The body in Fig. 10-37 is pivoted at O. Three forces act on it in the directions shown: FA = 13 N at point A, 6.4 m from O; FB = 17 N at point B, 3.6 m from O; and FC = 15 N at point C, 2.2 m from O. Taking the clockwise direction to be negative, what is the net torque about O?

Question

The body in Fig. 10-37 is pivoted at O. Three forces act on it in the directions shown: FA = 13 N at point A, 6.4 m from O; FB = 17 N at point B, 3.6 m from O; and FC = 15 N at point C, 2.2 m from O. Taking the clockwise direction to be negative, what is the net torque about O?

Answer 1

To find the net torque about point O, we need to calculate the torque produced by each force and then sum them up.

Torque is calculated using the formula:

Torque = Force * Distance * sin(angle)

Let's calculate the torque produced by each force and then sum them up:

1. Torque due to FA:
Torque_FA = FA * distance_FA * sin(angle_FA)
Given: FA = 13 N, distance_FA = 6.4 m
Since FA is perpendicular to the line from O to A, the angle is 90 degrees.
Torque_FA = 13 N * 6.4 m * sin(90°)

2. Torque due to FB:
Torque_FB = FB * distance_FB * sin(angle_FB)
Given: FB = 17 N, distance_FB = 3.6 m
Since FB is perpendicular to the line from O to B, the angle is 90 degrees.
Torque_FB = 17 N * 3.6 m * sin(90°)

3. Torque due to FC:
Torque_FC = FC * distance_FC * sin(angle_FC)
Given: FC = 15 N, distance_FC = 2.2 m
Since FC is perpendicular to the line from O to C, the angle is 90 degrees.
Torque_FC = 15 N * 2.2 m * sin(90°)

Now, sum up the torques to find the net torque about O:
Net torque about O = Torque_FA + Torque_FB + Torque_FC

You can now substitute the values, perform the calculations, and find the net torque about point O.

Answer 2

To calculate the net torque about point O, we need to find the torque created by each force and then sum them up. The torque created by a force is given by the formula:

Torque = Force * Perpendicular distance from the force to the pivot point

Let's calculate the torque created by each force:

For force FA:
Torque_FA = FA * Perpendicular distance_FA = 13 N * 6.4 m = 83.2 N*m (counterclockwise)

For force FB:
Torque_FB = FB * Perpendicular distance_FB = 17 N * 3.6 m = 61.2 N*m (counterclockwise)

For force FC:
Torque_FC = FC * Perpendicular distance_FC = 15 N * 2.2 m = 33 N*m (counterclockwise)

Now, let's calculate the net torque by adding up the torques:

Net Torque = Torque_FA + Torque_FB + Torque_FC
= 83.2 N*m + 61.2 N*m + 33 N*m
= 177.4 N*m (counterclockwise)

Since the clockwise direction is considered negative, the net torque about point O is -177.4 N*m.

Answer 3

The body in Fig. 10-37 is pivoted at O. Three forces act on it in the directions shown: FA = 13 N at point A, 6.4 m from O; FB = 17 N at point B, 3.6 m from O; and FC = 15 N at point C, 2.2 m from O. Taking the clockwise direction to be negative, what is the net torque about O?

To find the net torque about point O, we need to calculate the torque produced by each force and then sum them up.

To calculate the net torque about point O, we need to find the torque created by each force and then sum them up. The torque created by a force is given by the formula:

no figure.