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Example 5.2 “Readying a wrecking ball”
(a) RECOPY THE FBD. Work with force magnitudes. T1, T2, W = M g
What is a magnitude?

Identify the unknown magnitudes.
(b) Use trigonometry to write the X and Y components of each of the three forces. Get your
answers to 3 significant figures.
(c) Use Newton’s 2nd Law (for static equilibrium) to generate 2 equations in 2 unknowns. The
2 unknowns are T1 and T2.
(d) Solve the equations for part (c).
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Example 5.2 “Readying a wrecking ball” according to abe’s variation
(a) Review of tension force “T” in a rope (4 possible tension forces!!)
 Consider a vertically hanging wrecking ball on a rope.
What are the four tension forces associated with the rope?
Make a FBD for the just the wrecking ball. What is the magnitude of the tension force on the
ball? (2500. N)
(b) Let’s redraw the free body diagram for just the ball.
 Write down Newton’s 2nd Law equations for static equilibrium
Note the difference between “force magnitudes” and “force components”
Equation 1
X equation
- T1 + T2 sin + 0 = 0
Equation 2
Y equation
0 + T2 cos + (- 2500. N) = 0
We have 2 simultaneous equations. Solve (8th grade)
T2 = 2.66 x 103 N
T1 = 910. N
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IN CLASS PROBLEM
Example 5.3 Tension in towing a car at a constant velocity (zero
acceleration or “dynamic equilibrium”) Note: Fnet = 0 even though the car is in moving!!
(a) What is a magnitude?

Re-sketch the free body diagram using only force “magnitudes and angles.”
(b) Use trigonometry to write the X and Y components of each of the four forces. Get your
answers to 3 significant figures. What is the difference between a magnitude and a
component?
(c) Make a motion diagram for the car.
(d) Use Newton’s 2nd Law to generate 2 equations in 2 unknowns. The 2 unknowns are N and
T. Solve the equations for the unknowns.
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Example 5.3
Tension in towing a car at a constant velocity (zero acceleration or
“dynamic equilibrium”) Note: Fnet = 0 even though the car is in moving!!
(d) Solve the equations for part (c).
#1
X equation
T cos + (- 320. N) + 0 + 0 = 0
#2
Y equation
T sin + 0 + n + (- Wt) = 0
T = 341. N
n = 1.46 x 104 N < Wt
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IN CLASS PROBLEM
Example 5.5 Tension in towing a car with constant acceleration (not in equilibrium!)
(a) What is a magnitude? Re-sketch the free body diagram using only force “magnitudes and
angles.”
(b) Use trigonometry to write the X and Y components of each of the four forces. Get your
answers to 3 significant figures. What is the difference between a magnitude and a
component?
(c) Make a motion diagram for the car.
(d) Use Newton’s 2nd Law to generate 2 equations in 2 unknowns. The 2 unknowns are N and
T. Solve the equations for the unknowns.
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Find the tension T in the rope.
 Draw motion diagram
 Draw free body diagram
Note force magnitudes (+) T, f, n, Wt
Wt = 1500.· 9.80 = 1.47 x 104 N
Note force components (+) Tx, Ty, fx, fy, nx, ny, Wtx, Wty
#1
X equation
find aX = VX / t = 1.20 m/s2
T cos + (- 320. N) + 0 + 0 = M aX
T = 2.26 X 103 N
Note: If you have time, use the Y equation to find the normal force magnitude n.
#2
Y equation
T sin + 0 + n + (- Wt) = M aY
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
Calculate the "weight" and "mass" for each of the following. Give your answer
in both BE and SI units. [HRF3,5-15E]
(a) a 5.00 pound bag of sugar.
(b) a 240. lb fullback.
(c) a 1.80 ton automobile.

A man “weighs” 800. N.
(a) Determine his weight and mass in SI units.
(b) Determine his weight and mass in BE units.
(c) Determine his weight and mass in CGS units.
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Example 5.2 “Readying a wrecking ball”
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Example 5.2 “Readying a wrecking ball” according to abe’s variation
(c) Review of tension force “T” in a rope (4 possible tension forces!!)
 Consider a vertically hanging wrecking ball on a rope.
What are the four tension forces associated with the rope?
Make a FBD for the just the wrecking ball. What is the magnitude of the tension force on the
ball? (2500. N)
(d) Let’s redraw the free body diagram for just the ball.
 Write down Newton’s 2nd Law equations for static equilibrium
Note the difference between “force magnitudes” and “force components”
Equation 1
X equation
- T1 + T2 sin + 0 = 0
Equation 2
Y equation
0 + T2 cos + (- 2500. N) = 0
We have 2 simultaneous equations. Solve (8th grade)
T2 = 2.66 x 103 N
T1 = 910. N
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 You are walking on the floor in the positive X direction. Study the following
figure (4.35 a from MP) and then answer the questions.
(a) Make a diagram which shows all forces on
your body, the floor, and the earth as you are
walking.
(b) Make a FBD of your body as it is walking.
(c)
What is the origin of the force which causes you to move forward?
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 Example 5.15 Pushing two blocks. (VARIATION)
Figure 5.33 shows a 5.00 kg block A being pushed
with a 3.00 N force. In front of this block is a 10.0
kg block B; the two blocks move together. (Do all
calculations to 3 sig figs.)
(a)
Make an FBD of the combined system.
(b) Use N-2 to obtain the acceleration of the combined
system.
(c)
Make an FBD for just block A.
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CONTINUED Example 5.15 Pushing two blocks. Figure 5.33 shows a 5.00 kg block A being
pushed with a 3.00 N force. In front of this block is a 10.0 kg block B; the two blocks move
together. (Do all calculations to 3 sig figs.)
(d) Use N-2 to find the force magnitude Fon B by A on block B by block A?
(e)
Make an FBD for just block B.
(f)
Use N-2 to find the force magnitude Fon A by B on block A by block B?
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Example 5.3 Tension in towing a car at a constant velocity (zero acceleration or “dynamic
equilibrium”) Note: Fnet = 0 even though the car is in moving!!
 Sketch free body diagram
Note force magnitudes (+) T, f, n, Wt
Wt = 1500.· 9.80 = 1.47 x 104 N
Note force components (+) Tx, Ty, fx, fy, nx, ny, Wtx, Wty
#1
X equation
T cos + (- 320. N) + 0 + 0 = 0
#2
Y equation
T sin + 0 + n + (- Wt) = 0
T = 341. N
n = 1.46 x 104 N < Wt
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Example 5.5 Tension in towing a car with constant acceleration (not in equilibrium!)
Find the tension T in the rope.
 Draw motion diagram
 Draw free body diagram
Note force magnitudes (+) T, f, n, Wt
Wt = 1500.· 9.80 = 1.47 x 104 N
Note force components (+) Tx, Ty, fx, fy, nx, ny, Wtx, Wty
#1
X equation
find aX = VX / t = 1.20 m/s2
T cos + (- 320. N) + 0 + 0 = M aX
T = 2.26 X 103 N
Note: If you have time, use the Y equation to find the normal force magnitude n.
#2
Y equation
T sin + 0 + n + (- Wt) = M aY
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 A box of books is initially at rest on a floor. The mass of the box is 90.0 kg. The coefficients of static and kinetic
friction for the bottom of the box and the floor are µs = 0.700 and µk = 0.600 . Let the applied force Fapp on the box be
horizontal in the positive X direction.
Y direction
X direction
(a) On the above figure, sketch and label the forces (which only act) on the box.
(b) Calculate the magnitude of the normal force.
(c) For the following table, calculate the magnitude of the friction force and state whether the box is moving or not.
Fapplied
Friction Force
Does box move ? If so, what is the acceleration aX?
Show work on back if more space is required.
___________________________________________________________________________________
0N
___________________________________________________________________________________
50 N
___________________________________________________________________________________
100 N
___________________________________________________________________________________
250 N
___________________________________________________________________________________
400 N
___________________________________________________________________________________
600 N
___________________________________________________________________________________
650 N
___________________________________________________________________________________
700 N
___________________________________________________________________________________
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A box of mass M = 15.0 kg is motionless on a rough inclined plane at an angle of 35.0 degrees with respect to the horizontal. The coefficient of static friction for
the box and plane is .800. Hint: Use tilted coordinate system for this problem.
(a) Illustrate all forces acting on the box.
(b) On a diagram, show the x and y components of the gravitational force. Also, calculate these components. (84.4 N, -120. N)
(c) Use the equilibrium condition for y-components of force to calculate the magnitude of the normal force.
(d) Calculate the magnitude of the friction force.
(e) If the mass of the box is increased sufficiently, will it start to slide ?
(f) At what angle will the box start to slide?
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
You are given the earth and the moon.
(a) Make a diagram showing gravitational interaction forces between the earth and
the moon. Label the forces appropriately.
(b) Look up the masses of the earth and moon and also their separation distance.
Mearth =
Mmoon =
rseparation =
(c) Calculate the gravitational force between the earth and the moon.
(d) Calculate the “weight” force on a 70.0 kg person standing on the earth’s
surface. Use SI units and 3 sig figs.
 Use the standard method.
 Use Newton’s Law of Universal Gravitation to get the same answer! Hint: Look
up the radius of the earth first!
Challenge: How far from the center of the earth will the net force on the person
be zero?
(e)
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Two types of forces
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continued
 2. Action at a distance forces
force seems to act through “empty space”
o Gravitational force
Fg
Weight = “Wt” = M g
Wt  " weight of object "
o Electric and magnetic forces
Fundamental particles electron, proton, neutron
More fundamental
proton and neutron are composites of quarks
o Nuclear (strong) and nuclear (weak) forces
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Newton’s Law of Universal Gravitation
Fundamental Law of Gravity between any two objects in
the universe (includes Solar System and beyond. The force seems to have an infinite range. This is
an exact law with only tiny corrections from General Relativity.
Fon 2 by 1
Fon 1 by 2
M1
M2
r
Fon 1 by 2 = Fon 2 by 1 =
F = G M 1 M2 / r 2
G = 6.67 x 10-11 N m2 / kg2
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What is the relation between G and g = 9.80 m / s2 ?
Consider the gravitational force on a basketball at the surface of the Earth.
Fon B by E = MB g
Fon E by B
Mball
MEarth
r = R E + RB
RE
r
RB
r ≈ RE
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FEB = FBE = F = G ME MB / r 2
FBE = MB g = F = G ME MB / r 2
M B g = G ME MB
/ RE 2
g = G ME / RE 2
On the surface of the Earth
g = (6.67 x 10-11) (5.98 x 1024) / (6.38 x 106)2
g = 9.80 m / s2 (SI units)
g = 980. cm / s2 (CGS units)
g = 32.2 ft /s2 (BE units)
On the surface of the Moon
gMoon = (6.67 x 10-11) (7.35 x 1022) / (1.74 x 106)2
gMoon = 1.62 m / s2
Memorize:
Newton’s Laws and corresponding formulas; formulas for static and kinetic friction forces.
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MORE Optional HOMEWORK PROBLEMS WITH ANSWERS
All SI, BE, and CGS units defined until present assignment.
(1) A constant force of 3.75 N is applied to a hockey puck in the positive x-direction. (Note: A hockey stick remains in contact with the puck in order to apply the
constant force.) The puck is initially at rest at the origin; its mass is .150 kg. Assume there is no friction between the puck and the ice. (a) What is the
acceleration in the x-direction ? (b) At what time t will the puck have a speed of 5.20 m/s ? (c) How far has the puck moved at the time from part (b) ? [HRF3,54E] ( 25.0 m/s2, .208 s, .541 m )
(2) A 200 kg boat is being pulled in the water by two ropes. The first rope exerts a force magnitude F1 of 700. N in the positive x-direction; the second rope
exerts a force magnitude F2 of 900 N at an angle of 30.0 degrees CCL with respect to the positive x-direction. (a) Make a vector diagram which illustrates the two
forces and their resultant force FR . (You may wish to use the “parallelogram rule”.) (b) Calculate the x and y components of FR. (c) Calculate the x and y
components of the acceleration vector of the boat. (d) Calculate the magnitude and angle of the acceleration vector. (e) Assuming the boat starts from rest, find
where its coordinates after 8.00 s and the distance which it has moved. (1.48 . 10 3 N, 450. N, 7.40 m/s2, 2.25 m/s2, 7.73 m/s2, 16.9 degrees CCL WRT + x
direction, 237.,72.0, 247.)
(3) Given a 5 kg box which is resting on the surface of the earth. (a) Make a free body diagram of the box. Sketch and label all forces which act on the box. (b)
Calculate the magnitude of each force acting on the box. (c) Make a “cartoon” sketch of the box sitting on the earth. Illustrate action and reaction pairs.
(4) Given a 5 kg box which is hanging by a rope. (a) Make a free body diagram of the box. Sketch and label all forces which act on the box. (b) Calculate the
magnitude of each force acting on the box. (c) Sketch the box hanging from the ceiling. Illustrate action and reaction pairs.
(5) A woman is walking on the floor. What is the origin of the force which causes her to move forward? Make a sketch to illustrate.
(6) Calculate the "weight" and "mass" for each of the following. Use SI units for your answer. (a) a "5.00 lb" bag of sugar. (b) a "240 lb" fullback. (c) a "1.80 ton"
automobile. [HRF3,5-15E] (22.2 N, 2.27 kg, 1.07x103 N, 109. kg, 1.60x104 N, 1.63x103 kg)
(7) A man “weighs” 150 lbs. (a) Calculate his weight and mass in SI units. (b) Calculate his weight and mass in BE units. (c) Calculate his weight and mass in
CGS units.
WAIT (8) Sketch and calculate the magnitudes of the tension forces for each of the following ? Let M = 5 kg. Assume frictionless and massless pulleys. (49.0 N,
24.5 N, 24.5 N) Can you design a system which reduces the tension below 15 N. See “Solved Problem 1” in Cutnell and Johnson.
(9) An automobile which weighs 3800 lbs is accelerating at 12.0 ft/s 2 along the positive x-axis. (a) Calculate the mass of the automobile in British units. (b)
Calculate the force on the automobile in British units. (c) What is applying the force to move the automobile? [HRF3,5-20E] (119 slugs, 1.43x103 lbs, to be
discussed in class)
(10.0) Given two boxes stacked on the floor. The top box is 5 kg and the bottom box is 8 kg. Sketch the boxes and show all forces acting on the system. Which
forces are “action & reaction” pairs ? Where do the missing “reaction” forces act ? Make a free-body diagram for the bottom box. Calculate the magnitude of
each force on the bottom box.
(10.1) You are walking on the floor. What is the origin of the force which causes you to move forward? Which of Newton’s three laws applies? Make a sketch to
illustrate your answer.
(11.0) Two railroad cars are being pulled along the positive x-axis by an engine which exerts a force F of magnitude 6500 N. The two cars are connected by a
rope with tension FT . Assume there is no rolling friction. The masses are as follows: M1= 1200 kg and M2 = 2400 kg . Calculate the magnitude “ a” of the
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acceleration of the system of the two railroad cars and the tension “F T” in the rope. [HRF3,5-43P] (1.81 m/s2, 4.34 . 103 N) Suggested approach: set up a freebody diagram and free-body equations for each car; get two simultaneous equations in “a” and “FT”.
(11.1) See figure below. Given an Atwood’s machine (two masses hanging on a frictionless pulley). Let M 2 > M1. Show that the magnitude of the acceleration is
given by a = g (M2 - M1)/(M1 + M2) and that the tension in the string is
FT = 2 M1 M2 g/(M1 + M2). Hint: Set up free-body diagrams. What is the relation between a1y , a2y , and a ?
y axis
a
M1
a
FT
M2
FT
M2
x axis
M1
(11.2) See right hand figure above. Given the following simple machine. Let M 1 = 4 kg and M2 = 2 kg. Find the tension in the string and the magnitude of the
acceleration. Assume that M1 slides on a frictionless surface. Hint: Set up free-body diagrams. What is the relation between a1x , a2y , and a ? [HRF3,5-23E]
(13.1 N, 3.27 m/s2)
(11.3) See right hand figure above. Given the following system of two masses. The the surface on which M1 rests has a coefficient of kinetic friction equal to
0.150. Let M1 = 4 kg and M2 = 2 kg. Find the tension in the string and the magnitude of the acceleration of the system. Hint: Set up free-body diagrams and freebody equations. What is the relation between a1x and a , and, between a2y and a ? (15.0 N, 2.29 m/s2) Note: Do not use any formulas from text; start with
free body diagrams and free body equations (Newton’s Second Law) to arrive at answers.
WAIT (11.3) Given the following machine. Let a1 = 6 m/s2. Determine a2. What are the x and y components of a1 and a2 ?
(12) A 160 lb man stands on a bathroom scale in an elevator which is at rest. (a) When the elevator starts moving the scale reads 200 lbs. Find the accleration
(magnitude and direction) of the elevator. (b) The elevator is again at rest. But, when the elevator starts to move this time, the scale reads 120 lbs. Find the
accleration (magnitude and direction) of the elevator. (c) If the scale reads zero, should the man worry? Explain. (+8 ft/s 2, -8 ft/s2, Yes!....) (new)
(13) Given an acrobat of mass M hangs in the middle of a tightrope. Show that the tension in the rope is given by
FT = M g /(2 sin( )) . The angle  is the angle between the horizontal and the rope at the points where it is tied.
(14) Given a 3 kg metal puck which is sliding on ice. The coefficient of kinetic friction between the puck and the ice is 0.0255 . At t = 0 , the speed of the puck is
5 m/s. (a) Sketch all forces on the moving puck. (b) Calculate the magnitude of the force of kinetic friction on the puck. (c) Calculate the x-component of
acceleration. (d) Estimate when the puck will come to rest. (e) How far will the puck have moved? (0.750 N, -.250 m/s2,20.0 s, 50.0 m)
(15) A box of books is initially at rest on a floor. The mass of the box is 80 kg. The coefficients for static and kinetic friction between the bottom of the box and the
floor are 0.750 and 0.650. A horizontal force is applied to the box in the positive x-direction. (a) Sketch the box and show all forces acting on it. (b) Calculate
the magnitude of the force of friction and state whether the box moves for each of the following applied forces: 150 N, 450 N, 575 N, 720 N, 985 N. Hint: Do a
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quick calculation with the coefficient of static friction first. (150 N, No; 450 N, No; 575 N, No; 510 N, Yes; 510 N, Yes)
(16) A dockworker loading crates on a ship finds that a 20 kg crate, initially at rest on a rough horizontal surface, requires a 75 N horizontal force to set it in motion.
However, after it is in motion, horizontal force of 60 N is required to keep the crate moving with a constant speed. Find the coefficients of static and kinetic friction
between the crate and the floor. (.383,.306)
(18) To move a box of books along the floor, it is necessary for a rope to have a tension of 80.0 N at an angle of 35.0 degrees. The box of books has a mass of
25.0 kg and the coefficient of kinetic friction between the box and the floor is 0.300. (a) Make a free-body diagram of all the forces acting on the box. On your
diagram, illustrate the x and y components of the tension force of the rope on the box; also, calculate these components. (b) Calculate the normal force on the
box. (c) Calculate the magnitude of the kinetic force of friction on the box. (d) Use Newton's second law to calculate the horizontal acceleration of the box. [SF2,
Example 3.5] (65.5 N,45.9 N,199 N,59.7 N,0.232 m/s2)
(19) A man is applying a horizontal force to drag a 200. kg box of books on a floor whose static and kinetic coefficients of friction are 0.500 and 0.450. The box is
moving with a constant speed of 2 m/s. Calculate the magnitude of the force that the man applies to the box. (882. N) (new)
Given the following object of mass M = 15.0 kg which is hanging from the ceiling as shown.


M
(a) With a free-body diagram, illustrate the tension forces on the knot. Sketch the x and y components of tension forces T1 and T2 on the knot. (b) Apply the
equilibrium condition to the hanging mass in order to calculate the tension T3. (T3 = 147. N) (c) Specify the x and y components of the three tension forces on
the knot in terms of the magnitude of the tension force and, if necessary, the appropriate trigonometric function. (d) Using the equilibrium condition for forces,
write down the two x and y free-body equations for the knot. (e) Solve for the magnitudes of the tensions, T1 and T2 . (T1 = 147. N, T2 = 134. N)
(20) Does the following conversion equation make sense “physically” ? 1 lb = 2.20 kg
units and in terms of exclusively force units.
Explain how it should be corrected to read in terms of exclusively mass