Question

five forces act on a rod that is free to pivot at point p, as shown in the figure. wh of these forces is producing a counter-clockwise torque about point p?
horizontal rod with one end labeled p, with forces acting on it from five directions, (a) horizontal force acting directly towards the rod end opposite p; (b) force directed away from the top of the rod at an angle to the left; (c) force directed toward the top of the rod at an angle from the right; (d) force directed toward the rod bottom at an angle from the left; (e) force directed toward the rod bottom at an angle from the right.

Explanation:

Step1: Understand torque direction

Torque is calculated as $\tau = r\times F\times\sin\theta$, where $r$ is the distance from the pivot, $F$ is the force, and $\theta$ is the angle between $r$ and $F$. A counter - clockwise torque will tend to rotate the rod in the counter - clockwise direction around point $P$.

Step2: Analyze each force

• Force A: The line of action of force A passes through the pivot point $P$. So, $r = 0$ (since the distance from the pivot to the line of action of the force is zero). Using the torque formula $\tau = r\times F\times\sin\theta$, with $r = 0$, $\tau_A=0$.
• Force B: The force B, when considered about point $P$, will cause a counter - clockwise rotation. If we use the right - hand rule (where the position vector is from $P$ to the point of application of the force and the force vector is as given), the cross - product $r\times F$ (for torque $\tau = r\times F\times\sin\theta$) will result in a counter - clockwise torque.
• Force C: The force C will cause a clockwise torque. Using the right - hand rule (position vector from $P$ to the point of application of force C and the force vector), the cross - product $r\times F$ (for $\tau = r\times F\times\sin\theta$) gives a clockwise torque.
• Force D: The force D will cause a clockwise torque. Using the right - hand rule (position vector from $P$ to the point of application of force D and the force vector), the cross - product $r\times F$ (for $\tau = r\times F\times\sin\theta$) gives a clockwise torque.
• Force E: The line of action of force E passes through the pivot point $P$. So, $r = 0$ (distance from the pivot to the line of action of the force is zero). Using the torque formula $\tau = r\times F\times\sin\theta$, with $r = 0$, $\tau_E = 0$.

Answer:

Force B is producing a counter - clockwise torque about point $P$.