Chapter 15. tension, pulleys and problem solving
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high-risk skateboarding
It’s a bird... it’s plane...
...no, it’s... a guy on a skateboard?!
A new high-risk sporting event has come to town. The challenge?
Jump off an 11.0 m pier and hit a target floating in the sea 15.0 m
from the foot of the pier. Michael, a daredevil and skateboarding
fiend, plans to take home the first place prize.
Michael on a
skateboard.
Stack of masses ready
to be dropped off the
edge of the pier.
Michael wants to give himself a predictable
launch velocity so he can be sure of hitting the
target in the water. He’s attached a skateboard to
one end of a rope, put a large stack of masses
at the other end, and placed a pulley in between.
The problem is, Michael’s not so great at physics.
And that’s where you come in... can you help
Michael out?
Rope is attached to the
stack and the skateboard.
When the skateboard reaches
the end of the pier, Michael
continues with velocity v.
Here’s what SHOULD happen...
Skateboard is pulled along
the pier by the rope.
v
With the correct initial
velocity at the end of
the pier, Michael will
follow this trajectory
and hit the bullseye.
The competition takes
place when the tide is
in and the sea is 11.0 m
below the top of the pier.
The center of
the target is
15.0 m from the
foot of the pier.
11.0 m
The mass has just
hit the surface
of the water.
15.0 m
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tension, pulleys and problem solving
Always look for something familiar
This problem involves a skateboard, a person, a stack of
masses, a rope, a pulley, gravity, the height of the pier, and the
distance to the target. Plus it takes place in two dimensions.
It’s a complicated problem!
But you don’t have to start the
problem in the same place as
Michael, with the rope, pulley and
stack of masses. The best place for
you to start tackling a problem is
from a point where it looks like
something you’ve seen before....
You’ve not seen ropes and
pulleys before. But you
HAVE seen a situation
like this before!
Michael has an initial
horizontal velocity the
moment he leaves the
skateboard.
v
You can start at the point where
Michael flies through the air with
velocity v. That looks kinda familiar...
Break down a
complicated problem
into smaller parts,
then look for something
that’s LIKE what
you’ve seen before.
Suppose Michael is launched horizontally from the end of a pier. What velocity does he need to
possess in order to hit a target in the water 15.0 m from the foot of the pier, which is 11.0 m high?
Always start
with a sketch
to help you
figure a
problem out.
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look for the familiar
This part of the problem uses equations of motion. The key
to solving it is to realize that you can do it separately from
the rest of the problem, before you know anything about the
stack of masses or the pulley.
Suppose Michael is launched horizontally from the end of a pier. What velocity does he need to
possess in order to hit a target in the water 15.0 m from the foot of the pier, which is 11.0 m high?
Down is positive direction.
av = 9.8 m/s2
xv = 11.0 m
Get time from vertical components:
V0h = ?
V0v = 0 m/s
x0v = 0 m
x0h = 0 m
xh = 15.0 m
These terms
xv = x0v + v0vt + ẵvat2
are both zero.
ẵat2 = xv
2 ì 11.0
t = 2xv =
9.8 = 1.50 s (3 sd)
av
Get horizontal velocity from horizontal components:
x
v0h = h = 15 = 10.0 m/s (3 sd) left to right.
t
1.50
For this part of the problem, Michael’s initial
velocity v0h = 10 m/s. But for the other part, with
the rope etc, his initial velocity is zero, and his final
velocity is 10 m/s. So we need to be careful, right?
Watch out for what you call
your variables in the next part.
Breaking this problem down has made it
easier - which is great! But if you forget
to redraw your sketch and redefine your
variables for the next part, you’ll just
confuse yourself.
Your starting point this time around was
when Michael is launched from the pier,
where you’ve worked out that his initial
velocity is v0h = 10.0 m/s.
But the other part of the problem involves
him reaching this velocity of 10.0 m/s
from a standing start - where it’ll be his
final velocity. So be careful!
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Michael’s velocity at this
point must be 10.0 m/s.
Here, Michael’s initial
velocity is 0 m/s.
When you break
up a problem, you
might need to
redefine variables
as you move from
part to part.
tension, pulleys and problem solving
So if Michael’s velocity
is 10 m/s, he hits the
target, right?
Jim: Yeah, and we’ve got that stack of weights on the other end
of the rope to accelerate him with.
Frank: Yeah, the stack is falling, so it will accelerate at 9.8 m/s2
and drag Michael along behind it. So Michael accelerates at
9.8 m/s2, just like the stack. This is gonna be a piece of cake!
Joe: Um, I’m not so sure about that. The stack has to pull
Michael along, so I don’t think the stack will fall as fast as it would
if it wasn’t attached to the skateboard. I don’t think it would
accelerate at 9.8 m/s2.
Frank: But if something’s falling, its acceleration doesn’t depend
on its mass. Everything falls at the same rate, no matter what its
mass is (as long as air resistance isn’t a big factor).
Joe: But Michael isn’t falling - he’s travelling horizontally.
Jim: Oh yeah ... I guess if there was an elephant on the
skateboard instead, the board would hardly accelerate at all.
Frank: Yeah, the force is due to the weight of the falling stack
- but not Michael’s weight, as he isn’t falling.
Joe: This isn’t so straightforward after all.
BE the skater
Your job is to imagine you’re
Michael on the skateboard.
What happens to you as
the mass at the other
end of the rope falls?
How will making the mass
larger or smaller affect
what happens to you?
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be the skater
BE the skater - SOLUTION
Your job is to imagine you’re
Michael on the skateboard.
What happens to you as
the mass at the other
end of the rope falls?
How will making the mass
larger or smaller affect
what happens to you?
If there was no mass
here, you wouldn’t go
anywhere at all.
The mass pulls you
in this direction.
A larger mass
accelerates
more rapidly.
As the mass falls vertically, I accelerate horizontally because it pulls the skateboard.
If the mass is larger, I accelerate more quickly.
If the mass is smaller, I accelerate more slowly, and if the mass is a really small I
might not accelerate at all.
Michael and the stack accelerate at the same rate
Michael and the stack are joined together by the rope, so they both
accelerate at the same rate. This is because of the tension in
the rope - the rope is pulled tight. If the rope wasn’t there or wasn’t
pulled tight, there would be no tension and Michael wouldn’t
accelerate as the stack falls.
A stack with a larger mass on it will accelerate Michael more
rapidly, and a smaller mass will accelerate him more slowly. This is
something you can analyze by thinking about the tension in the rope.
Normal force
Net force on
Michael = T
Tension = T
Tension = T
A free body
diagram for
Michael.
m
Weight
Weight = mg
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A free body diagram
for the stack.
A rope pulled tight
can mediate a
TENSION force.
The stack is attached to Michael by a rope. As well as
his weight and the normal force (which add to zero),
he also experiences a net tension force from the
rope, T, which accelerates him to the right.
As the stack exerts a force on Michael (via the rope)
Michael must exert an equal-sized force on the stack
(in this case, via the rope) . So the stack experiences
two forces - its weight and the tension.
Net force on
stack = mg - T
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tension, pulleys and problem solving
How can the tension act horizontally on one
thing and vertically on the other? Doesn’t
Newton’s 3rd Law say that pairs of forces
need to act in opposite directions?
The pulley changes the direction that the
tension force acts in.
If a rope experiences a pulling force at both ends that makes
it tight (like the rope we have here) it’s said to be in tension.
But a pulley changes the direction that the tension force
acts in. The pulley is able to do this because it’s firmly
attached to the pier, which is able to provide a support force.
Otherwise, the rope would just get pulled straight.
If you just draw the free body diagrams for Michael and the
stack, it looks like you have a Newton’s 3rd Law force pair
that isn’t acting in opposite directions. But when you include
the free body diagram of the pulley, you can see that there
are horizontal and vertical pairs of tension force pairs.
The forces on the
pulley must add to
zero, as the pulley
isn’t accelerating.
Pulley’s free
body diagram.
Normal force
Support
force
Michael’s free
body diagram.
T
Support
force
T
T
Weight
If there is a pulley,
you have to include
it when working
out force pairs.
This is a
force pair.
T
This is a
force pair.
T
m
Weight = mg
Stack’s free
body diagram.
T
With no support force, the
pulley would accelerate in
this direction as the rope
pulls straight..
Because the rope is being
pulled tight, it exerts a
force on the pulley. But the
pulley’s not accelerating,
so the net force on it
must be zero. Therefore,
the pier that the pulley’s
attached to must provide a
support force.
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objects joined together
Q:
Why doesn’t the stack just accelerate at 9.8 m/s2,
like it usually would if you dropped it?
A:
It’s attached to Michael (and his skateboard) by a rope.
So the force of the stack’s weight has to accelerate Michael’s
mass as well as the stack’s own mass.
Q:
But don’t all falling objects accelerate at the same
rate, whatever their mass?
A:
Q:
A:
Not if they’re tied on to something else that isn’t falling!
So where does tension come in?
Tension is the name given to the force exerted at each
end of a rope. For example, if the stack was hanging from a
rope attached to the ceiling, the tension would be the same
size as the stack’s weight.
Q:
Is the tension in a rope always the same size as the
weight of the object it’s supporting?
A:
If an object is hanging straight down from a rope, then
its acceleration is zero and the net force must be zero. So
the tension in the rope must be the same size as the object’s
weight (in the opposite direction).
But if the object is accelerating downwards, (like the stack is
here), the object’s weight must be greater than the tension in
the rope to produce a net downwards force.
Q:
Is it OK to think of a tension force a bit like the
normal force in other problems - as it’s a support force?
A:
Q:
As long as you remember that the tension always acts
in the direction of the rope, that should be OK conceptually.
Do I have to take into account the mass of the rope
as well?
A:
Great question! In real life you would, but in practice
the mass of the rope has very little effect if it’s much smaller
than the masses it’s attached to. So we’re making the
approximation that the rope is massless.
Q:
How do you know that the two objects attached to
the rope both accelerate at the same rate?
A:
If the two objects are at either end of a rope which is
pulled tight, they must always move with the same speed
in order for the rope to remain tight. Therefore, they must
also both accelerate at the same rate (though they may
accelerate in different directions depending on how the rope
is positioned).
When two objects are
joined together and
move together, they
both accelerate at the
SAME rate.
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tension, pulleys and problem solving
Use tension to tackle the problem
Because the stack and Michael are joined together by a
rope under tension, they both accelerate at the same rate
.You need to work out what mass the stack should have
in order for both it and Michael to be going at 10.0 m/s
after the stack has fallen 11.0 m from the top to the
bottom of the pier.
Michael needs to
have a velocity
of 10.0 m/s at
this point.
10.0 m/s
If you draw separate free body diagrams for the stack
and for Michael, showing all the forces acting on them,
you can use these to work out the mass of the stack.
I guess we need to be careful about
how we define the directions of our forces?
Think about how the rope moves.
When you’re using a pulley, with forces
mediated by the tension in a rope, you
have to be very careful about defining the
direction of your force vectors.
As the objects are connected to each other
by the rope, it’s best to define one direction
of rope movement as the positive
direction, draw the free body diagram for
each object separately, and mark the positive
direction on each free body diagram with a
big arrow.
Make sure your
‘positive direction’
arrow is a
different style
from your force
vector arrows.
Positive direction
Normal
force
Positive
direction
Tension = T
Tension = T
Weight
m
Define one
direction of rope
movement as
positive and mark
it on your free
body diagrams.
Weight = mg
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michael’s moving
Michael on a skateboard, mass M, is attached
via a rope and a pulley to a stack of masses,
mass m, as shown in the picture. When the
stack is allowed to accelerate downwards in
a gravitational field, strength g, the rope has
tension, T and Michael also accelerates.
a. Draw separate free body diagrams of Michael on the skateboard and of the stack
b. Write down the size of the net force on Michael.
c. Write down the size of the net force on the stack.
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tension, pulleys and problem solving
d. Michael and the stack both accelerate with acceleration, a. Use Newton’s 2nd Law to write down a
separate expression for each of them that relates their mass, their acceleration and the net force on them.
e. The size of the tension in the rope is the same for both Michael and the stack. Make a substitution using
your equations from part d. and rearrange your answer so that you have an equation for m, the mass of the
stack, in terms of M, g and a.
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free body diagrams
Michael on a skateboard, mass M, is attached
via a rope and a pulley to a stack of masses,
mass m, as shown in the picture. When the
stack is allowed to accelerate downwards in
a gravitational field, strength g, the rope has
tension, T and the man also accelerates.
a. Draw separate free body diagrams of Michael on the skateboard and of the stack.
Normal force
Positive direction
of rope movement.
Tension = T
Positive direction
of rope movement.
Tension = T
Weight = mg
Weight = Mg
b. Write down the size of the net force on Michael.
c. Write down the size of the net force on the stack.
Normal force and weight add to zero.
Fnet = T
Drawing on the big arrows
helps you get the signs
correct in these equations.
Fnet = mg - T
d. Michael and the stack both accelerate with acceleration, a. Use Newton’s 2nd Law to write down a
separate expression for each of them that relates their mass, their acceleration and the net force on them.
Man has mass M.
Fnet = ma becomes
T = Ma
Stack has mass m.
Fnet = ma becomes
Make substitutions.
mg - T = ma
e. The size of the tension in the rope is the same for both Michael and the stack. Make a substitution using
your equations from part d. and rearrange your answer so that you have an equation for m, the mass of the
stack, in terms of M, g and a.
T = Ma
mg - T = ma
(1)
(2)
Substitute the expression for
T in (1) into (2)
mg - Ma = ma
Rearrange equation to say “m = ... ”
mg - Ma = ma
mg - ma = Ma
m(g - a) = Ma
Then you can divide
m = gMa
both sides by (g - a)
-a
Both the g and the a
are multiplied by the m.
to get “m = ...”
So you can introduce some
parentheses so that there’s only one
instance of m on the left hand side.
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