You pick up a board of length 2.60 m and mass 13.00 kg. To do this, you exert a

Question

You pick up a board of length 2.60 m and mass 13.00 kg. To do this, you exert a force upward with your left hand a distance LL=0.754 m from the end of the board as shown, and you need to exert a force with your right hand a distance LR=0.422 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

In the given case, the measurements are as follows:

Let us consider mass is uniformly distribured along the rod. Then, the Center of mass = 2.6 m / 2 = 1.3 m

Left Hand: 0.754 m and   Right Hand: 0.422 m from the board

In this case, the center of mass will cause the board to rotate clockwise around the left hand. It implies that the Right hand must push DOWN to balance that torque.

Take the moments about the Left Hand
Tcm = ( length ) ( Force )

Here, force is

F = ma = 13 kg . 9.8 m/s2 = 127.4 N

Now substitute in equation for moments

Tcm = (1.3 m - 0.754 m ) ( 127.4 N )  = 69.5604 N. m

Now equate torques with respect cneter of mass and with respect center of mass

Tcm = Trh

69.5604  N. m = (0.754 m - 0.422 m ) . F

Solve for force F

F = 209.5 N acts down

Total Force downward: 209.5 N + 127.4 N = 336.9 N

Round off the result to 3 sig. digits, total force = 337 N

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Thus, the left hand pushes force upward is same in magnitude of force downward, 337 N

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