# Non-zero torque for a part of rod in equilibrium? [closed]

Suppose a vertical rod of $L$ length has force $F$ acting towards right, on top and bottom positions. A force $2F$ acts towards left on the centre of rod. The body is equilibrium. Let's take the lower $\frac{1}{4}$th part of rod as my system. Now an internal force $F$ must act on this part towards left to create equilibrium. But due to this force there is a non-zero torque on it. How is this possible?

## closed as off-topic by John Rennie, Kyle Kanos, ZeroTheHero, Yashas, glSSep 22 '18 at 19:10

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – Kyle Kanos, ZeroTheHero, Yashas, glS
If this question can be reworded to fit the rules in the help center, please edit the question.

• Hint. Treat as a beam. Let the bar be horizontal. Treat the two end forces as upward reactions at supports and the center force as a downward load, an draw shear and moment diagrams – Bob D Sep 14 '18 at 11:32

See diagram below. Free body diagram shows internal vertical shear (V=F) and bending moment (M) anywhere to the right of the 2F load. The shear is constant from the center to each end along the beam. The bending moment varies linearly from a maximum at the center to zero at the ends. It is the bending moment that is responsible for maintaining rotational equilibrium.

Hope this helps.

Now an internal force F must act on this part towards left to create equilibrium.

That is not true as the object is at equilibrium even without considering that force. Consider the point 1/4th up the rod to be the pivot(marked as the red dot). Calculate the clockwise and anti-clockwise moments: Clockwise moments: $$\tau=\frac{3}{4}aF$$ Anti-clockwise moments: $$\tau=\frac{1}{4}a*2F +\frac{1}{4}aF=\frac{3}{4}aF$$ Hence you don't need to consider internal forces as they cancel out in a rigid body and here the external forces in this situation do indicate equilibrium my balance or torque and forces from any point.

Notice that for rigid body all parts of it are connected like for this rod, hence applying a force at any other point transmits internal forces to that point. Notice that here the force transmitted on the 1/4th point will be both to the right and left as there are both clockwise and anti-clockwise torques acting. Hence it indeed stays in equilibrium.

• The whole body is not my system. Just the 1/4th part is. To calculate torque for only that part why do I have to consider the forces on some other system? – jatin Sep 14 '18 at 11:48
• Notice that torque is a quantity that we take for a system there is no point in taking torques about just a point if there are no distances to take the perpendicular distance from. All of this is because if the object is a rigid body, you must consider torques acting on all parts of the body. If they cancel out, then only is the object in equilibrium(along with forces being cancelled out too). – Tausif Hossain Sep 14 '18 at 11:51
• Notice that for rigid body all parts of it are connected like for this rod, hence applying a force at any other point transmits internal forces to that point. Notice that here the force transmitted on the 1/4th point will be both to the right and left as there are both clockwise and anti-clockwise torques acting. Hence it indeed stays in equilibrium. – Tausif Hossain Sep 14 '18 at 11:54