Solving Two-Block Pulley System w/Expln

[keemosabi]

Homework Statement

As part a of the drawing shows, two blocks are connected by a rope that passes over a set of pulleys. One block has a weight of m1 = 312 N, and the other has a weight of m2 = 758 N. The rope and the pulleys are massless and there is no friction.

(a) What is the acceleration of the lighter block?
m/s2
(b) Suppose that the heavier block is removed, and a downward force of 758 N is provided by someone pulling on the rope, as part b of the drawing shows. Find the acceleration of the remaining block.
m/s2
(c) Explain why the answers in (a) and (b) are different.

Homework Equations

Fnet = ma

The Attempt at a Solution

758 - 312 = 446, the net force of the entire system. Then I plugged this into Fnet = ma, 446 = ma. To get the mass, I did 758 + 312 1070, and then divided by the acceleration due to gravity, 9.8, and got 109.184. I plugged in and got 446 = 109.184(a), solved for a and got 4.085 as the accelearation in part a. What I don't understand is how the acceleration in part b is different, since the same force is being applied? Or is it because I don't add the weight of the human to my mass measurement when finding out that acceleration? Why not?

 

[LowlyPion]

Homework Statement

As part a of the drawing shows, two blocks are connected by a rope that passes over a set of pulleys. One block has a weight of m1 = 312 N, and the other has a weight of m2 = 758 N. The rope and the pulleys are massless and there is no friction.

(a) What is the acceleration of the lighter block?
m/s2
(b) Suppose that the heavier block is removed, and a downward force of 758 N is provided by someone pulling on the rope, as part b of the drawing shows. Find the acceleration of the remaining block.
m/s2
(c) Explain why the answers in (a) and (b) are different.

Homework Equations

Fnet = ma

The Attempt at a Solution

758 - 312 = 446, the net force of the entire system. Then I plugged this into Fnet = ma, 446 = ma. To get the mass, I did 758 + 312 1070, and then divided by the acceleration due to gravity, 9.8, and got 109.184. I plugged in and got 446 = 109.184(a), solved for a and got 4.085 as the accelearation in part a. What I don't understand is how the acceleration in part b is different, since the same force is being applied? Or is it because I don't add the weight of the human to my mass measurement when finding out that acceleration? Why not?

What is the difference in rope tension between a and b?

 

[keemosabi]

What is the difference in rope tension between a and b?

Don't I need the acceleration in part B to find the rope tension in that part? I usually use Fnet = ma and I substitute in for m and a, and that gives me the net force, and I can easily then find out the tension.

 

[keemosabi]

Have I done something wrong?

 

[LowlyPion]

Don't I need the acceleration in part B to find the rope tension in that part? I usually use Fnet = ma and I substitute in for m and a, and that gives me the net force, and I can easily then find out the tension.

In part a) you have the masses m1 and m2 accelerating according to the difference in the 2 masses. Hence as you calculated the acceleration is based on that differences in m*g's, but it is applied to accelerating both the masses. Here the acceleration is expressed by g*(m2 - m1)/(m2 + m1).

In part b) the force is given as a constant 758N and that is the tension provided by the human. The tension 758N acts on the mass m1 and it's accelerated with the whole 758N minus just the 331 from its m1*g. The acceleration derived this Fnet then is g*(m2 - m1)/m1