Question

A man applies an upward force of 137 N to hold one end of a uniform board, leaving the other end resting on the ground. The board makes a 10.2-degree angle with respect to the ground. What is the magnitude of the normal force from the ground on the other end of the board?

Answer 1

The magnitude of the normal force from the ground on the other end of the board is approximately 22.42 N.

Step-by-step explanation:

To find the magnitude of the normal force from the ground on the other end of the board, we can use the concept of torque. Torque is the product of the force applied and the perpendicular distance from the point of rotation. In this case, the force applied is 137 N and the angle made by the board with respect to the ground is 10.2 degrees. Since the board is uniform, the distance from the point of rotation to the other end of the board is half of its length, which is 5.1 meters.

To calculate the torque, we multiply the force applied by the perpendicular distance:

Torque = force x distance = (137 N) x (5.1 m x sin(10.2))

This gives us a torque of approximately 114.33 N·m. In equilibrium, the torque exerted by the normal force must be equal and opposite to the applied torque. Since the normal force acts at the other end of the board, the perpendicular distance from the point of rotation to the normal force is also 5.1 meters.

Using the equation for torque, we can solve for the magnitude of the normal force:

normal force x (5.1 m) = 114.33 N·m

normal force = 114.33 N·m / (5.1 m)

normal force ≈ 22.42 N

Answer 2

To find the magnitude of the normal force from the ground on the other end of the board, we need to consider the forces acting on the board. By setting up an equation based on the sum of forces in the vertical direction, we can determine the normal force. The magnitude of the normal force from the ground on the other end of the board is 235 N.

Step-by-step explanation:

To find the magnitude of the normal force from the ground, we need to consider the forces acting on the board. The man applies an upward force of 137 N, and the board makes a 10.2-degree angle with respect to the ground. The gravitational force acting on the board can be calculated using the formula F = mg, where m is the mass of the board and g is the acceleration due to gravity. Since the board is uniform, the gravitational force can be considered to act at the center of mass of the board. In static equilibrium, the sum of all forces in the vertical direction is zero. Therefore, we can set up the following equation:

Normal force - Gravitational force - Man's upward force = 0

The gravitational force can be calculated as follows:

Gravitational force = mg = 10 N x 9.8 m/s^2 = 98 N

Substituting the known values into the equation, we can solve for the normal force:

Normal force - 98 N - 137 N = 0

Normal force = 98 N + 137 N = 235 N

Therefore, the magnitude of the normal force from the ground on the other end of the board is 235 N.