Lecture 6
Newton’s Laws of Motion
Exam #1 - next Thursday!
- 20 multiple-choice problems
- No notes allowed; equation sheet provided
- A calculator will be needed.
- CHECK YOUR BATTERIES!
- NO equations or information may be stored in your
calculator. This is part of your pledge on the exam.
- Scratch paper will be provided, to be turned in at the
end of the exam.
Cart on Track I
Consider a cart on a
horizontal frictionless
table. Once the cart has
been given a push and
a) slowly come to a stop
b) continue with constant acceleration
c) continue with decreasing acceleration
released, what will
d) continue with constant velocity
happen to the cart?
e) immediately come to a stop
Cart on Track I
Consider a cart on a
horizontal frictionless
table. Once the cart has
been given a push and
a) slowly come to a stop
b) continue with constant acceleration
c) continue with decreasing acceleration
released, what will
d) continue with constant velocity
happen to the cart?
e) immediately come to a stop
After the cart is released, there is no longer a force in the
x-direction. This does not mean that the cart stops
moving!! It simply means that the cart will continue
moving with the same velocity it had at the moment of
release. The initial push got the cart moving, but that
force is not needed to keep the cart in motion.
Force and Two Masses
a) ¾a1
A force F acts on mass m1 giving acceleration a1.
b) 3/2a1
The same force acts on a different mass m2
giving acceleration a2 = 2a1. If m1 and m2 are
c) ½a1
glued together and the same force F acts on this
d) 4/3a1
combination, what is the resulting acceleration?
e) 2/3a1
F
F
F
m1
a1
F = m1 a1
a2 = 2a1
m2
F = m2 a 2 =
m2 m1
a3
a3 = ?
Force and Two Masses
a) ¾a1
A force F acts on mass m1 giving acceleration a1.
b) 3/2a1
The same force acts on a different mass m2
giving acceleration a2 = 2a1. If m1 and m2 are
c) ½a1
glued together and the same force F acts on this
d) 4/3a1
combination, what is the resulting acceleration?
e) 2/3a1
F
F
m1
a1
m2
F = m1 a1
a2 = 2a1
F = m2 a2 = (1/2 m1 )(2a1 )
Mass m2 must be ( m1) because its
acceleration was 2a1 with the same
force. Adding the two masses
together gives ( )m1, leading to an
F
m2 m1
a3
acceleration of ( )a1 for the same
applied force.
F = (3/2)m1 a3 => a3 = (2/3) a1
Apparent weight:
Weight
Your perception of your weight is based on the
contact forces between your body and your
surroundings.
If your surroundings
are accelerating, your
apparent weight may
be more or less than
your actual weight.
In this case the “apparent weight” is the sum of the
gravitational attaction (actual weight) and the force
required to accelerate the body, as specified
Newton’s Third Law of Motion
Forces always come in pairs, acting on different
objects:
If object 1 exerts a force
object 2 exerts a force –
on object 2, then
on object 1.
These forces are called action-reaction pairs.
Some action-reaction pairs
Action-reaction pair?
a) Yes
b) No
Newton’s 3rd: F12 = - F21
action-reaction pairs are equal and
opposite, but they act on different bodies
Newton’s Third Law of Motion
Although the forces are the same, the
accelerations will not be unless the objects
have the same mass.
Q: When skydiving, do you
exert a force on the
earth? Does the earth
accelerate towards you?
Is the magnitude of the
acceleration of the earth
the same as the magnitude
of your acceleration?
Contact Force
Two blocks of masses 2m and m
a) 2F
are in contact on a horizontal
b) F
frictionless surface. If a force F0
c) ½F
is applied to mass 2m, what is the
d) 1/3F
force on mass m ?
e) ¼F
F
2m
m
Contact Force
Two blocks of masses 2m and m
a) 2F
are in contact on a horizontal
b) F
frictionless surface. If a force F0
c) ½ F
is applied to mass 2m, what is the
d) 1/3 F
force on mass m ?
e) ¼ F
The force F0 leads to a specific
acceleration of the entire system. In
order for mass m to accelerate at the
same rate, the force on it must be
smaller! For the two blocks together,
F0 = (3m)a. Since a is the same for both
blocks, Fm = ma
F
2m
m
Newton’s Third Law of Motion
Contact forces:
The force exerted by one
box on the other is
different depending on
which one you push.
Assume the mass of
the two objects scales
with size, and the
forces pictured are the
same. In which case is
the magnitude of the
force of box 1 on box 2
Two boxes sit side-by-side on a smooth horizontal surface. The lighter box has a mass
of 5.2 kg, the heavier box has a mass of 7.4 kg.
(a) Find the contact force between these boxes when a horizontal force of 5.0 N is
applied to the light box.
(b) If the 5.0-N force is applied to the heavy box instead, is the contact force between
the boxes the same as, greater than, or less than the contact force in part (a)? Explain.
(c) Verify your answer to part (b) by calculating the contact force in this case.
The boxes will remain in contact, so must have the same acceleration. Use
Newton’s second law to determine the magnitude of the contact forces
(a)
(b)The lighter box will require less net force than
the heavier box did (for the same acceleration). If
the external force is instead applied to the heavier
box, the contact force must be less than it was
before:
Collision Course I
a) the car
A small car collides with
a large truck. Which
experiences the greater
impact force?
b) the truck
c) both the same
d) it depends on the velocity of each
e) it depends on the mass of each
Collision Course I
a) the car
A small car collides with
a large truck. Which
experiences the greater
impact force?
b) the truck
c) both the same
d) it depends on the velocity of each
e) it depends on the mass of each
According to Newton’s Third Law, both vehicles experience
the same magnitude of force.
Forces in Two Dimensions
The easiest way to handle forces in two
dimensions is to treat each dimension separately,
as we did for kinematics.
y
x
mass of child + sled = 19 kg
Fmom,y
35o
Fmom,x
Σ Fy = 0
ay = 0
ax = Fx/m = 33N / 19 kg = 1.7 m/s2
Normal Forces
The normal force is
the force exerted by
a surface on an
object, to keep an
object above the
surface.
The normal force
is always
perpendicular to
the surface.
The normal force may be equal
to, greater than, or less than
the weight.
No vertical motion, so
Normal Force
Below you see two cases: a
physics student pulling or
pushing a sled with a force F
that is applied at an angle  . In
which case is the normal force
greater?
a) case 1
b) case 2
c) it’s the same for both
d) depends on the magnitude of
the force F
e) depends on the ice surface
Case 1
Case 2
Normal Force
Below you see two cases: a
physics student pulling or
pushing a sled with a force F
that is applied at an angle  . In
which case is the normal force
greater?
a) case 1
b) case 2
c) it’s the same for both
d) depends on the magnitude of
the force F
5) depends on the ice surface
Case 1
In case 1, the force F is pushing down (in
addition to mg), so the normal force needs
to be larger. In case 2, the force F is
pulling up, against gravity, so the normal
force is lessened.
Case 2
On an Incline
Consider two identical
a) case A
blocks, one resting on a
b) case B
flat surface and the other
c) both the same (N = mg)
resting on an incline. For
which case is the normal
force greater?
A
d) both the same (0 < N < mg)
e) both the same (N = 0)
B
On an Incline
Consider two identical
a) case A
blocks, one resting on a
b) case B
flat surface and the other
c) both the same (N = mg)
resting on an incline. For
which case is the normal
force greater?
d) both the same (0 < N < mg)
e) both the same (N = 0)
In case A, we know that N = W.
y
In case B, due to the angle of
the incline, N < W. In fact, we
N
f
can see that N = W cos( ).
A
B

W
 Wy
x
Newton’s Laws
(I)
In order to change the velocity of an object –
magnitude or direction – a net force is required.
(II)
(III)
Newton’s third law: If object 1 exerts a force on
object 2, then object 2 exerts a force – on object 1.
Ftop
A weight on a string with
identical string hanging from
bottom …
if I pull the bottom string
down, which string will
break first?
a) top string
Fbot
W
b) bottom string
c) there is not enough information to answer
this question
When you lift a bowling ball with a force of 82 N, the ball accelerates upward
with an acceleration a. If you lift with a force of 92 N, the ball’s acceleration
is 2a. Find (a) the weight of the bowling ball, and (b) the acceleration a.
When you lift a bowling ball with a force of 82 N, the ball accelerates upward
with an acceleration a. If you lift with a force of 92 N, the ball’s acceleration
is 2a. Find (a) the weight of the bowling ball, and (b) the acceleration a.
a)
b)
On vacation, your 1300-kg car pulls a 540-kg trailer away from a
stoplight with an acceleration of 1.9 m/s2
(a) What is the net force exerted by the car on the trailer?
(b) What force does the trailer exert on the car?
(c) What is the net force acting on the car?
A 71-kg parent and a 19-kg child meet at the center of an ice rink. They
place their hands together and push.
(a) Is the force experienced by the child more than, less than, or the
same as the force experienced by the parent?
(b) Is the acceleration of the child more than, less than, or the same as
the acceleration of the parent? Explain.
(c) If the acceleration of the child is 2.6 m/s2 in magnitude, what is the
magnitude of the parent’s acceleration?
Before practicing his routine on the rings, a 67-kg gymnast stands
motionless, with one hand grasping each ring and his feet touching the
ground. Both arms slope upward at an angle of 24° above the
horizontal.
(a) If the force exerted by the rings on each arm has a magnitude of 290
N, and is directed along the length of the arm, what is the magnitude
of the force exerted by the floor on his feet?
(b) If the angle his arms make with the horizontal is greater that 24°,
and everything else remains the same, is the force exerted by the
floor on his feet greater than, less than, or the same as the value
found in part (a)? Explain.
N
An archer shoots a 0.022-kg arrow at a target with a speed of 57 m/s.
When it hits the target, it penetrates to a depth of 0.085 m.
(a) What was the average force exerted by the target on the arrow?
(b) If the mass of the arrow is doubled, and the force exerted by the
target on the arrow remains the same, by what multiplicative factor
does the penetration depth change? Explain.