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Problem with friction

  1. Mar 15, 2004 #1
    !!!!!!!friction Problem!!!!help Needed!!

    I NEED THE SOLUTION OF THIS NUMERICAL FOR MA PHYSICS ASSIGNMENT SOMEONES PLEEZ SOLVE OT FOR ME..


    A 9.75 kg lead brick rests on a level wooden table. If a force of 46.4 N is required to slide the brick across the table at a constant speed, what is the coefficient of friction?
     
  2. jcsd
  3. Mar 15, 2004 #2

    Doc Al

    User Avatar

    Staff: Mentor

    You must at least give it a try if you want help.

    Hint: constant velocity implies that the net force is zero
     
  4. Mar 15, 2004 #3
    friction

    this problem is simple if u nail the concepts down.It usually helps to draw diagram for theze problems.

    We are given the mass of the block which is 9.75 kg. So we know that the weight of the object is 9.75 kg * 9.8 m/s^2 = 95.55 kg*m/s^2 or simply 95.55 newtons. Since, the block is lying on a level wooden table, we know that the table is pushing back on the block with equal force of 95.55 newtons [remember itz in tha opposite direction [3rd law]]. physicists usually call thiz the normal force N.

    ok. now we can use the formula F of Kinetic fric = normal force * coeff of kinetic fric. Therez a max force that is required to move the object that iz the static force of fric...after that u need to decrease your force in order to keep the object moving at a constant speed. lookin at how the question is phrased, we know that they are asking us for the kinetic force of fric. If u imagine a XY cordinate plane; the block iz being pushed in the pos X direction then the Force of Kin fric is pushing in the neg X direc. thinking in temrs of vectors, so we can say actual
    Force of block = Force applied [46.4N] - F of kin fric
    we are told that itz moving in constant speed so therez no accel. hence actual force =0. leaving us with >>>> 0=46.4-F of fric
    so kinetic force of fric = 46.4 N
    Now we can go ahead and plug in the numberz into the variablez.
    >> 46.4 N = Mu [coeff of kin fric] * 9.55 N [normal force]
    Mu= .4856 approx

    Hope this helpz.
    - Mr. Kamadolli
     
    Last edited: Mar 16, 2004
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