2. A thin rod of length 10m is in the x-y plane parallel to the x axis. Its centre is at the point (3,2) .A force of 100 N in the x-y plane is applied at its right end at an angle of 30o from the x-axis. An equal and opposite force is applied at its other end.

(i) Make a diagram showing the x-and y axes, the rod, and the forces on it.
(ii) Calculate the torque due to these forces with respect to the origin with respect to the centre to the rod, and with respect to each end of the rod.
(iii)Why all the torques are equal?

(i) Diagram:

To make a diagram showing the x- and y-axes, the rod, and the forces on it, we can draw a Cartesian coordinate system with the x- and y-axes labeled. Then, we can draw the thin rod parallel to the x-axis and label its center point at (3,2). Next, we can draw a vector representing the force of 100 N being applied to the right end of the rod at an angle of 30 degrees from the x-axis. Finally, we can draw an equal and opposite force vector applied to the other end of the rod.

(ii) Calculating the torque:
To calculate the torque due to the forces, we need to determine the lever arm (perpendicular distance from the point of rotation to the line of action of the force) and the magnitude of the force.

- Torque with respect to the origin:
To calculate the torque with respect to the origin, we need to find the distance between the origin (0,0) and the line of action of the force. Using the distance formula, we can find that the distance is 5 units (10m/2). The magnitude of the force is 100 N. The torque is calculated by multiplying the magnitude of the force by the lever arm. In this case, the torque with respect to the origin would be 100 N * 5m = 500 Nm.

- Torque with respect to the center of the rod:
To calculate the torque with respect to the center of the rod, we need to find the distance between the center of the rod (3,2) and the line of action of the force. Using trigonometry, we can find that the horizontal component of the force is 100 N * cos(30 degrees) = 86.6 N. The distance between the center of the rod and the line of action of the force is the distance in the x-direction, which is 10m/2 = 5m. The torque is calculated by multiplying the magnitude of the force by the lever arm. In this case, the torque with respect to the center of the rod would be 86.6 N * 5m = 433 Nm.

- Torque with respect to each end of the rod:
To calculate the torque with respect to each end of the rod, we need to find the distance between the respective end and the line of action of the force. The distance in this case would be the length of the rod divided by 2 (10m/2 = 5m). The torque is calculated by multiplying the magnitude of the force by the lever arm. In this case, the torque with respect to each end of the rod would be 100 N * 5m = 500 Nm.

(iii) Why are all the torques equal?
The torques are equal in this scenario because the forces being applied at each end of the rod are equal in magnitude and opposite in direction. The perpendicular distance between the line of action of the force and the point of rotation is the same for both ends of the rod. Therefore, the torque calculation (force x lever arm) results in the same value for both ends and the origin.