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Determine Mass of object for no motion on ramp

  1. Oct 17, 2012 #1
    1. The problem statement, all variables and given/known data
    Two blocks are joined together with a rope that runs over the pulley. The mass of m2 is 5kg and the incline is 35 degrees (m2 hangs off the edge of the ramp, pulling on m1). The coefficient of static friction between mass 1 and the inclined board is 0.2. Determine the largest mass for m1, so that both blocks remain stationary. (m1 does not slide down).


    2. Relevant equations
    f=ma
    Force friction=coefficient(Force normal)



    3. The attempt at a solution
    Force friction = 0.2(Fn)
    =0.2(m)(g)(cos35)
    Force Friction = 1.61m

    Because block 2 pulls with 49N of force [5(9.8)], then I assume the force of friction on block 1 is 49N. However, gravity pulls down on the block 1 as well (Fgx=m9.8sin35=5.62m). So:
    Ff=1.61m
    49N-5.62m=1.61m
    Gravity reduces the amount of friction required to hold block 1 in place.
    Therefore:
    m=49/7.23
    m=6.8kg

    The listed answer is 14kg.
    What am I doing wrong?

    Thank You
     
  2. jcsd
  3. Oct 17, 2012 #2

    cepheid

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    Staff Emeritus
    Science Advisor
    Gold Member

    Welcome to PF SinghTJay!

    Try to be systematic about this. Draw a free body diagram for each block (i.e. a diagram with just the block and the forces acting on that block, with all other elements of the system removed). For block 2, there are no horizontal forces acting on it, and the block is not accelerating, which means the net force must be 0 in the vertical direction. This means that the vertical forces sum to 0. The two vertical forces are tension T upward and weight m2g downward. So:

    T - m2g = 0

    T = m2g

    The force due to tension in the rope is therefore m2g.

    Now draw a free body diagram for block 1. Consider the two directions "perpendiculiar to the plane" and parallel to the plane. In the perpendicular direction two forces act: 1. the perpendicular component of the weight, downward, and 2. the normal force, upward. These two must balance, hence Fn = m1gcos(35°) as you have written.

    In the direction parallel to the plane, the forces must balance. Two forces act up the plane: 1. Tension, T, and 2. Friction, Ff, which tries to prevent the block from sliding down the plane. A third force 3. is gravity i.e. the component of the weight that is parallel to the plane. So the sum of forces is given by:

    T + μm1gcos(35°) - m1gsin(35°) = 0

    Can you take it from here? I would advice solving for m1 algebraically first before plugging in numbers. It just makes things cleaner.
     
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