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=0.775 m from the end of the board as shown, and you need to exert a force with your right hand a distance LR=0.271 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) Magnitude:

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

Explanation / Answer

Given:

On measuring from Left end ,

For unform mass distribution ,the center of mass of

board is lies exactly at the center of the board

Center of mass : 3.1 m / 2 = 1.55 m

Force exerting upward with your left hand a distance

LL=0.775 m from the (left end of the board )

Force exerting with your right hand a distance

LR=0.271 m from the (left end of the board)

Mass of the board , m = 15.50 kg

Solution:

In order to balance the torque , right hand must be push down ,

when the center of mass causes the board to rotate .

  weight of the board acting down ward ,

         W = mg

              = (15.5 kg)(9.8 m/s2)

              = 151.9 N

consider , moments about the Left Hand
we have ,

   torque , Tcm = (1.55 m - 0.775 m) ( 151.9 N)

                         = 117.72 N m        ......(1)

    torque, Tright =(0.775m - 0.271 m) Fr

                         = ( 0.504 Fr ) Nm    .......(2)

  For force does the right hand need to exert to keep the

board in static equilibrium equate eqn (1) = (2)

117.72 N m = ( 0.504 F ) Nm   

      thus,     Fr = 233.57 N  
                    Direction:(down ward)

Since , weight acting down ward direction ,

   the total force is acting in the downward direction is ,

             Fd= F + W

                 = (233.57 N )+ (151.9 N )

                 = 385.47 N

Hence , magnitude of the force that the left hand needs to

exert to keep the board in static equilibrium is,

             Fl = Fd

                 =385.47 N (left hand push up)

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