You pick up a board of length 1.80 m and mass 9.00 kg. To do this, you exert a f

Question

You pick up a board of length 1.80 m and mass 9.00 kg. To do this, you exert a force upward with your left hand a distance LL=0.378 m from the end of the board as shown, and you need to exert a force with your right hand a distance LR=0.147 m from the end of the board. Assume the board to be in static equilibrium and that the board is symmetrical with the mass evenly distributed.

a.) What force does the right hand need to exert to keep the board in static equilibrium? (you need to get both the magnitude and the direction)

b.)What is the magnitude of the force that the left hand needs to exert to keep the board in static equilibrium?

Explanation / Answer

a)

Measuring from the Left end:

Center of mass: 1.80 / 2 = 0.90 m (Uniform mass distribution)
Left Hand:0.378 m
Right Hand: 0.147 m (The hands are crossed since the Right Hand is closer to the left end than the left hand.)

===> The Center of mass will cause the board to rotate clockwise around the left hand ==> Right hand must push DOWN to balance that torque.

Taking moments about the Left Hand

Tcm = length * Force

F = MA = 9kg * 9.8 m/s^2 = 88.2 N

Tcm = (0.9 - 0.378) m * 88.2 N = 46.04 N*m

Tcm = Trh

46.04 N*m = (0.378 - 0.147)m * F
F = 199.31 N Down

b)

Total Force DOWN: 199.31+ 88.2 = 287.51 N

Left hand pushes UP: 287.51 N

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