EAMCET 2005 Question on Friction

The following MCQ appeared in EAMCET 2005 question paper:

A cubical block of mass ‘m’ rests on a rough inclined surface. ‘μ’ is the coefficient of static friction between the block and the surface. A force mg acting on the cube at an angle ‘θ’ with the vertical side of the cube pulls the block. If the block is to be pulled along the surface, then the value of cot(θ/2) is

(a) less than μ (b) greater than μ (c) equal to μ (d) not dependent on μ

The pulling force mg applied at an angle θ with the vertical can be resolved into a vertical (upward) component mgcosθ and a horizontal component mgsinθ as shown in the figure. It is the horizontal component which moves the block along the surface.

But the vertical component is effective in reducing the frictional force since it reduces the normal force from mg to (mg – mg cosθ) so that the frictional force is μ(mg – mg cosθ) = μmg(1– cosθ). So, the condition for the block to move along the surface is

mg sinθ > μmg(1– cosθ).

Or, sinθ > μ(1– cosθ).

But, sinθ = 2 sin(θ/2)cos(θ/2) and (1– cosθ) = 2sin²(θ/2).

Substituting these in the above condition, we obtain

cos(θ/2) > μ sin(θ/2), from which cot(θ/2) > μ

Do you know why it is easier to pull a lawn mower than to push it?

You have seen in the above problem that the upward component of the pulling force opposes the weight of the block and therefore reduces the normal force (N) on the surface. The frictional force μN is therefore reduced. If a pushing force is applied to move the block, the vertical component of the pushing force will be downwards and it will add with the weight of the block to produce a greater normal force and therefore a greater frictional force.