Determine the acceleration of block m1 in this wedge-pulley system

[Kaushik]

Homework Statement

The acceleration of block m2 with respect to m1 is 2m/s^2 upward along the inclination. The block m3 accelerates downward with 5 m/s^2.

Relevant Equations

a( of m2 wrt m1) = 2m/s2
a(of m3) = 5m/s2

The equations i got are attached below. Is it right? If yes what should we do after this. I tried solving the equations, but i did not arrive at the solution.

 

[jbriggs444]

What quantity are you trying to determine? If it is ##a_0## (the rightward acceleration of block ##m_1##) then you seem to be overthinking it.

 

[Kaushik]

What quantity are you trying to determine? If it is ##a_0## (the rightward acceleration of block ##m_1##) then you seem to be overthinking it.

Yes. I am trying to determine ##a_0##. So, do we have an easy way to solve this problem?

 

[jbriggs444]

Yes. I am trying to determine ##a_0##. So, do we have an easy way to solve this problem?

Yes.

Let us start with a simpler version of the problem. If blocks ##m_1## and ##m_2## were glued together, how rapidly would the pair need to be accelerating rightward so that ##m_3## would be accelerating downward at 5 m/sec²?

 

[Kaushik]

Yes.

Let us start with a simpler version of the problem. If blocks ##m_1## and ##m_2## were glued together, how rapidly would the pair need to be accelerating rightward so that ##m_3## would be accelerating downward at 5 m/sec²?

## 5 \frac{m}{s^2} ##?

 

[jbriggs444]

## 5 \frac{m}{s^2} ##?

Yes indeed.

If block ##m_1## were glued to the ground while block ##m_2## were once again free to slide, how rapidly would it need to be accelerating so that block ##m_3## would be accelerating downward at 5 m/sec²?

 

[Kaushik]

Yes indeed.

If block ##m_1## were glued to the ground while block ##m_2## were once again free to slide, how rapidly would it need to be accelerating so that block ##m_3## would be accelerating downward at 5 m/sec²?

## 5 \frac{m}{s^2} ##again? as the pulley attached to ## m_1 ## is not moving anymore?

 

[jbriggs444]

## 5 \frac{m}{s^2} ##again? as the pulley attached to ## m_1 ## is not moving anymore?

Yes. Now unglue all blocks and return to the original problem. Can you write an equation for the downward acceleration of ##m_3## in terms of the rightward acceleration of ##m_1## and the diagonal acceleration of ##m_2##?

[If it were me, I would label the accelerations of ##m_1##, ##m_2## and ##m_3## as ##a_1##, ##a_2## and ##a_3##]

 

[Kaushik]

Yes. Now unglue all blocks and return to the original problem. Can you write an equation for the downward acceleration of ##m_3## in terms of the rightward acceleration of ##m_1## and the diagonal acceleration of ##m_2##?

[If it were me, I would label the accelerations of ##m_1##, ##m_2## and ##m_3## as ##a_1##, ##a_2## and ##a_3##]

Is ## a_0 = 3 \frac{m}{s^2}##?

 

[jbriggs444]

Is ## a_0 = 3 \frac{m}{s^2}##?

Yes

 

[Kaushik]

Yes

Oh, thanks. But in my book the solution given was 2. So is the solution given in my book wrong?

 

[jbriggs444]

Oh, thanks. But in my book the solution given was 2. So is the solution given in my book wrong?

Yes, assuming the question is as stated then the book is wrong.

 

[Kaushik]

Yes

I used relative acceleration.

If the block ## m_3 ## is accelerating downwards with 5, then string that is horizontal should also accelerate right with 5. But the acceleration of that horizontal string with respect to the block ## m_1 ## should be 2.

Let the horizontal string be h.

## a_h = 5 \frac{m}{s^2} ##
## a_{h} ## ( with respect to ## m_1 ## ) ##= 5 - a_0##

But as ## a_h ##(with respect to ## m_1 ## ) ## = 2 \frac{m}{s^2} ##

we get ## a_0 = 3 \frac{m}{s^2} ##

 

[Kaushik]

Yes, assuming the question is as stated then the book is wrong.

Thanks for your help !