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Homework Help: Find the extra force required to maintain the speed of the belt

  1. Sep 16, 2010 #1
    1. The problem statement, all variables and given/known data

    coal is falling onto a conveyor belt at a rate of 540 tonnes every hour. the belt is moving horizontally at 2.0 ms-1. find the extra force required to maintain the speed of the belt. (1 tonne=1000kg.)



    2. Relevant equations

    F=m*a



    3. The attempt at a solution

    i tried to solve by saying that F=m*a. so to find force i would need the accelaration of the belt and multiply it by the mass of the coal. the acceleration is the rate of change of velocity so i would use the 2ms-1 as the final velocity and 0 ms-1 as the annitial subtracting 0 from two leaves 2 ms-1 and i divide by the rate of change of time which is 3600s.this leaves me with an accelaeration of 0.000556 ms/s^2 multiplying this by the mass of the coal 540000kg leaves 300 N. is this right
     
  2. jcsd
  3. Sep 16, 2010 #2
    Re: Mechanics

    I'm not too sure on that problem. What I think I can advise however is a few things.

    You are trying to maintain the speed, this means that
    final velocity and initial velocity are both 2 m/s which means that

    v-v_0=at
    2m/s - 2m/s = at
    0=at

    so there is no acceleration.

    If you take the 540,000kg/h and convert it into seconds you get 150kg/s . So now you know how much the mass is increasing by per second and the acceleration.

    Using dimensional analysis, we have 150 kg/s and 2.0 m/s , if we multiply these values, we get 300 kg.m/s^2 which is a newton , so I believe the answer is 300 N like you suspected. However I would also check with others as I am not certain on this.
     
  4. Sep 16, 2010 #3
    Re: Mechanics

    i just looked at it and found out another solution. in accordance to newton's second law force = rate of change of momentum. the innitial momentum that occured in 1 second would be 150kg *2 ms-1 (mass x velocity) which equals 300 kg ms-1 the final momentum that occured in 3600 seconds would be 540000kg *2 ms-1.( the final and innitial velocities are the same) which = 1080000 kg ms-1. change in momentum = final momentum - innitial momentum this equals 1079700 kg ms-1. the change in time = T2-T1 = 3600s-1s which gives 3599s. Now in accordance to Newtons second law the rate of change of momentum = force this implies that change in momentum(1079700 kg ms-1) divided by change in time(3599s) = 300 kg m/s^2 =300N
     
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