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)

Magnitude:

0 m/s ^2

To determine the force that the right hand needs to exert to keep the board in static equilibrium, we can use the principle of torque.

Torque is the measure of a force's tendency to cause an object to rotate around a certain point or axis. In this case, the torque produced by the left hand and the right hand should balance each other out to keep the board in equilibrium.

The formula for torque is given by (torque = force x distance), where force is the perpendicular force applied and distance is the perpendicular distance from the point of rotation.

Let's calculate the torque produced by the left hand:
TorqueL = forceL x distanceL

The force applied by the left hand is the weight of the board, which is given by:
forceL = mass x gravity

Plugging in the given values:
forceL = 13.00 kg x 9.81 m/s^2 (acceleration due to gravity) = 127.53 N

The perpendicular distance from the left hand to the pivot point is given as LL = 0.754 m. So:
distanceL = LL

Now we can calculate the torque produced by the left hand:
TorqueL = forceL x distanceL = 127.53 N x 0.754 m = 96.17 N·m (clockwise torque)

To keep the board in static equilibrium, the torque produced by the right hand should balance out the torque produced by the left hand. Since the board is symmetrical and the mass is evenly distributed, the distance between the right hand and the pivot point is equal to the distance between the left hand and the pivot point, i.e., LR = 0.264 m.

Now, let's calculate the force required by the right hand (forceR) to balance the torques:
TorqueR = forceR x distanceR

Since the torques are equal when the board is in equilibrium:
TorqueR = TorqueL

Substituting the given distances:
forceR x distanceR = forceL x distanceL

Rearranging the equation to solve for forceR:
forceR = (forceL x distanceL) / distanceR

Plugging in the values:
forceR = (127.53 N x 0.754 m) / 0.264 m = 364.70 N

Therefore, the magnitude of the force that the right hand needs to exert to keep the board in static equilibrium is 364.70 N.