Torque — Question 2

Question 2

Sliding a force along its line of action preserves torque

A force $(0,10)\,\mathrm N$ is applied at $(0.30,0.20)\,\mathrm m$ relative to a pivot. It is moved to $(0.30,-0.50)\,\mathrm m$ without changing its direction or magnitude.

1. Compare the torques about the pivot.

2. Prove the general result for moving the application point by a displacement parallel to the force.

Question 2 - Solution

Sliding a force along its line of action preserves torque

(a) Derivation and calculation

Evaluate both signed moments using the same origin: $$\tau_{1,z}=0.30(10)-0.20(0)=\boxed{3.0\ \mathrm{N\,m}},$$ $$\tau_{2,z}=0.30(10)-(-0.50)(0)=\boxed{3.0\ \mathrm{N\,m}}.$$ The perpendicular lever arm is 0.30 m in both cases.

(b) Result and interpretation

For any displacement along the force line, write $\vec r^{\prime}=\vec r+\lambda\vec F$. Then $$\vec\tau^{\prime}=(\vec r+\lambda\vec F)\times\vec F =\vec r\times\vec F+\lambda(\vec F\times\vec F)=\boxed{\vec\tau}.$$ External torque is preserved; internal stresses can still change when the application point moves.

Physical check

A displacement along the force adds no perpendicular moment arm.