Newton law's - Finding acceleration

[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.