When we are dealing with systems of bodies we apply $\mathbf{F}= m\mathbf{a}$ to the whole system.

How do I apply $\mathbf{F}= m\mathbf{a}$ to the whole system in the following example.?

($m_1 \lt m_2$)

Pulley System

Also, why is the following application of $\mathbf{F}= m\mathbf{a}$ wrong?

For the whole system,

$$\mathbf {F = ma \downarrow }$$ $$\mathbf {m_1g + m_2g = m_2a -m_1a}$$

  • $\begingroup$ You could explain how you get that equation. BTW, the equation you've written is not right. $\endgroup$ – Oswald Dec 24 '15 at 4:41
  • $\begingroup$ I know that it is wrong. I want to know why :) $\endgroup$ – jessij Dec 24 '15 at 5:09
  • $\begingroup$ if you could tell how you got that equation, we could point out where you went wrong. $\endgroup$ – Oswald Dec 24 '15 at 5:17
  • $\begingroup$ It would be helpful if you defined what you meant by "the whole system"... $\endgroup$ – DJohnM Dec 25 '15 at 1:25

The short answer:

Because the pulley exerts no force on the system, we can redraw it to understand it better:

enter image description here

Because the masses has the same acceleration, you can look at the system as a system of one mass of m1+m2, where the external forces are: m2g, m1g (T are internal forces). this forces try to increase the acceleration in different direction so they are opposite to each other. finally we get:

(m2 - m1)g = (m1 + m2)a

The long answer:

We can get the same answer by applying F=ma to each mass.

m1a = T - m1g
m2a = m2g - T

notice that I chose the acceleration as m2 going down, and m1 going up - as you did in your sketch. this is why m2a = m2g - T , and not m2a = T - m2g. (this led you to your wrong answer!)

We add both equations to get:

(m2-m1)g = (m1 + m2)a

The center of the pulley is fixed. There is force on the pulley from its axle, which is supplied by the support. You can write an equation saying the acceleration of the pulley is zero, and will find that the force upward from the support equals the force downward on the two masses due to gravity. If the pulley were free you would be correct and the pulley would be accelerating downward.

  • $\begingroup$ "will find that the force upward from the support equals the force downward on the two masses due to gravity" Is this true when one mass is larger than the other, and the masses are both accelerating? $\endgroup$ – DJohnM Dec 25 '15 at 1:34
  • $\begingroup$ @DJohnM: yes. You can just do a force balance. The only external forces are gravity on the masses and the support force on the pulley. The accelerations of the masses are altered by the tension in the rope, but that is internal. $\endgroup$ – Ross Millikan Dec 25 '15 at 1:38
  • $\begingroup$ But doesn't the tension in the rope, which applies the downward force to the pulley, go to zero as m1 goes to zero? $\endgroup$ – DJohnM Dec 25 '15 at 1:54

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