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 left end of the board and you need to exert a force with your right hand a distance LR=0.264 m from the left end of the board. Assume the board to be in static equilibrium and that the board is symmetrical with the mass evenly distributedWhat 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)

Question

Magnitude:

Answer 1

Let's call the force exerted by the left hand F_left and the force exerted by the right hand F_right. We can set up an equation for the forces and torques acting on the board. First, consider the forces: the sum of the forces in the vertical direction must be zero for the board to be in static equilibrium.

F_left + F_right - m*g = 0, where m = 13.00 kg (mass of the board) and g = 9.81 m/s^2 (acceleration due to gravity).

Next, let's consider the torques. Torques are calculated as the force times the distance from the center of mass. We can choose any point to calculate the torques, but for simplicity, let's choose the left end of the board. The torque due to the left hand will be zero since it is exerting force at the point we are calculating torques from (distance = 0).

Now, let's calculate the torque due to the right hand and the torque due to the weight of the board. The weight of the board acts at the center of mass, which is at the middle of the board (1.30 m from the left end).

Torque_right = F_right * LR
Torque_weight = m * g * (1.30 m)

The sum of the torques must also be zero for the board to be in static equilibrium:

F_right * LR - m * g * (1.30 m) = 0.

Now we have two equations and two unknowns (F_left and F_right):

1) F_left + F_right - m*g = 0
2) F_right * LR - m * g * (1.30 m) = 0.

Now we can solve these equations for F_right:

From the first equation, we can solve for F_left:

F_left = m * g - F_right.

Substitute this expression for F_left into the second equation and solve for F_right:

F_right * LR - (m * g - F_right) * (1.30 m) = 0
F_right * LR - m * g * (1.30 m) + F_right * (1.30 m) = 0
F_right * (LR + 1.30 m) = m * g * (1.30 m)
F_right = (m * g * (1.30 m)) / (LR + 1.30 m)

Now plug in the values:

F_right = (13.00 kg * 9.81 m/s^2 * (1.30 m)) / (0.264 m + 1.30 m)
F_right = (168.111 kg*m/s^2) / (1.564 m)
F_right = 107.47 N

So, the right hand needs to exert a force of 107.47 N in the upward direction to keep the board in static equilibrium.

Magnitude: 107.47 N
Direction: Upward