NAEST 2020: The direction of frictional force on a clamped s

Question: A wooden scale is clamped near one end. It stays in a horizontal position as shown in the video. The frictional force on the scale by the nut is

1. in the vertically upward direction at all points of contact
2. in the vertically downward direction at all points of contact
3. cannot be horizontal at any point of the contact
4. cannot be vertical at any point of the contact

Solution: The scale is in equilibrium. Hence, (i) the resultant force on the scale is zero, and (ii) the net torque on the scale about its centre of mass is zero.

The forces acting on the scale are (i) its weight (ii) frictional force by the clamp and (iii) the normal force by the clamp. The normal force acts in a direction perpendicular to the vertical plane containing the scale. Its only role is to control the magnitude of the frictional force.

The frictional force on the scale acts on both surfaces. On each surface, the frictional force acts on all contact points. The net frictional force on the scale acts in the vertically upward direction to balance the weight.

Let us see the torque about centre of mass. The torque due weight is zero as weight passes through the centre of mass. Thus, the net torque due to the frictional force must be zero. To understand this, let us divide the contact area by a vertical line passing through its centre.

The frictional force at contact points acts so that (i) its resultant is vertically upward and (ii) its torque about centre of mass is zero. This is possible only when a large vertically upward force acts at contact points closer to the centre of mass and a small vertically downward force acts at contact points farther away from the centre of mass. We encourage you to make a figure showing these forces.

The net force on the scale in the horizontal direction is zero. Also, the net torque about the centre of mass due to the horizontal forces, if any, must be zero. If a horizontal force acts at contact point A then there must be an equal and opposite horizontal force at another contact point B to make the net force zero. This pair of equal and opposite forces will produce a non-zero torque about the centre of mass (unless the contact points A and B lies on a horizontal line passing through the centre of mass.) Thus, the best possible correct option is (C).

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