Three Blocks Two Pulleys and a Table

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Homework Help Overview

The problem involves a system of two pulleys and three blocks, where block A hangs from a string connected to block B, which rests on a table, while block C hangs from another string connected to block B. The task is to determine the acceleration of block C and the time it takes to reach the ground, given the masses of the blocks and the coefficient of friction between block B and the table.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • The original poster attempts to calculate the net force on block B to find the system's acceleration, leading to a derived formula involving the coefficient of friction.
  • Some participants question whether the net force on block B should be divided by the total mass of all blocks to find the system's acceleration.
  • One participant suggests drawing a free-body diagram to clarify the forces acting on each block.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem setup and the calculations involved. There is a recognition of the need to consider the total mass of the system when calculating acceleration, but no consensus has been reached on the correctness of the original poster's approach.

Contextual Notes

The problem is framed within a homework context, and participants are encouraged to clarify their understanding without providing direct solutions.

RJVoss
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This was a problem on an exam I had the other day and I want to know if my attempt was correct.

Homework Statement



Please look at the attached image. You have a system of two ideal pulleys set up on a table. Block A is hanging from a string on the left side of the table which passes by a pulley connecting it to block B. Block B is resting on the table, and the kinetic coefficient of friction between the surface of the table and the block is 0.557. Block C is hanging from a string on the right side of the table which passes by a pulley and is attached to Block B. Block C is four times as massive as block A, and block B is three times as massive as block A. When the system is released from rest, block C accelerates downward until it reaches the floor, a distance of 105cm.

Find: The acceleration of block C and the time it takes to reach the ground.


Homework Equations



F=ma
FF=\mukN
y = y0 + v0yt + 1/2 at2
N = mg


The Attempt at a Solution



I started by trying to find the net force acting on block B in order to find the acceleration of the system. In this case,

Fnet= FC - (FA + FF)

so

Fnet= (4mAg) - 1mAg - (\mu3mAg)

simplify

Fnet= 3mAg(1-\mu)

because mB=3mA you can rephrase the above equation

Fnet= mBg(1-\mu)

so the acceleration of block B would be g(1-\mu), and plugging in the values I got the answer -4.34m/s2 rounded to three figures. This would also be the acceleration of block C.

Next I needed to find the time it took block C to hit the ground.

y = y0 + v0yt + 1/2 at2

so

y = 1/2 at2
t = sqrt(2y/a)
t = sqrt(2*-1.05m/-4.34m/s2)
t = 0.695s rounded to three figures

Did I do this problem correctly? And if not, can you please show me the correct way of doing a problem like this? Thanks.
 

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Now that I review my notes, I think I might have gotten this problem incorrect. Was I supposed to take the net force acting on block B and divide by the total mass of A + B + C to find the acceleration of the system?
 
It is better if you first draw the free-body diagram for all bodies, including the forces of tension in the ropes.

ehild
 
RJVoss said:
Now that I review my notes, I think I might have gotten this problem incorrect. Was I supposed to take the net force acting on block B and divide by the total mass of A + B + C to find the acceleration of the system?

what have you meant is correct.
To find the acceleration , you have to divide by total mass as they are all connected by a string, therefore sharing same a
 

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