You pick up a board of length 2.20 m and mass 11.00 kg. To do this, you exert a

Question

You pick up a board of length 2.20 m and mass 11.00 kg. To do this, you exert a force upward with your left hand a distance LL=0.506 m from the end of the board as shown, and you need to exert a force with your right hand a distance LR=0.152 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

if you consider the torque about the center of the board,

the force by the right hand has to be in downward direction to cancel the torque by left hand

so let force by right hand is Fr and force by left hand is Fl.

then writing torque balance equation:

Fl*(1.1-0.506)=Fr*(1.1-0.152)

==>Fl=1.6*Fr

writing force balance equation :

Fl=weight of the board+Fr

==>1.6*Fr=11*9.8+Fr

==>0.6*Fr=107.8

==>Fr=179.67 N

==>Fl=1.6*Fr=287.467 N

hence force exerted by left hand=287.467 N

force exerted by right hand=179.67 N

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