Homework Help: Newton law's - Finding acceleration

1. Apr 28, 2013

Saitama

1. The problem statement, all variables and given/known data
(see attachment)

2. Relevant equations

3. The attempt at a solution
I assumed that the acceleration of 5M is towards the right and is $a$. Moving to the frame of 5M, the other two blocks experience a pseudo force.
The forces acting on M are the tension in the string, pseudo force, its weight and the normal reaction from 5M.
$T+Ma=Ma'$
Similarly for 2M,
$2Mg-T=2Ma'$
But I still need one more equation to reach the answer.

Any help is appreciated. Thanks!

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2. Apr 28, 2013

ehild

The equation for the CM?

ehild

3. Apr 28, 2013

Saitama

$$a_{cm}=\frac{Ma'+2Ma'}{8M}=\frac{3a'}{8}$$
Is this correct? And how does this help me?

4. Apr 28, 2013

ehild

Does the CM accelerate in horizontal direction in the inertial frame of reference?
By the way , I would use inertial frame for all parts.

ehild

5. Apr 28, 2013

Saitama

Whoops, missed that.
$$a_{cm,x}=\frac{-a'}{8}$$
$$a_{cm,y}=\frac{-a'}{4}$$
I have assumed that the direction to right is positive x-axis and the direction vertically upwards is positive y-axis.

How will you do this using the inertial frame? I assume I will have to consider the motion of the 2M block in horizontal and vertical directions.

6. Apr 28, 2013

ehild

The cm accelerates in vertical direction, but is there any external horizontal force?

ehild

7. Apr 28, 2013

Saitama

No. Do you mean I should have $a_{cm,x}=0$? I think my expression for $a_{cm,x}$ is incorrect.

8. Apr 28, 2013

ehild

I mean the CM of the whole system in the inertial frame of reference.

ehild

9. Apr 28, 2013

Saitama

Sorry, I still don't get it. Can you elaborate a bit more? :)

10. Apr 28, 2013

haruspex

If you're not comfortable thinking about the horizontal accn of CM, you can get two more equations by introducing one more unknown, the normal reaction between 5M and 2M.

11. Apr 28, 2013

ehild

You need the acceleration of the big block with respect to the ground. Of course, the acceleration is zero with respect to itself, so you can not avoid using the inertial frame of reference.

ehild

12. Apr 28, 2013

Saitama

I have to do this in the frame fixed to 5M or inertial frame?

I will make equations again for the inertial frame of reference.
For M, the forces acting on it are tension, weight & normal reaction from 5M.
$T=Ma$ (I feel this is not correct)

For 2M, it performs two motions, one in horizontal and the other in vertical direction.
For horizontal direction,
$N=2Ma'$ (where N is the normal reaction on 2M by 5M)
For vertical direction,
$2Mg-T=2Ma$

For 5M, it perform horizontal motion, the forces acting on 5M in the horizontal direction are the tension and the normal reaction due to 2M but how would should be the value of tension? Should it be T?

13. Apr 28, 2013

ehild

The tension is the same along a single rope if the pulley has zero mass.

ehild

14. Apr 28, 2013

Saitama

I should have used the proper words. I meant the direction of force of tension on 5M.

15. Apr 28, 2013

ehild

Actually there are three tensional forces on the big block...

16. Apr 28, 2013

Saitama

But I need to consider only one, the other two are in vertical direction. Right?

17. Apr 28, 2013

ehild

I meant three horizontal forces of tension.

18. Apr 28, 2013

Saitama

I still don't see how there are three horizontal forces of tension.

19. Apr 28, 2013

ehild

Two at the left pulley and one at the right one.

20. Apr 29, 2013

sankalpmittal

Its all Newton's THIRD Law.

At the right pulley, there is a string passing over it. That pulley, due to strands of the string exerts a force on the big block rightwards due to newton's third law. (Why ?) Can you find that force ?

Also, at the left pulley, there are two forces acting on the big block and they are horizontal. (Find their directions as a hint.)

Also, on observing carefully the right pulley, there is one strand of the string pushing the big block downwards. (That's Newton's third law. Do not consider that as tension.)

21. Apr 29, 2013

Saitama

Looks like I reached close to the answer but its still not correct.

For 5M, it performs horizontal motion, hence $T-N=5Ma'$ ...(i)
$T=Ma$ ....(ii)
$2Mg-T=2Ma$ ....(iii)
$N=2Ma'$ ....(iv)

From (i) and (iv), $T=7Ma'$, equating this with (ii), $a=7a'$
From (ii), $2Mg=3Ma \Rightarrow 2g=21a' \Rightarrow a'=2g/21$ but this is incorrect.

22. Apr 29, 2013

ehild

The 2M mass moves both in vertical and horizontal direction, with different accelerations.

ehild

23. Apr 29, 2013

Saitama

Yes, I even took care of that. I expressed the vertical acceleration(a') in terms of thorizontal acceleration(a).

24. Apr 29, 2013

ehild

The vertical acceleration of 2M is not the same as the horizontal acceleration of M.

25. Apr 29, 2013

SammyS

Staff Emeritus
Why not?