1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Three Blocks Two Pulleys and a Table

  1. Oct 24, 2009 #1
    This was a problem on an exam I had the other day and I want to know if my attempt was correct.

    1. The problem statement, all variables and given/known data

    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.


    2. Relevant equations

    F=ma
    FF=[tex]\mu[/tex]kN
    y = y0 + v0yt + 1/2 at2
    N = mg


    3. 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 - ([tex]\mu[/tex]3mAg)

    simplify

    Fnet= 3mAg(1-[tex]\mu[/tex])

    because mB=3mA you can rephrase the above equation

    Fnet= mBg(1-[tex]\mu[/tex])

    so the acceleration of block B would be g(1-[tex]\mu[/tex]), 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.
     

    Attached Files:

  2. jcsd
  3. Oct 25, 2009 #2
    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?
     
  4. Oct 25, 2009 #3

    ehild

    User Avatar
    Homework Helper
    Gold Member

    It is better if you first draw the free-body diagram for all bodies, including the forces of tension in the ropes.

    ehild
     
  5. Oct 25, 2009 #4
    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
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Three Blocks Two Pulleys and a Table
Loading...