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 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:
To solve this problem, we need to analyze the forces acting on the board in order to keep it in static equilibrium.
First, let's identify the forces acting on the board:
1. Weight (W): This is the gravitational force acting downward on the board. It can be calculated as the product of the mass (m) and the acceleration due to gravity (g). Considering the symmetry of the board, we can assume that the weight acts at the center of the board.
2. Normal force (N): This is the force exerted by the support surface (such as the ground or your hand) perpendicular to the board to counteract the weight. It acts upward and is equal in magnitude to the weight.
3. Force exerted by the left hand (FL): This force acts upward and is located distance LL from the left end of the board.
4. Force exerted by the right hand (FR): This force acts upward and is located distance LR from the left end of the board. This is the force we need to determine.
Since the board is in static equilibrium, the sum of all vertical forces must be zero. Therefore, the magnitude of the force exerted by the right hand can be calculated using the following equation:
FR + FL + N - W = 0
Since the weight is equal to the normal force (W = N), the equation simplifies to:
FR + FL + N - N = 0
FR + FL = 0
To find the magnitude of FR, we need to apply the concept of torques. Torque can be calculated as the product of the perpendicular distance from the point of rotation (pivot) and the force. In this case, the pivot is the left end of the board.
Since the board is in equilibrium, the net torque about any point on the board must be zero. Therefore, the sum of the clockwise torques must be equal to the sum of the counterclockwise torques. Mathematically, this is expressed as:
Sum of clockwise torques = Sum of counterclockwise torques
The clockwise torque due to the force exerted by the left hand (FL) can be calculated as FL * LL, and it acts in the clockwise direction.
The counterclockwise torque due to the force exerted by the right hand (FR) can be calculated as FR * LR, and it acts in the counterclockwise direction.
Since the board is symmetrical and evenly distributed, the torque due to the weight and normal force is zero because they act at the center of the board, which is the pivot point.
Using the equation for torque equilibrium:
FL * LL = FR * LR
Now we can solve for the magnitude of FR by substituting the given values.