Block and Pulley System: Acceleration and Tension Homework Solution

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In a block and pulley system with frictionless surfaces, the acceleration of Block 2 (a2) can be expressed in terms of the accelerations of Blocks 1 (a1) and 3 (a3). The correct relationship is a2 = (a1 + a3) / 2, indicating that the acceleration of Block 2 is the average of the accelerations of the other two blocks. The initial attempts to express a2 in terms of a1 and gravitational force were incorrect. Understanding the kinematic relationships between the blocks is crucial for deriving the correct formula. The discussion emphasizes the importance of recognizing how the movements of the blocks are interconnected.
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Homework Statement



The diagram shows a block and pulley
system where all the surfaces are
frictionless, and all the pulleys are
massless and frictionless.

Express the magnitude of the acceleration of Block 2, a2, in terms of a1 and/or a3.

nyinvm.jpg


Homework Equations



F=ma

The Attempt at a Solution


FBD for block 1
m1 * a1 = T
a1 = T

fbd for block2

m2 * a2 = 2T - m2*g

a2 = [2(a1) -m2*g]/m2

a2 = a1 - g

But i am wrong for some reason..

It looks like the answer is a2=3/2 * a3 or 3/4*a1



Any help would be awesome
 
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Any ideas
 
The answer is a2=3/2 * a3 or 3/4*a1 which is the solution, I'm not sure how that was achieved.
 
Last edited:
NoobeAtPhysics said:
a2 = [2(a1) -m2*g]/m2

a2 = a1 - g

But i am wrong for some reason..
Hi NoobeAtPhysics. There is a bit of mind reading involved, but you haven't got to the intended answer only because you have expressed a2 in terms of a1 and g. You are to find it in terms of a1 alone.

There is a relationship you have overlooked, viz., the linking rope dictates that the motion of block 2 is tightly tied (no pun intended :smile:)[/color] to the motions of blocks 1 and 3.

The speed of descent of 2 is the average of the speeds of 1 and 3. The distance through which 2 descends is the average of the distances moved by 1 and 3. Likewise, the magnitude of the acceleration of 2 is related to the accelerations of 1 and 3...

It looks like the answer is a2=3/2 * a3 or 3/4*a1
I agree.
 
Hmm, I still don't comprehend why mathematically the solution is what it is
 
NoobeAtPhysics said:
Hmm, I still don't comprehend why mathematically the solution is what it is
Can you figure out the simple equation relating the acceleration of 2 to the accelerations of 1 and 3? There is a strong logical hint in the way I worded this:
The speed of descent of 2 is the average of the speeds of 1 and 3. The distance through which 2 descends is the average of the distances moved by 1 and 3. Likewise, the magnitude of the acceleration of 2 is related to the accelerations of 1 and 3...

:wink:
 
The first step is to establish the kinematics of the motion. Suppose mass 3 is held fixed, and mass 1 moves to the right by 1 unit. How much does mass 2 move downward? Then, suppose that mass 1 is subsequently held fixed, and mass 3 moves to the left by 1 unit. How much does mass 2 move downward? From this, you should be able to conclude that a2=(a1+a3)/2.

Chet
 
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