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Chapter 15. tension, pulleys and problem solving

# 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

604   Chapter 15

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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.

you are here 4   605

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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!

606   Chapter 15

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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.

610   Chapter 15

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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.

you are here 4   613

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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.

614   Chapter 15

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