You pick up a board of length 3.10 m and mass 15.50 kg. To do this, you exert a

Question

You pick up a board of length 3.10 m and mass 15.50 kg. To do this, you exert a force upward with your left hand a distance LL=0.682 m from the end of the board as shown, and you need to exert a force with your right hand a distance LR=0.327 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)

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

Explanation / Answer

the mass should be applied at the centre of mass

so

Fh + F r= 15.50*9.8 --------------{1}

equating torque at the centreof mass located at 1.55 from both the ends

so the equation =

Fl * (1.55-0.682)= F r* (1.55- 0.327)

= Fl = 1.408 F r ----------------{2}

so putting 2 in equation 1

we get

2.408 F r = 15.5*9.8

Fr = 63.08 N

so F l = 1.408 * 63.08 =88.81 N

so the two forces must be in the same direction and the direction should be opposite to mg force that is upward direction

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