Atwood machine calculation for launching a pebble

[dyn]

Homework Statement

Massless pulley . massless string with block of mass m1 at one end . block of mass m2 at other end. If pebble of mass m is placed on block of mass m2 , what is the force exerted by the pebble om block of mass 2 ?

Relevant Equations

Newton's 2nd law ; F=ma

Assume m₂>m₁ and take positive direction as downwards. String is inextensible so acceleration of block up = acceleration of block +pebble down =a
I used the following 2 equations to obtain a ;
-m₁a = m₁g - T , (m₂+m)a = (m+m₂)g - T
which gives a = g(m+m₂-m₁)/(m+m₁+m₂)

I then constructed a free-body diagram for the pebble which has its weight acting down and the normal reaction from the block acting upward.
Using ma= mg - N ; i arrived at N = 2m₁mg/(m+m₁+m₂
I then said that the force exerted by pebble on m₂ is equal and opposite to the normal reaction , N

Where am i going wrong ? Thanks

 

[vela]

Looks right to me.

 

[dyn]

Hi,
The answer given in Physics for Scientists & Engineers by Tipler & Mosca is g( m² + m₁² + m₂²) / (m+m₁ + m₂ )

I have no confidence in my answer because Atwood's machine questions totally confuse me

 

[vela]

Consider the limiting case where $m_1 \to 0$. Then $m$ and $m_2$ would be in free fall, and the masses would not exert a force on each other. Your answer gives that result; the one from the book doesn't.

 

[Steve4Physics]

I have no confidence in my answer because Atwood's machine questions totally confuse me

(Edited to correct typo'.)

Hi. I agree with your final answer and, as Vela observed (#4), the book-answer can't be correct.

Many students find Atwood machine calculations confusing. I think it's because there are 2 different directions of acceleration and errors with signs creep in.

One ‘trick’ that might help you for simple problems is to note that it is the difference in weights which drives the acceleration.

For a simple example I’ll use |g| = 10m/s². Say the masses are 1kg (weight = 10N) and 3kg (weight = 30N). The weight difference is 30N – 10N = 20N
.
***We have 20N accelerating a mass of 4kg as if we had a simple 1D arrangement.***
$$|a| = \frac{|F_{net}|}{m_{total}} = \frac{20}{1+3} = 5m/s^2 $$We didn’t even need to write equations involving tension (an internal force)!

For the smaller mass alone, tension is greater than weight, so the accelerating force is |T| - 10 upwards
|F| = m|a| gives |T| -10 = 1*5 .

For the larger mass alone, weight is greater than tension, so the accelerating force is 30 - |T| downwards.
|F| = m|a| gives 30 - |T| = 3*5.

In your problem (using your assumption that m₂+m> m₁) the weight difference is (m₂g + mg) – m₁g and the total mass is m₁ + m₂ + m. So we can immedately write:
$$|a| = \frac{|g|(m_2 +m – m_1)}{m_1 + m_2 + m} $$

Note we are taking each object's positive direction to be the direction of its acceleration. So when we look at the pebble (accelerating down) we know the net force is m|g|- N, giving m|g|- N = m|a|

 

[dyn]

I then constructed a free-body diagram for the pebble which has its weight acting down and the normal reaction from the block acting upward.
Using ma= mg - N ; i arrived at N = 2m₁mg/(m+m₁+m₂
I then said that the force exerted by pebble on m₂ is equal and opposite to the normal reaction , N

Where am i going wrong ? Thanks

Thanks for your help everyone. Can i just check one last thing ? The normal reaction from the block acting upward on the pebble which is numerically equal ( due to Newton's 3rd Law ) to to the force exerted by the pebble on the block. Are these 2 forces an example of an action-reaction pair ? The other relevant action-reaction pair being the gravitational force downwards on the block and pebble and the gravitational force upwards on the Earth due to the mass of the block and pebble. Is that all correct ?

 

[haruspex]

Thanks for your help everyone. Can i just check one last thing ? The normal reaction from the block acting upward on the pebble which is numerically equal ( due to Newton's 3rd Law ) to to the force exerted by the pebble on the block. Are these 2 forces an example of an action-reaction pair ? The other relevant action-reaction pair being the gravitational force downwards on the block and pebble and the gravitational force upwards on the Earth due to the mass of the block and pebble. Is that all correct ?

Yes.

 

[dyn]

I'm going to post a related question in the Classical Physics forum if anyone is interested