29. If the rope is pulling the ball upwards at a rate of 1.8 m/s2, what would be the tension in
the rope? Draw a FBD.
S Fy = T – Fg = may
S Fy = T – Fg = may
Fg = 29.4N
m = 3.0kg
a = 1.8m/s2
T = Fg + may
T = 29.4 + 3.0(1.8)
T = 34.8N or 35N
#
!$
30. If the ball is being lowered, such that the tension in the rope is only 20.0 N, at what rate is
it accelerating? Draw a FBD. Write the net force equation without numbers.
S Fy = T – Fg = may
S Fy = T – Fg = may
Fg = 29.4N
m = 3.0kg
T = 20.0 N
ay= ?
ay = (T - Fg )/m
a = [20.0 - 3.0(9.8)]/3 = -9.4/3 = -3.13 ~ -3 m/s2 (down)
#
!$
31. If the ball is being lowered with a constant velocity of 0.50 m/s, what is the tension in the
rope? Draw a FBD. Write the net force equation without numbers.
11
12
Read Section 4.4 page 90 C&J Physics
The equation for Net force now looks like this: Σ! = O + $
Because we can think of the weight as being made up of two components, we can write it like this:
Σ! = !? + !∥ + $
Forces can be added to form a resultant, and they can be resolved into x and y components.
If you look at the second drawing, you can see that F^ and the Normal force are equal in
32. And now, the famous…Box on a ramp!!! (INCLINED PLANES)
Draw a free body diagram of a box on a ramp.
Þ Show all of the forces with solid arrows.
Þ On a second diagram - resolve weight into components that are parallel
and perpendicular to the ramp. Draw these vectors as dashed arrows.
Þ Write the formula for calculating the components
So, Σ! = 2<J,-Θ = (54kg) S9.8 /+T J,- 15 = 137$
H. More on Vector Nature of Newton’s Second Law
magnitude but are acting in opposite directions, so they cancel each other out. Therefore, in this
situation,
Σ! = !∕∕ .
We calculated (in the box problem above) that the parallel component,
!∕∕ = 2<J,-Θ.
6
Now we know Σ! = 23, so 3 =
B4
6
(CDE
= 97>$ = 2.537 ~ 2.5 m/s2
Given vH8 = 0, 3 = 2.5 2⁄* ) , 3-] ^ = 20, _8 3'8 *&`H,-< %&' H'
H' ) = H8 ) + 23^ so H' = aH8) + 2 3^
= a0 + 2 (2.5)(20) = 10.07 ~ cd e/f
Note: friction is ignored in
this solution; if you included
it, it should be directed
uphill in this case.
More Problems
(On a separate sheet of paper - draw FBD, write net force equations, solve for unknown, compute
the unknown) (PGUESSS……)
33. Horace is sledding down an ice covered hill inclined at an angle of 15o
to the horizontal. If Horace and the sled have a combined mass of 54kg
and the hill is 20 meters long, how fast will he be going when he reaches
the bottom of the hill?
34. In a clever attempt to foil the enemy, James Bond (mass 85 kg) skis down a hill that makes a
23.0° angle and is 15 meters in length. He then skis on a level surface for 10 meters until he
skis off of a cliff that is 6 meters high and lands in a moving snowmobile. The snowmobile
travels at a constant rate of 2 m/s in a direction parallel to James’ path of descent. If James and
the snowmobile start at the same point in time, how far from the base of the “catch point” must
the snowmobile start?
This problem is all about time.
If you can figure out how much time it takes
James Bond to get to the “catch point” you can
use the rate equation to calculate the distance
from the point that the snowmobile must be.
Break James’s story into 3 different parts. The
hill, the level stretch and the projectile portion.
Part 1: Hill Given:
Mass (which is irrelevant)
Θ = 23.0o
x = 15m
vo=0
From the box on a ramp problem we know that it is only the downhill component we care about,
so
Since Friction is not mentioned here, we assume there is none (Gotta love that friction free ice!)
Free Body Diagram looks like this:
ΣF//=mgSinθ. And ΣF=ma so,
B4
6$G";H
6
3 = 6 = 6 = <*,-g = (9.8 /+ )(J,- 23o) = 3.8 m/s2
Now we use kinematics to find the amount of time he is on the slope as
Assuming he starts at rest, vo = 0 so
1
^ = H3 h + 3h )
2
Next I break the weight vector, Fg into two
perpendicular components, one parallel to the hill
)*
(F// ) and one perpendicular to the hill (F^)
)((96)
Solving for t, i % = h, h = iC.L 6
M/ +
13
=2.8 seconds
14
In order to do the second part, we need to find the velocity that James is traveling at the bottom of
the hill. Again, bring back the kinematics…
2
2
H = H3 + 3h = 0 + S3.8 ) T (2.8*) = 10.6
*
*
Now onto part 2: The level strip
Given:
vo=10.6 m/s
x = 10 m
No friction so a constant velocity situation
^ = Hh *& … h =
^
102
=
= 0.94*
H 10.6 6
/
Part 3: The drop. James is basically a projectile here. All we care about is his time for fall, and,
all object fall at the same constant velocity (remember?) so just like the projectile problem in your
lab, all that really matters here is the y component or the height of the hill.
Given:
h = 6.0m
vo = 0
So,
The free body diagram is above. The force pulling her forward can be resolved into its x and y
components. (Diagram 2) Because Penny is accelerating in a forward direction, not the y direction
the vertical forces, must sum to zero such that, ΣFy=Fg + N + Fy = 0. So, for the purposes of this
problem, we don’t need to worry about the Y component. However, DO look at the equation of
how they are calculated.
It is only the force in the x direction that is accelerating her, so from diagram 2 we can see,
ΣFx=Fx + Ff. so, ΣFx = F1Cosθ + Ff . Carefully watching our signs (forward is + and backwards
is - ) we substitute in such that, Fx=(500N)(Cos35o) + (-50N) = 360N.
Sorry! Penny has no mass in this problem. You will need to leave your answer as an expression,
!( m&*+ − !'
3=
2
36. A barge is being pulled along the two tug boats pulling on tow ropes that make and angle of
60.0o with each other. The Barge has a mass of 1.5 x 108 kg and each boat pulls with a force
of 1.5 x 105 N. In addition, the barge’s engine provides a forward thrust of 7.5 x 104 N and the
water provides a backwards resistance of 4.0 x 104N. Find the rate of acceleration of the
barge.
Given: θ = 30.0o,
M=1.5x108kg,
T1=T2=1.5x105N,
F = 7.5 x 104 N
R = 4.0 x 104N
1
k = H3 h + 3h )
2
2k
2 (6.02)
h= l =l
= 1.05*
3
9.8/6+
In this situation, the weight of the barge will be equal and opposite to the Normal force, so they
will cancel. We will not draw anything in the direction coming in-out of the paper. We draw a top
view and focus on the forces up and down in the plane of the paper (x and y).
Now we can calculate how long James’s whole trip took
t = (2.8s) + (0.94s) + (1.05s) = 4.79 sec or, to 2 sig figs, 4.8 seconds.
Below is the free body diagram and diagram 2, where the tensions in the ropes have been resolved
into the x and y components.
Final step: Calculate the distance where the snowmobile must start from
Given
vo = 2.0 m/s, t = 4.8 sec
x=vt = (2.0m/s)(4.8s) = 9.6m
35. Penny the Penguin is riding on a sled as shown in the picture above to the right. She is being
pulled forward by a rope that makes an angle of 35o with the horizontal. The force on the rope
is 500N and the force of friction is 50N. At what rate does Penny accelerate?
FBD
Diagram 2
FBD
Diagram 2
Look at diagram 2. By resolving T1 and T2 into x and y, we now have only forces in the two
perpendicular directions. We can deal with them separately.
The sum of the forces in the y direction can be quantified in an equation like this:
ΣFy = T1y + T2y = 0
The sum is zero because there is no motion in the y direction.
The sum of the forces in the X direction (which is the total story here..) can be written as…
ΣFx = F + T1x + T2x + R
ΣFx = F + 2(Tcosθ) + R
Taking direction into account (forward is + and backwards is -)
ΣFx = (7.5 x 104N) + 2(1.5 x 105N(cos 30o)) + (-4.0 x 104)N) = 2.95 x 105N
15
16
Now the easy part… F =ma so 3 =
4
6
=
).N9 × (8, ,
(.9 × (8. >$
= 1.97 × 105C
6
/+
or (to 2 sig figs,) 2.0 x 10-3 m/s2
Friction Problems! On a separate sheet of paper - draw FBD, write net force equations, solve
for unknown, compute the unknown.
41. A 38 kg sled and rider are at rest on a flat patch of snow. The coefficient of static friction
on the snow, μs, is 0.350. How hard can the sled be pushed, horizontally, before the sled
begins to move?
I. Friction Force & Coefficient of Friction
Read Section 4.9, C&J Physics
ΣFy = may
ΣFy = FN – Fg = 0
ΣFx = F + 2(Tcosθ) + R
37. What is Friction? What are the various types of friction:
Friction is a parallel component of force that opposes motion. Types of
friction include static and kinetic friction.
38. What causes friction?
The force of a surface acting on the object resting or moving on the
surface.
39. What would life be like without friction?
We wouldn’t have cars, tug-o-war… it would be nearly impossible to keep anything in one place
(put a book on a table, or table on the floor, and it will be hard to stay still)… what else?
40. Write the equation and identify each term in it that is used to calculate….
a. Kinetic Friction
Fk = μkFN
Force of kinetic friction = coefficient of kinetic friction * normal force
ΣFx = max
ΣFx = Fapp– Ff = 0
By setting a=0, we can find the limit of
the Fapp beyond which it will accelerate
FN
!'
!%&&
!$
Unknown: Fapp = ?
Givens: μs = 0.350
M = 38kg
From ΣFy = FN – Fg = 0, we know FN = Fg and Fg = mg, so FN = mg
So, FN = (38)(9.8)= 372.4N
For the equation:
ΣFx = Fapp– Ff = 0
Fapp= Ff and Ff = μsFN
So, Fapp = μsFN
Solve:
Fapp= (0.350)(372.4) = 130.34N ~ 130.N
b. Static Friction
Fs = μsFN
Force of static friction = coefficient of static friction * normal force
*this equation solves for the maximum possible force
of static friction; Fs can be anything from zero to this
calculated force.
42. The alarm at a fire station rings and an 86-kg fireman, starting from rest, slides down a pole to
the floor below (a distance of 4.0m). Just before landing his speed is 1.4 m/s. What is the
magnitude of the kinetic frictional force exerted on the fireman as he slides down the pole?
To solve this one, we will need to use our FBD and net force equation (y dimension only) to derive
an equation for the force of kinetic friction. We will also need to solve for the acceleration.
c. Fluid Friction Force of friction opposition motion through a
fluid (liquid or gas), such as water.
ΣFy = Fg– Ff = may
**We will call down positive since we know the acceleration is downward
Ff
Ff = Fg - may
We know Fg = mg, so we can sub in and get Ff = mg - may
Caution: Two concepts (compare & contrast)
• force of friction
• coefficient of friction
17
Now to solve for ay, with our kinematics PGUESS:
Givens:
v0 = 0
vf = 1.4m/s
∆y = 4.0 m
Unknown: a = ?
Equation & Solve for a:
H' )
H' ) = H8 ) + 23^, so 3 =
2Δk
Now we can sub in this expression for a in our Ff expression:
H' )
!' = 2< − 23 → !' = 2< − 2
2Δk
and Ff = (86)(9.8) – (86)(1.4)2/8.0 = 821.87 ~ 820N
!$
18
43. Remember our box on a ramp? Now let’s add friction. A 2.5 kg wooden
box slides down a wooden ramp (μk = 0.30). The angle that the
ramp makes with the horizontal is 25o. What is its rate of
acceleration?
!'
ΣFy = N - Fg = may = 0, so N = Fg
ΣFx = - Ff = max
!$*
!$+
Perpendicular component (using y subscripts)
ΣFy = N - Fgy = may = 0, so N = Fgy = mgcosθ
46. A 92 kg baseball player slides into second base. The coefficient of kinetic friction between
the player and the ground is μk = 0.61.
$
!$
a. What is the magnitude of the frictional force?
Ff = µkmg = (0.61)(92kg)(9.8) = 550N
Parallel component (using x subscripts)
4/0 54'
ΣFx = Fgx - Ff = ma so 3 = 6
We know Ff = μkN and N = mgcosθ so we can say Ff = μkmgcosθ
b. If the player comes to rest after 1.2s, what is his initial speed?
From our horizontal net force equation, we have -Ff = max
So, ax = -Ff / m = -µkmg / m = -µkg
Givens:
vf = 0
t = 1.2 s
a = µ kg
We are solving for v0 (unknown)
!! = !" + $%
!" = !! − $% = !! − (−)# *%)
since vf = 0, !" = )# *%
And we can simplify Fgx = Fgsinθ = mgsinθ
Therefore, 3 =
4/0 54'
6
=
6$/";H5P1 6$=3/H
6
and cancelling out m, a = gsinθ - μkgcosθ
Now we just need to sub in our values
a = gsinθ - μkgcosθ = 9.8sin(25) -0.30(9.8)cos(25) = 1.4771 ~ 1.48 m/s2
Note that the mass of the box does not affect the final answer!
v0 = (0.61)(9.8)(1.2) = 7.174 ~ 7.2 m/s
44. Now let’s do it without numbers. Write a mathematical expression for the acceleration of an
object down a ramp in terms of θ and μk.
Since I solved 43 above by first deriving an expression for a, we have already done this one! Using
the same free body diagram and force equations, we get:
FN
For the FBD and force equations we will use the same diagram from #41 but remove
the applied force.
ΣFy = N - Fg = may = 0, so N = Fg
ΣFx = - Ff = max
Ff = µkmg
a. What is the force of kinetic friction?
Ff = μkN and since it is horizontal, N = Fg = mg, so Ff = μkmg = 0.05*38*9.8 = 18.62 ~ 19N
b. What is the NET force on the sled horizontally?
ΣF = Ff = µkmg = (0.050)(38kg)(9.8) = 18.6N~19N
c. What is the acceleration of the sled?
∑ ! = 23* (1)
∑ ! = !' = q> 2< (2)
Putting (1) and (2) together
∑4
P 6$
3* = 6 = 16 = q> < = (0.05)(−9.8) = −0.49 2/* )
d. How far did the sled travel?
H') = H8) + 23∆^
=
85(S
)(58.7N)
=
(S
8.NL
3=
−!'
550 $
=
= 5.98 2/* )
2
92
!" = !! − $%
!" = 0 − (−5.98)(1.2) = 7.2 4/6
When the sum of the forces acting on an object is zero the object is in an inertial state. That is, it
is either at rest or traveling at a constant velocity in a straight line.
!'
)%
Alternatively, using our answer to part
a, solve for acceleration and then plug
directly in to the kinematics equation:
Read Section 4.11 C&J Physics
45. The (38kg) sled begins cruising at 4 m/s, horizontally, until it stops. How far did it travel
before stopping? (qk = 0.050)
R'+ 5 R2+
!$
J. Equilibrium Applications:
a = gsinθ - μkgcosθ
∆^ =
FN
!'
!$
47. Give 3 examples of objects in equilibrium that are at rest.
Answers will vary.
48. Give 3 examples of objects in equilibrium that are not at rest.
Answers will vary. Object must be at constant non-zero velocity to be in equilibrium and not at
rest.
For an object at rest, we can break down the forces into those acting in the x direction and those
acting in the y direction. If there is no motion, the sum of the forces in the x direction must be
zero AND those in the y directions will sum to zero as well.
= 16.32 2 ~ 16 m
19
20
50. A 3.5 kg sign hangs from two wires. Each wire makes an angle of 37o with the
ceiling.
Draw the free body diagram
Find the tension in each side of the wire supporting the sign
Problem Set: Mechanical Equilibrium (Statics)
49. Ginny the gymnast hangs from a variety of positions below. Her weight is 300N when she is
standing on the ground. In each of these situations Ginny is at rest – The forces acting on her
are balanced. For each of the situations, write in the reading of the spring scales.
370
0
T1y 53
370
T1
T2
T2y
T1x
T2x
:34
= *,- 37
:3
Fg
(4)
#(*
= )&* 37
#(
Since the sign is at rest net force on it is zero.
∑ !* = #)* − #(* = 0
(1)
∑ !+ = #)+ + #(+ − !$ = 0 (2)
#)+ = #(+ (3)
because the angles are equal, so T1 = T2 and their y-components are equal also
Substituting for T2y in (2)
#(+ + #(+ − !$ = 0
Fg = (3.5)(9.8) = 34.3 N
2#(+ − 34.3 = 0
2#(+ = 34.3
#(+ = 17.15$
From eqn (4)
T1 = T1y/sin 37 = 17.15/sin 37 = 29 N
T1 = T2 = 29N must be applied at an angle of 370 to support the sign
51. Pippi has a tire swing that hangs from a tree branch in her yard. She and the swing have a
combined weight of 808N. Her brother, Pepe pulls Pipi back far enough for her to make and
angle of 30.0 degrees with the vertical.
a) Draw the Free body diagram that illustrates the forces acting on Pippi.
b) Find the tension in the rope supporting Pippi
<300 N
150 N if her
weight is
evenly
balanced
between the
ring and the
ledge.
T1y
Since the sign is at rest net force on it is zero.
∑ !* = !%&& − #(* = 0
(1)
T1
300
600
T1x
Fapp
Fg
:34
:3
= *,- 60
(3)
∑ !+ = #(+ − !$ = 0 (2)
Fg = 808 N
#(+ − 808 = 0
#(+ = 808
From eqn (3)
T1 = T1y/sin 60 = 808/sin 60 = 940 N
#(*
= )&* 60
#(
21
22
54. A 110 kg object is supported by two ropes attached to the ceiling. What is the tension, T in the
right-hand rope?
52.
A Flower pot with a weight of 48 N is hung above a window by three ropes, each
making an angle of 15.0 degrees with the vertical as shown. What is the tension in each
rope supporting the flower pot?
T1y
Since the sign is at rest net force on it is zero.
∑ !* = 0
(1)
T1 = T2 = T3 because each rope makes the same 15 degree angle with the vertical
∑ !+ = 3#(+ − !$ = 0 (2)
Fg = 48 N
3#(+ − 48 = 0
3#(+ = 48
48
#(+ =
= 16$
3
T1
150
750
T1x
Fg
:34
:3
= *,- 75
250
25o
T2
600
T1y
60o
T1
Right hand rope
T2y
T1x
T2x
Fg
From eqn (3)
T1 = T1y/sin 75 = 16/sin 75 = 16.67N ~ 17N
(3)
#(*
= )&* 75
#(
#(+ = #( *,- 60
#(+ = 0.87 #(
#)+ = #) *,- 25
#)+ = 0.42#)
#(* = #( )&* 60
#(* = 0.5 #(
#)* = #) )&* 25
#)* = 0.9#)
CORRECT DIAGRAM
Since the sign is at rest net force on it is zero.
∑ !* = #)* − #(* = 0
(1)
#)* = #(*
0.9#) = 0.5 #(
(0.9#) )/0.5 = #(
53. Two forces, 12 N west and 5.0 N north, act on an object. What is the direction of a third force
that would produce static equilibrium?
Resultant = √12) + 5) = 13
k
+ = #3-5( S T = #3-5( (5/12) = 22.88
^
+
To produce static equilibrium the third force must be
equal but opposite to the resultant. So it must be 13N
at an angle of 22.8 degrees below x-axis – the
GREEN arrow.
∑ !+ = #)+ + #(+ − !$ = 0 (2)
#)+ + #(+ = !$
0.42#) + 0.87 #( = 1078 $
0.9#)
0.42#) + 0.87
= 1078 $
0.5
0.42#) + 1.566#) = 1078 $
1.986#) = 1078 $
#) = 1078/1.986 = 542.8 N ~ 540 N
Frames of Reference:
Read Section 4.2, C&J Physics
55. What is an inertial frame of reference? A frame of reference in which Newton’s 1st law
applies. The opposite would be an accelerated frame of reference.
56. What was Galileo’s thought experiment that demonstrated that all motion is relative?
Take your pick: http://practicalphysics.org/galileos-thought-experiment.html or
http://physicscentral.com/explore/plus/galilean-relativity.cfm
23
24
Forces Packet Page 11 #s 57-65
K. Newton’s Third Law: Answer Table Next Page
Whenever one object exerts a force on a second object, the second object exerts an equal and
opposite force on the first body.
CLAIM
57.
On a separate sheet of paper answer the following questions using CER:
• Claim: a statement about the solution to a problem - what you think you know
• Evidence: scientific data, principle, law, equation, graph to support your claim
• Reasoning: justification which shows why the data/principle supports your claim
EVIDENCE
The reaction force to our
weight is the force with
which we pull the earth.
REASONING
Newton’s 3rd Law:
For every action, there is an equal and opposite reaction and they
act on different bodies. Following Newton’s Third Law, since
weight is the earth’s pull on your body, your body pulls on the
earth.
The Normal force is NOT the reaction force because the normal
force is created by the supporting surface pushing on you.
57. Your weight is the result of the gravitational force of the earth on your body. Describe the
corresponding reaction force.
58. If you step off a ledge, you accelerate noticeably toward the Earth because of the gravitational
interaction between you and the earth. Does the Earth accelerate towards you as well? Explain.
Person
59. When a high jumper leaves the ground, what is the source of the upward force that accelerates
her? What force acts on her once her feet are no longer in contact with the ground?
60. A diver dives off of a raft - what happens to the diver? The raft? How does this relate to
Newton's Third Law?
Earth
58.
61. A tennis racquet hits a tennis ball. Why doesn't the racquet swing backwards when the
ball hits it? (Shouldn't it swing back because of action-reaction forces?)
U!"#$%& !())$ %& "*#+,
− WℎYZ [\]^_ _`YZWZ _a_b cℎ_b
d\e f]_ ZeZg_bh_h Yb Wℎ_ fY]
!$
62. What action-reaction forces are involved when a rocket engine fires? Why doesn't a
rocket need air to push on?
63. What forces are acting on a book sitting on a table? Are action-reaction forces involved
in this situation?
If you accelerate toward the
earth due to the
gravitational interaction
between you and the earth,
that same attraction causes
the earth to accelerate
toward you, even though
you do not feel it.
If you did this,
you’d see that
you do, in fact,
accelerate
down toward
the earth.
64. If two people each standing on a scooter board push off of each other what happens
(relate to Newton's Third Law)? How would the distance moved by the scooter boards
compare if one person had a lot more mass than the other person? If a person standing
on a scooter board pushes off of a wall, what happens? Can this situation be explained in
terms of Newton's Third Law (action-reaction)?
65. How is shooting a shotgun related to Newton's Third Law? Why does a rifle have less
"kick" than a shotgun?
Newton’s 2nd Law:
∑ ! = 23
Newton’s 3rd Law:
For every action, there is an equal and opposite reaction and they
act on different bodies
!$ = −!&-#/3; &T../ 01- -%#01
∑!
3-%#01 =
2-%#01
−!&-#/3; &T../ 01- -%#01
=
2-%#01
−!$
=
2-%#01
Since mass of earth is very large acceleration is very small
59.
Apply as many of Newton’s Laws as you to explain the following:
66. How do seatbelts keep you safe?
67. Explain why whiplash occurs most often in rear-end auto accidents.
25
The high jumper pushes
down on the ground which
pushes back with an equal
but opposite force.
When she
leaves the
ground, she
has an upward
velocity.
Something
must initially
accelerate her
upward.
Newton’s 3rd Law
For every action, there is an equal and opposite reaction and they
act on different bodies and they act on different bodies
When the high jumper leaves the ground she is pushing down of
the ground with her feet and the ground is pushing up on her feet
at the same time, pushing her into the air. Once she is in the air,
the only force that acts on her is gravity which attracts her to the
earth and air resistance which slows her down from Newton’s
Second Law
26
60.
The diver will propel
The diver will
forward while the raft will
propel forward
float backwards because the while the raft will
forces between the two are float backwards.
balanced.
Newton’s 3rd Law
For every action, there is an equal and opposite reaction and
they act on different bodies.
When you jump off the raft, you apply a force to the raft
which exerts an equal but opposite force back on you.The
raft moves "backwards" in the water because the jumper
pushes on the raft. The action force is the diver pushing off
of the raft, and the reaction force is the raft pushing back on
the diver (causing the diver to go forward and into the
water).
61.
The ball and racquet do
form an action-reaction pair
but there must be another
force that prevents the
racket from swinging
backward.
Newton’s 3rd Law
For every action, there is an equal and opposite reaction and
they act on different bodies
62.
The action-reaction pair
must be the gases produced
by the engine and the
rocket.
The racket
continues moving
in the same
direction even
after hitting the
ball. You can feel
your muscle
tighten as you hit
the ball.
A rocket being
launched involves
the rocket moving
upward while its
exhaust is sent
downward from
the rocket.
64.
Newton’s 2nd Law
∑ ! = 23
both act on the book. They
do cancel each other out
and hence the book is
stationary
does it move.
Two forces act on the book - Normal Force and Gravity.
Evidence for its
normal force could 3 = ∑ 4 = 0 *,-)8 $ = ! on a flat surface
$
6
be by putting it on
a scale; evidence
for its weight
(force of gravity)
could be collected
by measuring its
weight.
The two people move in
exactly opposite directions
with accelerations inversely
proportional to their mass
because their forces are
equal.
The two people
move in exactly
opposite
directions; the
person with
greater mass won’t
accelerate as
quickly nor move
as far.
The racquet does not accelerate backward because the force
of your arm prevents it from doing so. The action force is
the ball hitting the racquet, which your arm absorbs; the
reaction force is the racquet pushing back on the ball. If your
arm is not strong enough there is a recoil of the tennis
racket. The acceleration of the racquet is small because it’s
mass is large.
!<%=> 3; <"..
Ja
Jill
!<".. 3; <%=>
Newton’s 3rd Law
For every action, there is an equal and opposite reaction and
they act on different bodies
The engine producing the gases is the action force and the
gases pushing back on the rocket are they reaction force.
Since this action-reaction pair exists and propels the rocket
forward,
1. A space rocket obviously doesn't go anywhere unless you start its engine. As
Newton said, still things (like rockets parked on launch pads) stay still unless
forces act on them (and moving things keep moving at a steady speed unless a
force acts to stop them).
2. Newton said that when a force acts on something, it makes it accelerate (go
faster, change direction, or both). So when you fire up your rocket engine, that
makes the force that accelerates the rocket into the sky.
3. Rockets move upward by firing hot exhaust gas downward, rather like jet
planes—or blown-up balloons from which you let the (cold) air escape. This is
an example of what's often called "action and reaction" (another name for
Newton's third law of motion): the hot exhaust gas firing down (the action)
creates an equal and opposite force (the reaction) that speeds the rocket up. The
action is the force of the gas, the reaction's the force acting on the rocket—and
the two forces are of equal size, but pointing in opposite directions, and acting
on different things (which is why they don't cancel out).
65.
When you fire a shot, the
bullet exerts an equal but
opposite force directed
toward your shoulder,
resulting in a “kick” that
your shoulder absorbs.
(action-reaction pair).
When you fire a
shotgun, you feel
it push back on
your shoulder. The
same applies with
a rifle or a pistol,
but the kick is less.
Newton’s 3rd Law
For every action, there is an equal and opposite reaction and
they act on different bodies
Newton’s 2nd Law
∑ ! = 23
3<%=> =
!<".. 3; <%=>
2<%=>
3<".. =
!<%=> 3; <"..
2<"..
Even though an action-reaction pair exists in this situation,
the two people will travel different distances depending on
their masses. According to Newton’s 2nd Law, as the mass
of an object increases, its acceleration decreases. So a person
with a larger mass will have a small acceleration, resulting
in a shorter distance traveled. If the person pushes off of a
wall, he/she will move away from the wall.
Newton’s 3rd Law
For every action, there is an equal and opposite reaction and
they act on different bodies.
The rifle exerts a force on the bullet which exerts an equal
but opposite force back on the rifle (toward your shoulder).
The same happens with the shotgun, but since there is more
mass in the shot, the reaction force providing the kick is
greater.
#ℎ'9*h = !&')8 i-0 /0#-%6 3; #3=>-0
63.
Weight and Normal force
do not make an action
reaction pair because they
The books sits on
the table. It does
not accelerate nor
Newton’s 2nd Law
∑ ! = 23
27
28
66. Atwood Machine
a. Find the expression for acceleration in terms of m1 and m2
b. Find T, the tension in the string in terms of m1 and m2
c. Find force ceiling must exert on pulley in terms of m1 and m2
d. Now let m1 = 5 kg and m2 = 7 kg
e. Find acceleration
f. Find T, the tension in the string
g. Find force ceiling must exert on pulley
M. Application of Newton’s Laws to Multiple Objects:
105°
70°
140°
25
kg
#%
#&
32
kg
We define the system to include both the masses, pulley and rope – mass of system 2( + 2) (we
assume rope and pulley are massless)
2/+/ = 2( + 2)
The pulley can be straightened out. External forces on the system are 2) < and 2( <
a. Define the system – Circle it.
b. What are the external forces on the system?
c. Apply Newton’s second law specifically to this system.
The system accelerates
∑ !/+/ = 2/+/ 3/+/ ;
2) < − 2( < = 2/+/ 3/+/
. !/+/0-6 = 2/+/0-6 3/+/0-6
SYSTEM ANALYSIS to calculate acceleration
•
•
•
•
•
•
•
•
Straighten out pulley systems
Draw what is “in the system”?
What is the mass of the entire system?
What is the net EXTERNAL force on the system (Draw a FBD and figure it out)?
If necessary, set-up a coordinate system (which direction is X and which is Y?). Consider FNET in X and Y
directions (or: ^ and //)
Don’t forget Newton’s Second Law can be applied to the whole system for SYTEM ANALYSIS to find
acceleration ujkjlmn
Ø ∑ !/ = 2/ 3/
Newton’s Second Law can be applied to each individual object in the system- INDIVIDUAL OBJECT
ANALYSIS TO FIND INTERNAL FORCES
Ø ∑ !( = 2( 3/ !! (or ∑ !( = 0 $) in a different direction…
Remember that each object in the system moves with the same acceleration (magnitude) if connected by an
ideal rope
a) 3/+/ = (2) < − 2( <)/(2( + 2) )
INDIVIDUAL OBJECT ANALYSIS to calculate tension and the ceiling force on pulley
b) Draw FBD of 2)
Applying Second Law to 2)
. ! = 2) 3)
Acceleration 3) = 3/+/
2) < − # = 2) 3)
# = 2) < − 2) 3)
c.) Draw FBD of v9``8k;
Tension T down on left side; tension T down on right side, and Fapp up from ceiling
Applying Second Law to v9``8k
. ! = 2& 3& = 0
!=-".";$ − # − # = 2& 3& = 0
!=-".";$ = 2# = 2(2) < − 2) 3) )
e) a = g/6 = 1.635 m/s2
f) T = 57.2 N
g) F
=2T = 114.5 N
pulley
29
30
67. What is the acceleration of the cart, 2( in this problem? Is it g, less than g, or greater than g?
Justify your answer.
a. What is “the system” in the picture below? Given masses 2( and 2)
We define the system to include both the masses, pulley and rope – mass of
system 2( + 2) (we assume rope and pulley are massless)
68. Train
a. Derive an expression for the acceleration of the train in terms of the mass of the engine
2-;$";- and mass of the coach 2=3%=1
b. Derive an expression for the tension in the cable connecting the cars.
c. Calculate the acceleration, given 2-;$";- = 10000 ;< and 2=3%=1 = 2000 ;<
d. Calculate the tension in the cable connecting the cars
2,000 kg
b. What is the expression for the total mass of the entire system?
2/+/ = 2( + 2)
10,000 kg
9000 N
SYSTEM ANALYSIS: We define the system to include the train – not the tracks or the earth
a) 2/+/ = 2( + 2)
c. Derive an expression for the net force on the system (Draw a FBD and figure it out).
The pulley can be straightened out. External forces on the system are 2) <, 2( <, and !,
The pulley can be straightened out. External forces on the system are 2) <, 2( <, and !,
We can show !, = 2( < as follows by applying second law in Y direction on m1
∑ !+ = 2( 3+ ; 3+ = 0
. !/+/ = 2/+/ 3/+/
Assuming friction is Zero - only external force is 9000 N
The weights are canceled by normal force because acceleration in the y-direction is ZERO.
Thus we are left with ∑ !/+/ = 2) <
d. Derive an expression for the acceleration, a, of the system.
SYSTEM ANALYSIS to calculate acceleration
∑ !/+/ = 2/+/ 3/+/ ;
2) < = (2( + 2) ) 3/+/ ;
3/+/ = 2) </(2( + 2) )
!%&& = (2-;$";- + 2=3%=1 ) 3/+/
3/+/ = !%&& /(2-;$";- + 2=3%=1 )
!,
b) INDIVIDUAL OBJECT ANALYSIS on 2000 kg coach
∑ !+ = 2)888 3+ ; 3+ = 0
Hence !, = 2)888 <
e. Now let 2( = 300g and 2) = 100 g; calculate the acceleration of the system.
8.((N.L)
3/+/ = 8.7 = 2.45 2/* ) to the left/down/counterclockwise
#
2)888 <
. !* = 2)888 3*
# = 2)888 3/+/
Derive an expression for tension T1 in the rope in terms of 2( and/or 2) and constants
such as g
g. Calculate tension T in the rope given masses in part e.
f.
OR
INDIVIDUAL OBJECT ANALYSIS on 10000 kg engine
∑ !+ = 2(8888 3+ ; 3+ = 0
Hence !, = 2(8888 <
INDIVIDUAL OBJECT ANALYSIS to calculate tension
f. Draw FBD of 2)
Applying Second Law to 2)
. !* = 2(8888 3*
. ! = 2) 3)
!,
!%&&
2(8888 <
!%&& − # = 2(8888 3/+/
# = !%&& − 2(8888 3/+/
Acceleration 3) = 3/+/
2) < − # = 2) 3)
# = 2) < − 2) 3)
#
c) 3/+/ = !%&& /(2-;$";- + 2=3%=1 ) = 9000N/(12,000kg) = 0.75 m/s2
6
g. # = 0.1(9.8 − 2.45) = 0.735$
d) # = 2)888 3/+/ = (2000;<) S0.75 /+ T = 1500 $ i.e. force of engine on coach
OR
# = !%&& − 2(8888 3/+/ = 9000N – (10000)(0.75) = -1500N i.e. force of coach on engine
31
32
69. In the figure shown, assume that the pulley is frictionless and massless.
(a) (i)
From System Analysis:
∑ ! = !$) − !$( *,-+ = (2( + 2) )3/
F
FT
$' =
F
m2
mgcosθ
θ
Fg1
Fg2
mgsin
a. If the surface of the inclined plane is θ
frictionless, determine what value(s) of θ will cause the
box of mass 2( to
i.
Accelerate up the ramp;
ii.
Slide up the ramp at constant speed.
b. If the force of friction between the surface of the incline plane and the box of mass 2( is f,
derive (but do not solve) an equation satisfied by the value of θ, which will cause the box of
mass 2( to slide up the ramp at constant speed.
#&
#%
=
." 1) .# 1'"+,
(.# /." )
$' =
$" ! $# !
)
'"+,
$# ! $# !
($# &$" )
$# !
=
#
$"
) '"+,
$#
($# &$" )
$# !
To accelerate up the ramp uj > 0 hence ; eo /ep > fwxy on the RHS above OR fwxy <
eo /ep
(u) (ww) Constant speed uj = d;
fwxy = eo /ep
(b) With Friction {
From System Analysis in x direction OR parallel to the ramp:
Net Force Equations:
We define the system to include 2( , 2) , pulley, and rope. The ramp and earth are outside the
system.
FN
Fg1sinθ
(.# /." )
Since we are asked to find an expression for values of & that allow the block to accelerate up
the ramp we simplify the above equation to get '()& by itself.
Dividing the RHS of the expression below by #% * (an algebra trick……..if we divide both
numerator and denominator by the same number it is like multiplying the RHS by 1 which
. 1
DOES NOT change the equation because .# 1 = 1)
m1
f
(!" ) (!# '"+,
. !/+/ = !$) − !$( *,-+ − % = (2( + 2) )3/
!$) − !$( *,-+ − %
2) < − 2( <*,-+ − %
=
(2( + 2) )
(2( + 2) )
At constant speed 3/ = 0, ℎ8-)8 from the equation above
2) < − 2( <*,-+ − %
=0
(2( + 2) )
OR 2) < − 2( <*,-+ − % = 0
eo } − {
|wxy =
ep }
Fg2
3/ =
2( <)&*+
For the System
∑ ! = !$) − !$( *,-+ = (2( + 2) )3/
In case the problem asks you to derive an expression for tension we use Individual FBD as
shown below
(1)
3/ =
!$) − !$( *,-+
(2( + 2) )
For Block #1
ΣFperpendicular = maperp i.e. applying the second law perpendicular to the ramp in the direction of the normal force
ΣFperpendicular = FN - mgcosθ
aperp = 0 because the block is at rest (in equilibrium) in perpendicular direction (2)
Hence FN = mgcosθ
ΣFparallel = FT - mgsinθ = maparallel = ma (3)
For Block #2
ΣF = FT - mg = -ma (3) For block 1 to accelerate up the ramp (4)
!: = m(g-a) (5)
FN
FN
FT
m1
m1
m2
Fg2
33
FT
FT
Fg1
Ff
Fg1
34
70. Three children are holding hands on friction free ice. The children’s masses are
25 kg, 35 kg and 30 kg respectively. The 30 kg child grabs onto a branch and
pulls with a force of 250N at an angle of 35o to the horizontal. At what rate do
they accelerate? Find the tension between the hands of each of the children.
71. Two buckets are strung over a pulley. One has a mass of 3.4 kg, the other a mass of 2.8 kg. If
the 3.4 kg bucket is 1.2 meters above the ground, how long will it take the bucket to hit the
ground?
3/ = (
This is just like the train problem with an added TWIST. The applied external force is at an angle
of 35 o to the horizontal.
System: 3 kids
System Diagram:
$)
$(
#(
Larry
2(
)<=
C.75).L
C.7q).L
=
8.S
S.)
= 0.95 2/* )
Δk = 1.2 2; H8 = 0
Δk = H8 h +
(
)
3h ) ;
h= i
)r+
%
+
72. Three objects in the drawing are connected by strings that pass over massless and
friction-free pulleys. The objects move, and the coefficient of kinetic friction
between the middle object and the surface of the table is 0.100. What is the
acceleration of the three objects? What is the tension in each of the two strings?
!$C
!$)
!$(
3/ =
Joe
2C
6+ q63
!%
$C
#)
Moe
2)
6+ 563
∑ !/
2/
. !/ = !% )&*+
3/ =
∑ 4#
6#
=
45 =3/H
63 q6+ q66
)98 =3C9
= )9qC9qC8 = 2.28 2/* )
Individual FBD for 2( to find #(
$(
#(
2(
#)
!% )&*+
:
$)
#( = 2( 3( = 25(2.28)
Individual FBD for 2) or 2C to find #)
System: 3 masses, rope, pulley
System Diagram: External forces are weight of 2( , weight of 2C , and force of friction acting on
2)
$C
!$(
2C
!$C
!$(
10kg
2(
#(
80 kg
2)
#)
25 kg
2C
!$C
%
!$)
3/ =
∑ !/
2/
∑ !/ = !$C − !$( − % = 2C < − 2( < − q$)
35
36
3/ =
∑ 4#
6#
=
66 $563 $5P,+
63 q6+ q66
=
66 $563 $5P6+ $
((9
=
)985(8858.((L88)
((9
= 70/115 = 0.6 m/s2
74. A 200.0 kg mass is to be dragged across the floor by an applied force. The coefficient of
static friction is 0.60, the coefficient of kinetic friction is 0.45.
a. What is the normal force acting on the mass?
b. How much force must be applied just to get the mass moving?
c. How much force must be applied to keep the mass moving?
d. How much force must be applied to give the mass an acceleration of 5.0
m/s 2 ?
Individual FBD for 2( to find #(
#( − 2( < = 2( 3(
#( = 2( < + 2( 3( = 10(9.8+0.6) = 104N
Individual FBD for 2) or 2C to find #)
2( <
#(
2C < − #) = 2C 3C
#)
2C <
$
!%
%
2<
a. ∑ !+ = 23+ ; 3+ = 0 because there is no motion in the y-direction
2C
#) = 2C < − 2C 3C = 25(9.8-0.6)= 230N
. !+ = $ − 2< = 0
$ = 2<
b. ∑ !* = 23* ; 3* = 0 because there is no motion in the x-direction before it starts moving
. !* = !% − %/6%* = 0
!% =
N. More Problems… More Fun!
#* = 180 )&*30 = 156$; #+ = 180 *,-30 = 90$
2 = 10.0 ;<; O8,<ℎh = !$ = 2< = 100 $
$ + #+ = 2< so solve for N; $ = 2< − #+ = 100 − 90 = 10$
Max Static Friction %/6%* = q/ $ = (0.4)(10) = 4$
Kinetic Friction = %> = q> $ = (0.2)(10) = 2$
Net Horizontal force = ∑ !* = #* − %> (if the mass is stationary use %/ )
∑4
: 5'
(9S5)
(97
3* = 60 = 0 6 1 = (8 = (8 = 15.4 $
. !* = !% − %> = 0
!% = %> = q> $ = (0.45)(2000) = 900$
d. ∑ !* = 23* ; 3* = 5 2/* ) ; 23* = (200)(5) = 1000
∑ !* = !% − %> = 1000N
!% = %> + 1000 = q> $ + 1000 = (0.45)(2000) + 1000 = 1900$
75. A 40.0 kg mass rests at the edge of a board that is 2.0 meters long. The coefficients of
friction are 0.80, and 0.40. The board is slowly raised until the mass just begins to slide.
!,
a. At what angle does the mass begin to slide?
b. What is its acceleration as it slides down the ramp?
2<*,-+
c. How long does it take (seconds) to reach the bottom of the ramp?
#+
$
+
#*
%
2<
37
= q/ $ = (0.60)(2000) = 1200$
c. ∑ !* = 23* ; 3* = 0 because there is no acceleration in the x-direction just keep it moving
at constant velocity
73. A 10.0 kg mass is being pulled by a rope. The rope makes an angle of 30.0 degrees to the
floor, and an applied force of 180 N is exerted on the rope. The coefficient of static friction is
0.40, and the kinetic coefficient is 0.20.
a. What are the two components of the rope’s applied force (Tx, Ty)?
b. What is the weight of the mass?
c. What is the normal force?
180 N
d. What is the maximum static friction on the mass?
e. What is the kinetic friction on the mass?
308
f. What is the net horizontal force on the mass?
g. What is the acceleration of the mass?
a.
b.
c.
d.
e.
f.
g.
%/6%*
a. Just before the object starts sliding ∑ !&%#%..-. = 0 because a = 0
∑ !&%#%..-. = 2<*,-+ − %/ = 0
%/ = q/ !,
∑ !&-#& = !, − 2<)&*+ = 0
!, = 2<)&*+
%/ = q/ 2<)&*+
Just before it slides
∑ !&%#%..-. = 2<*,-+ − %/ = 0
∑ !&%#%..-. = 2<*,-+ − q/ 2<)&*+ = 0
2<*,-+ = q/ 2<)&*+
*,-+ = q/ )&*+
Dividing *,-+ by )&*+
#3-+ = q/
+ = #3-5( (q/ ) = #3-5( (0.8) = 39 ]8<'88*
%/
2<)&*+
2<
2<*,-+
38
77. In the design of a supermarket, there are to be several ramps connecting
different parts of the store. Customers will have to push grocery carts up
the ramps, as shown above, and it is obviously desired that this not be
too difficult.
a. Calculate the force necessary for the customer to push the cart at
constant speed along the ramp.
b. An engineer has done a survey and found that almost no one
complains if the force required to push the cart at constant speed is
no more than 50.0 N. Will a slope of θ= 5.0° be too steep, assuming
a 30.0 kg grocery cart?
Assume the coefficient of friction is μ = 0.10.
b. When it is sliding ∑ !&%#%..-. = 23*
∑ !&%#%..-. = 2<*,-+ − %> = 23*
%/ = q> !,
∑ !&-#& = !, − 2<)&*+ = 0
!, = 2<)&*+
%> = q> 2<)&*+
2<*,-+ − q> 2<)&*+ = 23*
<*,-+ − q> <)&*+ = 3*
!%&& = 2<*,-+ + %
= 2<*,-+ + q$
= 2< *,-+ + q2<)&*+
FN
10(Sin 39) – (0.4)(10)(cos39) = 6.24 – 3.11 =
3 m/s2
%
Fgpara
Fgperp
θ
c. 3* = <*,-+ − q> <)&*+ = 3 m/s2
2
H8 = 0
*
θ
Fg
Δ^ = 2.0 2
1
3 h)
2 *
Part b:
If m=30 kg, θ= 5.0, g = 10 m/s2; !%&& = 30(10)J,-5 + 0.10(30)(10))&*5 = 26.15 +
29.89 = 56 $
So θ= 5.0° is too steep
Δ^ = H8 h +
)r*
h = i % = a2(2)/3 = 1.2 *
0
76. A block of weight W is pushed and held against a rough, fixed wall by a force P. The
frictional force between the block and the surface is f, and the normal force is N.
(a) Draw a free body diagram of the block
(b) Which of the following is equal to the magnitude of the force of friction acting on the
block? Explain.
(A) P (B) W
(C) N
(D) N + P
(E) N – P
P
78. A 150 N block sits on an inclined plane, as shown below. (Note that newtons are units of
weight.)
a) Make a free body diagram showing all forces acting on the block.
b) Find the normal force on the block.
150N
c) Find the angle of the plane, θ.
FN
%
fs
N
O
Fgpara
θ
Fgperp
Fg
∑ !&-#& = 23&-#&
!, − !$ )&*+ = 0
!, = 2<)&*+
q = 0.30
. !&%#% = 23&%#%
%/ − !$ *,-+ = 0 (just before it starts sliding)
%/6%* = !$ *,-+ = 2<*,-+
%/ = μs!, = μs2<)&*+
%/ = 2<*,-+
Hence 2<*,-+ = μs2<)&*+
Tan θ = μs ;.
39
40
79. Diagram the forces acting on the 5kg block.
Calculate normal force on 5 kg block due to the 7 kg block.
3kg
FN7
. !+ = 0 = 0
5kg
7k
g
!,D − !$9 − !$C = 0
!,D = !$9 + !$C
!,D = 50 + 30 = 80 $
80. A 43.8 kg sign is suspended by two wires, as the drawing shows.
Find the tension in wire 1 and in wire 2.
670
##
T2
T1y
Fg3
a) T1 = 2W1
b) W1 > T2
c) T1 > W2
d) T1 <T2
e) T1 = T2
f) T1 >T2
O( = ( <
O) = ) <
#) = ) <
#( = ( < + #)
#( = ( < + ) <
#$
67
Fg5
81. The masses of M1 and M2 are equal. They are suspended from the ceiling as shown in the
diagram at right. Ignore the mass of the string. Which of the following statements about the
tensions in string T1 and T2 are true? (More than one is true! List all that are true).
370
82. A block of mass 3m can move without friction on a horizontal table. This block is attached to
another block of mass m by a cord that passes over a frictionless pulley, as shown above. If the
masses of the cord and the pulley are negligible, what is the magnitude of the acceleration of
the descending block?
a) Zero
b) g/4
c) g/3
d) 2g/3
e) g
370
T1
T2y
T1x
T2x
Fg
#(+ = #( )&*67
#(+ = 0.4 #(
#)+ = #) )&*37
#)+ = 0.8#)
#(* = #( *,-67
#(* = 0.9 #(
#)* = #) *,-37
#)* = 0.6 #)
Since the sign is at rest net force on it is zero.
∑ !* = #)* − #(* = 0
(1)
#)* = #(*
0.9#( = 0.6 #)
#( = (0.6 #) )/0.9 = 0.67 #)
∑ !+ = #(+ + #)+ − !$ = 0 (2)
#(+ + #)+ = !$
0.4#( + 0.8 #) = 430 $
0.4(0.67)#) + 0.8#) = 430 $
0.268#) + 0.8#) = 430 $
1.068#) = 430$
#) = 403$ N
#( = (0.67)#) = 270$ N
From Problem 69:
6 $
3/+/ = − 6 q+ 6
.1
= 2. =
1
3
+
2
83. A 9.0 kg bowling ball is suspended from the rope. If the rope provides a tension to accelerate
the ball upwards, the tension in the rope must be:
a) 9.0kg
b) 89 N
c) less than 89N
d) more than 89N
# − !$ = 23
# = !$ + 23
T
!$
84. A force applied to a rocket gives it an upward acceleration equal to 2 times the acceleration of
gravity, g. The magnitude of the force is equal to
a. the weight of the rocket
!% − !$ = 23 = 2(2<) = 22<
b. twice the weight of the rocket
c. three times the weight of the rocket
!% = !$ + 22< = 2< + 22< = 32<
d. four times the weight of the rocket
!%
!$
41
42
85. A boy exerts a force F pulling a sled by means of a cord making an angle of θ with the
horizontal. If the boy continues to pull the cord with the same amount of force, but increases
the angle that the cord makes with the horizontal,
a. both the horizontal and vertical components of F will increase
b. the horizontal component of F will increase while the vertical component decreases
c. The vertical component of F will increase whlle the horizontal component of F decreases
d. Both the horizontal and vertical components will decrease
e. only the vertical component will increase
86. A crate of mass M is being pushed down a plank that makes an angle of θ with the horizontal.
The frictional force slowing it down is f. and the force pushing it forward in a direction down
the ramp is P. Which expression best represents the net force acting
on the crate?
Like problem 79
a) W+P-f
b) W + P + mgCosθ + mgSinθ - f
c) mgSinθ
d) P – f
e) P + mgSinθ - f
FN
fs
P
θ
Fgpara
Fgperp
Fg
87. An expression for the normal force (sum of all the supporting forces,) experienced by the crate
would be
a) -W
b) mg Cosθ
c) mg Cosθ – mgSinθ
d) mg (1+Sinθ)
e) 0
88. A block of mass M equal to 50 kg is suspended from the ceiling by means of rope AB (fig.).
Another block of mass m equal to 10 kg is suspended from the 50 kg block, by means of rope
CD. The ropes AB and CD have the same breaking strength F. If the 10 kg block is pulled
suddenly with a force slightly less than F, what will happen?
a. Rope CD will break
b. Rope AD will break
c. Both AB and CD will break
d. Neither AB nor CD will break
e. Either AB or CD will break
A horse pulls a cart along a flat road. Consider the following four forces that are in this situation.
1. The force of the horse pulling on the cart
2. The force of the cart pulling on the horse
3. The force of the horse pushing on the road
4. The force of the road pushing on the horse
89. Which two forces form an action-reaction pair that obeys Newton’s third law?
a. 1 and 4
b. 1 and 3
c. 2 and 4
d. 3 and 4
e. 2 and 3
90. Suppose that the horse and cart have started from rest; and as time goes on their speed
increases in the same direction. Which one of the following conclusions is correct concerning
the magnitudes of the forces mentioned above?
a. Force 1 exceeds force 2
b. Force 2 is less than force 3
c. Force 2 exceeds force 4
d. Force 3 exceeds force 4
e. Force 1 and 2 cannot have equal magnitudes
A Physics student in a hot air balloon ascends vertically at constant speed. Consider the following
four forces that arise in this situation:
• F1 = the weight of the hot air balloon
• F2 = the weight of the student
• F3 = the force of the student pulling on the Earth
• F4 = the force of the hot air balloon pulling on the student
91. Which of the two forces form an action reaction pair that obeys Newton’s third law?
a. F1 and F2
b. F2 and F3
c. F1 and F3
d. F2 and F4
e. F3 and F4
92. Which of the following relationships concerning the forces or their magnitudes is true?
a. F4 > F2
b. F1 > F2
c. F4 > F1
d. F2 = -F4
e. F3 = -F4
When you pull suddenly there is not
enough time to overcome the inertia
of the big mass M. In a sudden pull
we only manage to overcome the
inertia of the little block m and hence
rope CD breaks. If you pull the rope
real slowly rope AB will break.
REMEMBER THE DEMO
43
44
93. A block of mass M is held motionless on a frictionless inclined plane by means of a string
attached to a vertical wall as shown in the drawing. What is the magnitude of the tension T in
the string?
a. 0 N
b. Mg
c. Mgcosθ
d. Mgsinθ
e. Mgtanθ
45 − $+-..3s = 5(3)
$+-..3s = 45 − 15 = 30 $
$2.T- = 10(3)
$2.T- = 30 $
94. In the problem above, if the string breaks what is the magnitude of acceleration of the block as
it slides down the inclined plane?
a. 0 m/s2
b. g
c. gcosθ
d. gsinθ
e. gtanθ
95. A lift of mass 500 kg starts moving down from rest. The variation of its velocity for 8 seconds
is shown in the adjoining velocity–time graph. What is the tension in the supporting wire
during the 7th second?
a. 5500 N
b. 5000 N
c. 4500 N
d. 6000 N
e. 4000 N
∑ !+ = # − !$ = −23+ ; when
elevator is going down
Slope of the graph gives
3+ ,- hℎ8 7hℎ *8)&-]
3+ =
T
!$
# − 2< = −2 3 # = 2< −
23=
−2
2
= −1 )
2
*
500(9.8 –(-1)) = 5500 N
45N
96. A 5.0-kg and a 10.0-kg box are touching each other. A 45.0-N horizontal force is
applied to the 5.0-kg box in order to accelerate both boxes across the floor. Ignore
friction forces and determine the acceleration of the boxes and the force acting
between the boxes.
SYSTEM ANALYSIS:
∑4
79
∑ !/ = 2/ 3/ OR 3/ = # = = 32/* )
6
(9
#
$2.T$+-..3s
5 kg
45 N
10 kg
45
46