Question

A board with a mass of 6 kg, which is 4 meters long, rests on two supports. Support one is 0.14 meters to the right of the left side of the board and support two is some distance to the left of the right side of the board. The normal force from support 1 is 1/4 of the board's weight. If a force is applied 0.22 meters to the right of second support vertically downwards, how much force is required, in Newtons, to cause the board to lift off support one and rotate around the second support?

Answer 1

the normal forces say N₁ and N₂ are in upper sides

so N₁ + N₂ = mg

Mg /4 + N₂ = M g

N₂ = 3 M g / 4

N₁ x ( 2 - 0.14 ) - N₂ x ( 2 - X ) = 0

1.86 N₁ = N₂ x ( 2 - X )

1.86 x M g / 4 = 3 M g /4 x ( 2 - X )

1.86 = 3 x ( 2 - X )  

2 - X = 0.62

X = 1.38 m

using the equilibrium

6 x 9.8 x ( 2 - X ) - F x 0.22 = 0

36.456 - F x 0.22 = 0

6 x 9.8 x ( 2 - 0.62 ) - F x 0.22 = 0

81.144 - F x 0.22 = 0

F = 81.144 / 0.22

F = 368.83636 N

so the force is required,

to cause the board to lift off support one and rotate around the second support is F = 368.83636 N.