Newton law's - Finding acceleration

[Saitama]

Homework Statement

(see attachment)

Homework Equations

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!

 

[ehild]

The equation for the CM?

ehild

 

[Saitama]

The equation for the CM?

ehild

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

 

[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

 

[Saitama]

Does the CM accelerate in horizontal direction in the inertial frame of reference?

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.

By the way , I would use inertial frame for all parts.

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.

 

[ehild]

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

ehild

 

[Saitama]

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

ehild

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

 

[ehild]

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

ehild

 

[Saitama]

I mean the CM of the whole system.

ehild

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

 

[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.

 

[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

 

[Saitama]

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.

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

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

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?

 

[ehild]

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

ehild

 

[Saitama]

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

ehild

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

 

[ehild]

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

 

[Saitama]

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

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

 

[ehild]

I meant three horizontal forces of tension.

 

[Saitama]

I meant three horizontal forces of tension.

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

 

[ehild]

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

 

[sankalpmittal]

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

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.)

 

[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)
I had the following equations:
$T=Ma$ ...(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.

 

[ehild]

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

ehild

 

[Saitama]

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

ehild

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

 

[ehild]

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

 

[SammyS]

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

Why not?

 

[Saitama]

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

Then how am I going to make one more equation for that? And why they aren't same?

 

[ehild]

Sammy, do you ask me? I wish Pranav to find it out. ehild

 

[ehild]

Then how am I going to make one more equation for that? And why they aren't same?

Why should they be the same? Remember, you are in the inertial frame of reference for all masses.

 

[Saitama]

Why should they be the same? Remember, you are in the inertial frame of reference for all masses.

I haven't got any idea about it but is this because of the motion of 5M?

 

[ehild]

Yes.

 

[Saitama]

Yes.

But how should I form the equations now? I need one more equation.

 

[SammyS]

Sammy, do you ask me? I wish Pranav to find it out.

ehild

My mistake.

Yes, I see now.

 

[rl.bhat]

Hear is a different approach.
Consider x- co-ordinates of M, 5M and 2M. When M moves through a distance x1 towards x- axis,
the whole system moves through a distance x2 away from the x-axis.
Hence the net change in the x co-ordinate of CM = M(x2 - x1) + 5Mx2 + 2Mx2 = 0.
Replace x1 and x2 by a1 and a2. You have already found a1. Find a2.

 

[ehild]

But how should I form the equations now? I need one more equation.

How is the acceleration of M related to the vertical acceleration of 2M and the acceleration of 5M in the inertial frame of reference? The length of the string can not change. And M moves with respect to 5M.

Start a new page, denote the horizontal acceleration components by a1,a2,a5, the vertical acceleration of 2M by a2y.
Decide a positive x and positive y direction and write up all equations for a1, a2, a5, and a2y.

ehild

 

[ehild]

Hear is a different approach.
Consider x- co-ordinates of M, 5M and 2M. When M moves through a distance x1 towards x- axis,
the whole system moves through a distance x2 away from the x-axis.
Hence the net change in the x co-ordinate of CM = M(x2 - x1) + 5Mx2 + 2Mx2 = 0.
Replace x1 and x2 by a1 and a2. You have already found a1. Find a2.

rbhat: I would like that Pranav figure out the relation between the accelerations. He does not need a different approach. And he haven't found a1 yet.

ehild

 

[Saitama]

How is the acceleration of M related to the vertical acceleration of 2M and the acceleration of 5M in the inertial frame of reference? The length of the string can not change. And M moves with respect to 5M.

Start a new page, denote the horizontal acceleration components by a1,a2,a5, the vertical acceleration of 2M by a2y.
Decide a positive x and positive y direction and write up all equations for a1, a2, a5, and a2y.

ehild

The direction to the right is positive x-axis and vertically upwards is y-axis.
Equation for motion of M: $T=Ma_1$

Equation for motion of 2M:
In horizontal direction: $N=2Ma_2$
In vertical direction: $2Mg-T=2Ma_{2y}$

Equation for motion of 5M: $T-N=5Ma_5$

I only have $a_2=a_5$ but I still need one more equation to relate $a_1$ and $a_{2y}$. I still can't figure out what I am missing. :(

 

[SammyS]

The direction to the right is positive x-axis and vertically upwards is y-axis.

Equation for motion of M: $T=Ma_1$

Equation for motion of 2M:

In horizontal direction: $N=2Ma_2$

It would be clearer to call this $\ a_{2x}\$ rather than $\ a_{2}\ .$

In vertical direction: $2Mg-T=2Ma_{2y}$

Equation for motion of 5M: $T-N=5Ma_5$

I only have $a_2=a_5$ but I still need one more equation to relate $a_1$ and $a_{2y}$. I still can't figure out what I am missing. :(

With the above change, you have $\ a_{2x}=a_5\$


Notice that if block 2M moves distance, d, down, then block M moves distance, d, to the left with respect to block 5M. That should tell you how $\ a_1\$ is related to $\ a_{2y}\$ and $\ a_{5}\ .$

 

[ehild]

I only have $a_2=a_5$ but I still need one more equation to relate $a_1$ and $a_{2y}$. I still can't figure out what I am missing. :(

You miss two things:

i. The length of rope does not change. When the hanging block goes down the vertical piece of rope becomes longer by dL (shown in blue). The horizontal piece becomes shorter by dL, so the block 1 slides to the left by dL. The speeds of both blocks dL/dt are the same with respect to the big block. The velocity components: negative velocity of block 2 involves negative velocity of block 1. If block 1 would move with speed v₁' to the right, block 2 would move upward with the same speed, so again, v₁'=v_2y '- with respect to the big block.

ii. We are in the frame of reference fixed to the ground. In that frame of reference the big block moves with velocity v₅ in horizontal direction. With respect to it, the horizontal velocity of block 2 is zero, so v_2x=v₅.
If block 1 has velocity v₁' with respect to block 5, moving with velocity v₅, block 1 moves with v1=v5+v1' with respect to the ground. (Consequence of Galilean transformation http://psi.phys.wits.ac.za/teaching/Connell/phys284/2005/lecture-01/lecture_01/node5.html) The same is true for the accelerations, dv/dt.

ehild

 

[Saitama]

You miss two things:

i. The length of rope does not change. When the hanging block goes down the vertical piece of rope becomes longer by dL (shown in blue). The horizontal piece becomes shorter by dL, so the block 1 slides to the left by dL. The speeds of both blocks dL/dt are the same with respect to the big block. The velocity components: negative velocity of block 2 involves negative velocity of block 1. If block 1 would move with speed v₁' to the right, block 2 would move upward with the same speed, so again, v₁'=v_2y '- with respect to the big block.

ii. We are in the frame of reference fixed to the ground. In that frame of reference the big block moves with velocity v₅ in horizontal direction. With respect to it, the horizontal velocity of block 2 is zero, so v_2x=v₅.
If block 1 has velocity v₁' with respect to block 5, moving with velocity v₅, block 1 moves with v1=v5+v1' with respect to the ground. (Consequence of Galilean transformation http://psi.phys.wits.ac.za/teaching/Connell/phys284/2005/lecture-01/lecture_01/node5.html) The same is true for the accelerations, dv/dt.

ehild

Thanks ehild, great explanation!

$a_{2y}=a_1+a_5 \Rightarrow a_{2y}=8a$

I had $2Mg-T=2Ma_{2y} \Rightarrow 2Mg=23Ma \Rightarrow a=2g/23$

Thank you!

 

[ehild]

I am not sure if I understand what you did, but the result is OK.

ehild

 

[Saitama]

I am not sure if I understand what you did, but the result is OK.

ehild

The acceleration of 2M is same in both the reference frame. The acceleration of M in the reference frame fixed to 5M is $a_{2y}$. $a_{2y}=a_1-(-a_5)$, this is how I derived the first equation in my previous post.