Pulley System (Physics C Dynamics)

Oct 2016
2
0
I've been stuck on this problem for a while. Any help is appreciated!

Two masses are connected in a pulley system as shown in the figure. Assume both pulleys are frictionless, and also assume that the strings and pulleys are of negligible mass. If M1 = 67 kg and M2 = 5 kg, what is the acceleration of block 1 in the y direction?
What is the acceleration of block 2 in the y direction?

Figure:


I have a lot of attempts:


I've been assuming that the downwards acceleration of M1 = the upwards acceleration of the pulley on the right. And because of that, am1 would also equal am3.

(in case images don't load, Imgur: The most awesome images on the Internet & https://imgur.com/a/aJatQ )
 
Oct 2016
2
0
Below is a diagram based off the OP. I hope it provides some more clarity, and is not just a repeat of information:
https://www.dropbox.com/s/x3lfg90f15ust1f/011016_Pulley System.gif?dl=0

Q: We want to solve for acceleration of the two masses

Answer:

Sign convention
The sign convention is:
Moving in the positive direction means (as shown by image in OP): up from the right hand side and around and then down toward the ground on the left hand side
System 1 (LHS)
Let a be the net acceleration of the pulley system 1.

System 2 represents everything on the right hand side of pulley 1, and system 1 represents everything on the left hand side of pulley 1.

The net force for system 1 is given by:
EQUATION 1: mass1*a = T1 – (Mass1)*g

System 2 (RHS)
We have T2 = (mass2)*g, since the chord for pulley 2 is fixed at one end. We have, following the same sign convention as pulley 1:
(Mass2) – T2 = 0
T2 = Mass2
T2 - Freaction = 0
And (Mass2) – Freaction = 0
(Mass2) = Freaction

This implies that the total weight for system 2 is:
W2 = T2 + T2 = 2*(mass2)*g

And the net force is (let a be the net acceleration of the pulley system 1):
EQUATION 2: (mass2)*a = W2 – T1 = 2*(mass2)*g – T1
NOTE: The acceleration of all objects within system 2 is 0. Imagine a bus with all things inside bolted down to cancel inertial forces (or inertial dampeners if you watch Stargate). The entire system 2 can be accelerating whilst its contents do no accelerate. So pulley 1 causes system 2 to accelerate (this is possible because pulley 2 itself should be able to move up and down).

Total system
Net force on pulley system 1 = (Mass1 + Mass2)*a = EQUATION 1 + EQUATION 2 = 2*(mass2)*g - (Mass1)*g; (the tensions cancel out)
We can now solve for acceleration:
a = [2*(mass2)*g - (Mass1)*g]/(Mass1 + Mass2)
[2*(5 kg)*(9.81 m/s^2) – (67*9.81 m/s^2)]/ 72 kg = -7.76625 m/s^2 (Do your rounding after all calculations complete to minimise error; use this value for calculations)

Therefore acceleration equal -7.77 (m/s^2). The sign convention means that the masses move in the same direction (up from the right hand side and around and then down toward the ground on the left hand side).

Extra notes
To find T1 substitute the acceleration into equation 1 and solve for T1. This should give 136.93 N.
 
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