A uniform board with a mass of 24 kg and length of 3.9 meters is placed on two s
ID: 1461925 • Letter: A
Question
A uniform board with a mass of 24 kg and length of 3.9 meters is placed on two supports. Support 1 is located 0.35 meters from the left of the board and support 2 is located 2.22 meters from the left end of the board. A force is applied in the -y direction to the board 2.86 meters from the left end of the board. What is the minimum force this force needs to be in Newtons such that the board starts to lift off the support 1? Hint: Start this problem off assuming nothing is rotating at first and that the board is still on support 1. Use torques to find the normal force produced by support 1 and then think what happens to the normal force as the board lifts off of support 1.
Explanation / Answer
when the board starts to lift off the support 1, the force exerted by the support 1 on the board is zero.
As the board is in equilibrium, net force and net torque acting on the board must be zero.
Apply net torque about right support = 0
m*g*(2.22 - 3.9/2) - F*(2.86 - 2.22) = 0
F*0.64 = m*g*0.27
F = m*g*0.27/0.64
= 24*9.8*0.27/0.64
= 99.225 N
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