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=0.986 m from the end of the board as shown, and you need to exert a force with your right hand a distance LR=0.325 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

let left end of the board be at x=0

then right end is at x=3.4 m

center of mass is at x=3.4/2=1.7 m

now, left hand is holding at x=0.986 m

and right hand is holding at x=3.4-0.325=3.075 m

let force exerted by left hand is F1 , in upward direction

let force exerted by right hand is F2, in upward direction

weight of the board is acting at its center of mass and in vertically downward direction

weight=mass*g=17*9.8=166.6 N

balancing forces in vertical direction:

F1+F2=166.6 N...(1)

balancing torque about center of mass:

F1*(1.7-0.986)=F2*(3.075-1.7)

==>F1=1.92577*F2...(2)

using equation 2 in equation 1:

1.92577*F2+F2=166.6

==>2.92577*F2=166.6

==>F2=56.9422 N

then F1=1.92577*F2=109.6577 N

hence force exerted by right hand is F2=56.9422 N, in upward direction

force exerted by left hand is F1=109.6577 N , in upward direction

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