You pick up a board of length 3.40 m and mass 17.00 kg. To do this, you exert a

Question

You pick up a board of length 3.40 m and mass 17.00 kg. To do this, you exert a force upward with your left hand a distance LL=0.748 m from the end of the board as shown, and you need to exert a force with your right hand a distance LR=0.396 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.

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)
Magnitude:  
Upward
Downward

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

Explanation / Answer

board of length 3.40 m and mass 17.00 kg.

left hand a distance LL=0.748 m

right hand a distance LR=0.396 m

Center of mass: 3.4 / 2 = 1.7 m

Taking moments about the Left Hand

Tcm = length * Force

F = MA = 17kg * 9.8 m/s^2

= 166.6 N

Tcm = (1.7 - 0.748) m * 166.6 n

= 158.60 N*m

Tcm = Trh

158.60 N*m = (0.748- 0.396)m * F

F = 450.56 N Down

Total Force DOWN is

450.56 N + 166.6

= 617.16 N

Left hand pushes UP: 617.16 N

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