# Wordy Physics on Forces, Pulleys, and Friction!

1. Oct 10, 2012

### capn awesome

Hi. I'm new here, and decided to sign up now that physics is actually difficult for me in first year university... Hope to find the help I need, and help those that I can. Anyway..

I'm having trouble with these wordy physics problems...

1) A light rope passing over a light frictionless pulley is fastened to a platform and a man on the platform holds the other end of the rope. The man pulls on the rope with sufficient force to give himself and the platform an upward acceleration of 0.91m/s^2. Find the tension in the rope and the reaction between the man's feet and platform on which he stands. The mass of the man and the platform are 72kg and 36 kg respectively.

So I tried drawing my FBD, where acceleration points upwards from the platform, and tension points upwards on each side of the pulley. Fg points downwards from the platform. Finally there is an applied force, Fp, opposing tension on the side with the man. I stated that F = ma, added the two masses together to find the force, and divided that by two to get the tension. Am I on the right track...? And then what is it asking me to do for the "reaction between the man's feet and the platform"?

Fp = a(m1+m2)
Ftension = Fp/2

2) A long plank weighing 142N slides on a level frictionless surface with initial speed = 7.3m/s moving the direction of its length. A block weighing 35.5N with initial speed = 0m/s is set down on the plank. If coefficient of kinetic friction between the plank and block is 0.5, what is the acceration of the plank and block?

Also, what is the length of the skid mark made by the block on the plank?

So far, I have found the masses of each using the given weights, found the force of kinetic friction by multiplying the coefficient with Fn. Where would I go with that to find the deceleration of the plank? I assume the masses will have to combine, but what does net force equal in order to solve for acceleration?

mass of plank = m1 = 14.5kg
mass of block = m2 = 3.6kg
kinetic friction = μFn = 17.8N

2. Oct 11, 2012

### Simon Bridge

Welcome to PF;
The purpose of these problems is to get you used to extracting the important information from a written or spoken description. It is to train up your judgement.

For (1)
The applied force of the man pulling on the rope is what creates the tension - as far as the man-mass is concerned, there is just the tension pulling up.

When the man pulls on the rope, the platform rises, and presses against his feet.
This is the reaction force they are talking about - you need to include it on your free body diagrams. You should end up with two equations with T (tension) and R (reaction) as your unknowns.

For (2)
... never assume - you will just make an *** out of u and me ;)
Did you try drawing the free-body diagrams for the different masses?

For both of them: you have been combining the masses too soon.
Write the expressions for ƩF=ma for each mass separately.
Let the math tell you how the masses combine.

3. Oct 11, 2012

### capn awesome

"The applied force of the man pulling on the rope is what creates the tension - as far as the man-mass is concerned, there is just the tension pulling up."

What exactly do you mean by this? If the man and platform were stationary instead of accelerating, would there be two different forces of tension? In other words, is there only one tension force because he is applying a force instead of just countering the force of gravity?

Even still, wouldn't the tension simply oppose the applied pull on the rope, working out to the same equation just not halved? The man and the platform move as one unit with the same upward acceleration... I must really be missing something crucial here, because I am not seeing how the forces being calculated separately do me any good.

4. Oct 11, 2012

### Simon Bridge

I mean that when you do a free body diagram you only consider the forces on the body.
The applied force is not applied to the man (it actually contributes to the reaction force at the feet via the rope.)

Tension can be tricky to think about: if you and a mate are having a tug of war, and you both pull on your end of the rope with force 50N - what is the tension in the rope?
(a) 0 (b) 50N (c) 100N (d) none of these

Did you try it? You get two equations with two unknowns.

5. Oct 11, 2012

### capn awesome

Is it 100N..? Conceptually, if I was pulling on the rope myself with the other end attached to a fixed point, there would be 50N of tension pulling the other way. If someone on the other end is pulling with the same force, there is still tension in the rope.

For the pulley question, I calculated F = ma for the man and the platform separately to get 32.8N and 65.5N.. Is this what you meant? I don't see how those values help me when you can add them together to get an almost equal force by adding the masses and solving.

For question 2)

I calculated the force of friction between the plank and the block, as it should be the only thing affecting both of their accelerations.

Ff = μFn
Ff = μ(m-block)(g)
Ff = (0.5)(3.62)(9.8)
Ff = 17.7N

Then dividing that value by each mass to get the acceleration of each separate item.

Friction is causing the plank to slow down in the opposite direction of its motion, so it is negative.

a-plank = -17.7/14.5 = -1.2m/s^2

Friction is causing the block to accelerate on the board.

a-block = 17.7/3.62 = 4.9m/s^2

Did I do this correctly? And if so, how would I find the length of the skid mark left by the block? I know it would be the elapsed time between the block hitting the plank to the time where static friction takes over for kinetic friction.. Not sure what formulas I could use here.

6. Oct 11, 2012

### Simon Bridge

And here is your problem ... it is 50N.
Proof - you can fix the rope to a post stuck in the ground anywhere along it's length without changing the physics. Each person is still pulling at 50N and each feels a 50N tension pulling back.
Dunno - I didn't crunch the numbers. That's your job. But there should be a summation sign before the F.

If the man has mass M, and the platform mass m, the tension is T, the reaction force is R, show me the equations from the free body diagrams.

For (2)
I still cannot tell if you got it right because you did not show enough of your working.
I'll give you an example of what I'm expecting:

The plank (weight W) is moving under the influence of only one force, friction (f) which acts against the motion of the plank so:
$\sum F=m_pa_p \Rightarrow -f=Wa_p/g$ ... eq.(1)

Friction is given by the weight (w) of the block so $f=\mu w$ ... eq.(2)
... and you need a third equation which you get from the other fbd.
You seem to have got the essential lesson that friction does not act opposite to the speed.

The "length of the skid mark" part is about your understanding of how friction works.