A light rigid rod XY has an object of weight W fixed at one end. The rod is in equilibrium, resting on a roller at Z and a vertical wall at X.

Question 4

A light rigid rod XY has an object of weight W fixed at one end. The rod is in equilibrium, resting on a roller at Z and a vertical wall at X. The roller exerts a force R on the rod as shown. The diagram shows the directions, but not the magnitudes, of the forces R and W.

What is the direction of the force on the rod at X?

Reference: Past Exam Paper – June 2016 Paper 12 Q12

Solution:

Answer: D.

The system consists of 3 force vectors: weight W, reaction R and a vector starting at point X and we need to find out its direction. Let’s call this unknown vector ‘vector Y’.

For equilibrium of 3 vectors, all the vectors should form a closed triangle when placed next to each other. BUT, since the question specifies that the diagram shows only the direction, and NOT the magnitude, we cannot rely on this method.

Another consequence of equilibrium of the 3 vectors is that the lines of action of the vectors should pass through a similar point. Thus, by extending vectors R and W, we can identify this point where both vectors cross. Vector Y can be drawn by drawing a line connecting point X on the wall to the point where all the vectors cross.

Another approach is to examine the situation in terms of moments.

For equilibrium,

sum of clockwise moment (about a point) = sum of anticlockwise moment

Moment = Force × perpendicular distance from line of action of the force to the pivot

Taking moments about X provides evidence that R must be very much larger than W since the distance of X from the pivot (X) is much smaller compared to the distance of W from pivot X.

W causes a clockwise moment, while R causes an anticlockwise moment.

Hence, since weight W is larger than reaction R, the value of the force at X must be larger than W, and with a component in a direction away from the wall (to provide an additional anticlockwise moment).