You pick up a board of length 2.10 m and mass 10.50 kg. To do this, you exert a

Question

You pick up a board of length 2.10 m and mass 10.50 kg. To do this, you exert a force upward with your left hand a distance LL=0.504 m from the end of the board as shown, and you need to exert a force with your right hand a distance LR=0.257 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

Figure is not visible.

But I can imagine the situation.

Let FL and FR are forces exerted by left hand and right hand respectively

As the board is in equilibrium net force and net torque acting on the board must be zero.

Apply, Fnety = 0

FL + FR - m*g = 0

FL + FR = m*g

FL + FR = 10.5*9.8

FL + FR = 102.9 N -------(1)

Apply, Net torque about center = 0

FR*(L/2 - 0.504) - FL*(L/2 - 0.257) = 0

FR*(2.1/2 - 0.504) = FL*(2.1/2 - 0.257)

FR*0.546 = FL*0.793

FR = 1.452*FL

FL = 0.689*FR ---(2)

from equations (1) and (2)

0.689*FR + FR = 102.9

FR = 102.9/(1+0.689)

= 60.9 N <<<<<<<----------Answer

from equation 2

FL = 0.689*60.9

= 41.9 N <<<<<<<----------Answer

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