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Homework Help: 1 Dimensional Motion with Pulley

  1. Sep 23, 2009 #1
    1. The problem statement, all variables and given/known data
    I uploaded the image with for the problem.

    The bucket is full of bricks and when the man unties the rope to bring the bucket to the ground, he gets pulled into the air while holding the rope. He goes all the way to the top (taking place in 2.91 seconds). The bricks fell out of the bucket when it hit the ground (when he hit the top) and the bricks fell out (now weighing only 18.8 kg). He then fell back to the ground and the bucket went back up, but slowed his fall.

    mass of bricks = 124 kg
    mass of bucket = 18.8 kg
    mass of man = 73.3 kg
    time of ascent = 2.91 s
    gravity = 9.8 m/s^2

    A. What was the initial upward acceleration by the man? (in m/s^2)
    B. What is the height of the building? (in m)
    C. How many seconds did it take (after starting his decent) for the man to hit the ground? (in s)

    2. Relevant equations
    x = vot + 1/2at^2

    3. The attempt at a solution

    A. mass of bricks + mass of bucket = 142.8 kg

    142.8 (9.8) = 1399.44

    73.3 (9.8) = 718.34

    1399.44 - 718.34 = 681.1

    681.1 = 73.3a
    a = 9.29195

    I don't think I'm doing it right though. I got some help from someone else and was told to multiply 142.8 by 9.8, but is this right? I'm not getting the right answer.

    x = vot + 1/2at^2
    x = 0 + 1/2(9.3)(2.91^2) = 39 m

    C. x = vot + 1/2at^2
    39 = 0 + 1/2(9.8)t^2
    t = 2.82 sec


    Attached Files:

    Last edited: Sep 23, 2009
  2. jcsd
  3. Sep 23, 2009 #2

    Doc Al

    User Avatar

    Staff: Mentor

    You found the net force (along the direction of motion) on the entire mass, not just on the man.

    You might be better off analyzing the forces on the man and the forces on the bucket+bricks separately. Get two force equations then combine them to solve for the acceleration.
  4. Sep 23, 2009 #3
    Oh I see now. I added the masses together to get 216.1 then used that for the mass. So 681.1/216.1 = 3.15 m/s^2 which is correct!

    I used that to get the height of the building, 13.34 meters.

    What I can't figure out is the last question; how many seconds did it take (after starting his decent) for the man to hit the ground?

    x = vot + 1/2at^2
    13.3448 = 0t + 1/2 (9.8)t^2
    t = 1.65 s

    That is incorrect...
  5. Sep 23, 2009 #4

    Doc Al

    User Avatar

    Staff: Mentor

    You need to figure out his acceleration. It's not 9.8! (Use the same method as in part A.)
  6. Sep 23, 2009 #5
    Yes! Thank you
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