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Rigid object in equilibrium

  1. Dec 4, 2007 #1
    1. The problem statement, all variables and given/known data
    A stick with a mass of 0.217 kg and a length of 0.435 m rests in contact with a bowling ball and a rough floor, as shown in the figure below. The bowling ball has a diameter of 21 cm, and the angle the stick makes with the horizontal is 30°. You may assume there is no friction between the stick and the bowling ball, though friction with the floor must be taken into account.


    (a) Find the magnitude of the force exerted on the stick by the bowling ball.

    (b) Find the horizontal component of the force exerted on the stick by the floor.

    (c) Repeat part (b) for the upward component of the force.


    2. Relevant equations
    for part a
    sum FX=Gx(the Ground)-P(force at BBall) = 0 Gx=P
    sum FY=Gy(the Ground)-Ws(stick weight) = 0 Gy=Ws
    sum of the torques is zero


    3. The attempt at a solution


    .217kg*9.8m/s = 2.1266N
    Ws=2.1266Ncos 30 = 1.8417 = -Wstick*Lstick
    P = .435m*sin 30 = .2175

    so sum of torques is -Wstick*Lstick+PLp

    P = 2.1266N*.2175m(mid stick)*cos30 / .435m*sin 30 or .4006/.2175 = 1.8418N



    is the diameter number extra info or am I missing something here? This is a first attempt
    for me at a problem like this, I am probably off-base?
     
  2. jcsd
  3. Dec 4, 2007 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Some things to consider: The force that the ball exerts on the stick is perpendicular to both (figure out the angle it makes with the horizontal). Use the diameter of the ball to figure out where the ball makes contact with the stick (it's not at the end).
     
  4. Dec 6, 2007 #3
    Stick & Bowling Ball

    The angle of the stick with the floor is given at 30 degrees

    The angle of the stick resting on the bowling ball is 60 degrees

    The diameter of the bowling ball is given at 21cm so the radius is half that at 10.5cm

    (.217kg)*9.8m/s = 2.1266N

    .217kg * cos 60 = .1085kg

    .217kg * cos 30 = .1879kg

    .217kg * sin 30 = .1085kg

    since the torque = 0 the sum of the forces should = 0

    What do I do next?
     
  5. Dec 6, 2007 #4

    Doc Al

    User Avatar

    Staff: Mentor

    OK.

    Not sure what you mean here. If you draw a line from where the stick touches the ball to the center of the ball, that line will be perpendicular to the stick.

    OK

    That's the weight of the stick. Where does it act?
    Not sure what you're doing here.
    Yes, the sum of the torque = 0 and the sum of the forces = 0.

    The first thing you need to do is draw yourself a diagram showing all the forces acting on the stick:
    -- force of the ground, which has a vertical and horizontal component
    -- force of gravity: Where does it act? What torque does it produce?
    -- force of ball, which has a vertical and horizontal component: Where does it act? What angle does it make with the horizontal? What torque does it produce?
     
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