4.6 Newton’s 3rd Law (N3)
Q: What is N3?
Q: What does it mean?
For every action-reaction pair, there are two forces acting. If they cancel one
another out, Fnet = 0.
Q: What if they don’t?
Look at Figure 4.19, p 101. Without reading any text, what can you conclude
about the diagram shown (aside from the fact that that is one BIG table!)?
Q: Is this a complete diagram for all the objects depicted in the diagram? Is it
a proper FBD? Which masses / forces are being represented? Which are not?
4.7 Problems with Two or Morr Objects
Q: What is tension? Can there be a true force opposed to tension?
A: Tension is nothing more than the force exerted by a rope, string, wire, or
spring. Many things are capable of both exerting a push and a pull. The
oppositional force exerted by a rope very well may be at the other end of the
same rope.
DEF: Tension = the magnitude of the force that one segment of the rope exerts
on the neighboring segment.
The tension throughout a rope can vary throughout the same rope.
Because we still get to live in a “perfect physics world,” at least some of the
time, variations in the masses of the ropes, when comparing one part of the
rope to another, are assumed negligible. Essentially, this allows us to assume
that the tension along the rope is the same everywhere, providing it isn’t
moving over a pulley.
Movement of a rope over a pulley requires a difference in tension to change the
rate of rotation of the pulley.
See Ex 4-11, p 102
A horse refuses to pull a cart. The horse reasons, “according to N3, whatever
force I exert on the cart will be exerted equal and opposite to me, so the net
force will be zero and I will have no chance of accelerating the cart.” What is
wrong with this reasoning?
It’s not a true action-reaction pair. We are looking for the acceleration of the
cart, not the horse. The reaction force, F’, is exerted on the horse not on the
cart. What will accelerate the horse is the force of the pavement on the horse.
(That is why / how we are able to move when we walk.) The force of the
pavement on the horse is greater than the reaction force from the cart on the
horse, F’, thus the horse accelerates the cart.
See Ex 4-12, p 104.
Paul accidentally falls off the edge of a glacier as shown in Fig 4-21 (p 103). He
is tied by a long rope to Steve, who has a climbing ax. Before Steve sets his ax
to stop them, he slides without friction along the ice, attached by the rope to
Paul. Assume no friction between the rope and the glacier. Find the
acceleration of each person and the tension in the rope.
Note: The Tensions are equal in magnitude, but opposite in direction, as
shown on p 103.
Q: Should they be?
A: Yep. That’s what N3 is all about!
Q: If the tensions are equal, but opposite in direction, what will be true of the
accelerations? Why?
A: They will be equal but opposite, too. T is just a force. F ~ ma, which
means F is ALWAYS ~ a, as m is just a scalar. That means “a” is in the same
direction as the F, which is “T” in this case.
We will use “S” for Steve and “P” for Paul.
S:
∑ Fx : T + msg sinθ = msax
∑ Fy : Fn – msg cosθ = ms · 0
P:
∑ Fy : T – mpg = may
Note: The T’s are equal but opposite, which means the
same is true of the a’s. So, what was the acceleration in
x for Steve becomes the equal but opposite acceleration
in y for Paul.
P:
∑ Fy : T – mpg = may → T – mpg = mp(-ax)
T = mpg - mpax
Since Ts = - Tp, and you have already replaced ax with
– ax, substitute for T, and solve for ax.
∑ Fx : T + msg sinθ = msax
∑ Fx : mpg - mpax + msg sinθ = msax
mpg + msg sinθ = msax + mpax
mpg + msg sinθ = (ms + mp) ax
mpg + msg sinθ = (ms + mp) ax
(ms + mp)
(ms + mp)
g(mp + ms sinθ) = ax
(ms + mp)
Now that you have ax, substitute that back into T = mpg - mpax
T = mpg – mpg  ms sin θ + m p 


ms + m p


T = - msmpg sin θ – mp2g + mpg
ms + mp
Cross-multiply by m1 + m2 to get
it out of the way for now. Note
that the last mpg is not over ms + mp
so, you need to multiply through by
ms + mp on top and bottom to
move the quantity ms + mp to
the other side for now. That
will give you msmp + mp2 g. ☺
(ms + mp)T = - msmpgsin θ – mp2g + msmpg + mp2g
(ms + mp)T = (-mssin θ – mp + ms + mp) mpg
(ms + mp)T = (-mssin θ + ms) mpg
(ms + mp)T = msmpg - msmpg sin θ
T = msmpg (1 – sin θ)
(ms + mp)
See Ex 4-13, p 105
You are an astronaut constructing a space station, and you push on a box of
mass m1 with a force of FA. The box is in direct contact with a second box of
mass m2. a) What is the acceleration of the boxes? B) What is the magnitude
of the force exerted by one box on the other?
Note: All of the forces are in x only. ☺
Box 1: FA – F2,1 = m1a1,x
Box 2: F1,2 = m2a2,x
a2,x = a1,x = ax
F2,1 = F1,2 = F
ax = __FA__
m1 + m2
This is the same result you’d get if FA acted on a single mass
that was equal to the sum of the two masses.
Q: Should that be the case?
A: Yep. The boxes have the same acceleration, they act as a
single object.
F = _FAm2_
m1 + m2