# Homework Help: Coefficient of friction pulley problem

1. Aug 25, 2011

### wallism

This question this struck me as fairly simple until I repeatedly tried and failed to get the right answer so any help would be greatly appreciated.

1. The problem statement, all variables and given/known data

A block of mass 30 tonnes is to be dragged across the floor by means of the smaller body (mass 5 tonnes) acting through the pulley system as shown (see attachment). If the coefficient of friction is 0.2 calculate the acceleration of the large block.

Ans 0.78ms^-2

2. Relevant equations

F=ma
F=uR

3. The attempt at a solution

Firstly looking at the 30t block

Using F= u R

0.2 x 9.81 x 30000kg = 58860N

So this is the force needed to be overcome for the block to slide.

Now focus on the affect the smaller block is having.

Weight = 49050N

The pulley system doubles this force on the larger block so that it has a force pulling it to
the right of 98100N?

Which would mean the block would slid with a force of 98100-58860=39240N

Use f/m = a, 39240N/30000kg = 1.308 ms^-2 which is not right

I seem to be having a fair few problems with questions like this. Is there something fundamental that I have missed here?

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Last edited: Aug 25, 2011
2. Aug 25, 2011

### tiny-tim

welcome to pf!

hi wallism! welcome to pf!

(have a mu: µ )
you should always do a separate F = ma for each moving block

i don't think you've done it for the little one

3. Aug 25, 2011

### Staff: Mentor

If you draw Free Body Diagrams for both masses you'll see that both of them have (related) accelerations; the force exerted on the 30T mass will only be 2 x 5T*g if the system is somehow held static.

4. Aug 26, 2011

### wallism

I’m sorry, I feel as thick as two planks, I’ve been trying for hours to do this now to no avail.

I’ve done each an individual free body diagram. For the smaller block downwards I have F=(m1)(a1). And tension upwards as equal to that so t1=(m1)(a1).

For the larger block there’s the vertical reaction force and weight. There’s a force to the left f=mu x R = -58860N (because it’s to the left). And to the right is t2. This t2 is twice as big as t1 due to the pulley system so 2t1=t2.

This means t2 = 2 x m1 x a1 = 10000(a1)

Use this to find the resultant f on the larger block = 10000(a1)-58860

Sub into the formula f/m = a gives [10000(a1)-58860]/30000 = a2

Simplifies to 1/3(a1)-1.962 = a2

And that’s it. I don’t know how to go about solving this, is this even right?

Thanks

5. Aug 26, 2011

### tiny-tim

sorry, wallism, i can't follow this

start again …

call the tension T (it will be the same in all three sections of the rope)

call the acceleration of the smaller mass a …

1) then what is the acceleration of the larger mass?

2) what is the F = ma equation for the smaller mass?

3) how many forces are there on the larger mass?

4) what is the F = ma equation for the larger mass?​

6. Aug 26, 2011

### wallism

Sorry, very frustrated at the moment.

1) then what is the acceleration of the larger mass?

A=f/m so A = (2T-µR)/m

2) what is the F = ma equation for the smaller mass?

force downwards = ma - T

3) how many forces are there on the larger mass?

One resultant to the right, which is 2T
Friction (µR to the left)
Reaction force and weight

4) what is the F = ma equation for the larger mass?

Not sure, presumably something different from the top one?

Is that any clearer?

7. Aug 26, 2011

### tiny-tim

1) no, i meant what is the acceleration of the larger mass in terms of the acceleration of the smaller mass (a) ?

3) it'd be clearer to say that there are two forces to the right, of T each

2) and 4) why have you not used g ?

8. Aug 27, 2011

### wallism

1) So the acceleration of the larger mass.
(2T-muR)/m = a

2) The F = ma equation for the smaller mass
F = T - mg

4) The F = ma equation for the larger mass on terms of the acceleration of the smaller mass (what you asked for in part one!?)

(2(f + mg)-muR)/30000 = a

Anywhere close? Still need an f to solve the final equation though...

Thank you again

9. Aug 27, 2011

### tiny-tim

hi wallism!

(what happened to that µ i gave you? )
yes, but what is R?

and i really meant how much bigger is the acceleration (A) of the larger mass than the acceleration (a) of the smaller mass …

ie, A = (?)a
the F = ma equation is mg - T = ma

carry on from there

10. Aug 27, 2011

### wallism

sorry it's here -> µ

R is just the normal reaction force, gm, 9.81 x 30000

So if I use mg-T = ma in my equation I get

(2mg - 2ma - µR )/30000 = A

And I can substitute in all the numbers to leave me with

(98100 - 10000a - 58860)/30000 = A

which simplifies to

39240 - 10000a = 30000A

then

1.308 - (1/3)a = A

Is that what you meant?

Thanks again

Last edited: Aug 27, 2011
11. Aug 27, 2011

### tiny-tim

i knew you had it somewhere!
i haven't checked the numbers, but yes that's the method …

you start with F = ma at one end, and work your way through to the other end

now all you need to do is to find the value of A/a (this is geometry, rather than physics)

12. Aug 27, 2011

### wallism

Okay thank you very much. So that equation is the relationship between the accelerations.

What will A/a give me? And I don't really know how to get started on that either...

13. Aug 27, 2011

### tiny-tim

You already have an equation in A, and an equation in a …

you need to know A/a to solve them simultaneously

let the vertical position of the smaller mass be x, and the horizontal position of the larger mass be X …

then x'' = a and X'' = A …

so you need to find X/x, in other words:

if you pull the smaller mass down one cm, how far does the larger mass move?

14. Aug 27, 2011

### Lobezno

Are you assuming that the body is already in motion? If not, then you'll need both the static and kinetic coefficients of friction.