1. Not finding help here? Sign up for a free 30min 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!

Mechanics problem. Please clear my doubts regarding forces.

  1. Aug 23, 2008 #1
    1. The problem statement, all variables and given/known data
    In the flgure below, the incline has mass M and is fastened to the stationary horizontal tabletop. The block of mass m is placed near the bottom of the incline and is released with a quick push that sets it sliding upward. It stops near the top of the incline, as shown in the flgure, and then slides down again, always without friction. Find the force that the tabletop exerts on the incline throughout this motion.

    http://i359.photobucket.com/albums/oo31/tanzl/PHYS1A_ProbSet2_08-09.jpg


    2. Relevant equations



    3. The attempt at a solution
    My answer is (M+m)g since the resultant downward force should be the sum of the weight of incline plane and the weight of the mass. But, it looks too simple to be correct.
    Please clear my doubts.
    1)Although there is a normal force of magnitude mg*cos(theta) that will eventually balance the weight of the mass m since the mass is not moving downward o upward.
    2)The initial push on the mass m is not necessary to be considered in the calculation because the force supplied is not constant and so the only force acting on mass m which parallel to the incline is mg*sin(theta).
    3)Throughout the motion of mass, only two force were involved. The weight of mass m and the normal force of incline exerted on the mass.
     
  2. jcsd
  3. Aug 23, 2008 #2

    Shooting Star

    User Avatar
    Homework Helper

    You are neglecting the horizontal component of the normal reaction.
    >
    1)Although there is a normal force of magnitude mg*cos(theta) that will eventually balance the weight
    >
    The normal force N is equal to mgCosθ. I don’t quite understand what you mean by saying that it will eventually balance the weight.

    The inclined wedge is in equilibrium due to three forces –the reaction R due to the tabletop, its own weight Mg, and the normal reaction N at the point of location of mass m. You know N in terms of m and θ. Draw a free body diagram and give it another try now. Resolve all the forces along the horizontal and vertical directions.

    (Hint: R will not be in the vertical direction.)
     
  4. Aug 24, 2008 #3
    This is my free body diagram for mass m.
    http://i359.photobucket.com/albums/oo31/tanzl/freebodydiagram1.jpg
    Fnormal = -Fg cos(theta)


    This is my free body diagram for incline plane.
    http://i359.photobucket.com/albums/oo31/tanzl/freebodydiagram2.jpg
    I let Fx and Fy be the resolved vector of Ftable.
    Fx=-Fg cos(theta) sin(theta) Fy=W-Fg [cos(theta)]^2
    |Ftable^2|=|Fx|^2+|Fy|^2
    By solving the equation, it yields

    Ftable = g * sqrt {m^2 * [cos(theta)]^2 + M^2 + 2Mm [cos(theta)]^2}
     
  5. Aug 24, 2008 #4

    Shooting Star

    User Avatar
    Homework Helper

    Right. As a check, note that when θ=0, the reaction is simply (m+M)g, as is to be expected.
     
  6. Aug 24, 2008 #5
    Thank you so much.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Mechanics problem. Please clear my doubts regarding forces.
Loading...