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Angle to topple a box

  1. Oct 3, 2016 #1
    1. The problem statement, all variables and given/known data
    As shown in the figure below, a cuboid with uniform density is tilted by hand so that one edge remains in contact with a roughly surfaced floor. The angle of tilt, θ , is gradually increased, and when becomes larger than θ0 , the cuboid leaves the fingers and rotates. As indicated in the figure, the lengths of the two edges are a and b. The edge in contact with the floor does not slide. What is tan θ0?
    Untitled%201_zpsnogepyor.png


    2. Relevant equations
    ∑ torque = 0
    torque = F . perpendicular distance

    3. The attempt at a solution
    I am stuck from the beginning. I try to draw the free body diagram. There are three forces acting on the box; weight, normal force and upward force from hand.

    Taking the vertex of the box where it touches the floor as pivot, the normal force won't produce torque and when the cuboids leaves the fingers and rotates, the upward force from the hand also zero. Maybe weight also won't produce torque because when the box rotates, the line of force of weight passes through the pivot. I also don't know the value of the upward force from the hand.

    How to set up the equation?
     
  2. jcsd
  3. Oct 3, 2016 #2

    gneill

    User Avatar

    Staff: Mentor

    That's the key point: When the box's center of mass is directly over the pivot the box exerts no torque in either direction around the pivot point. The box is then in an unstable equilibrium and could fall either way... :wink:
     
  4. Oct 3, 2016 #3
    I get it. We don't need to use equation of torque, just simple trigonometry. The answer is a/b

    Thanks a lot
     
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