Question

A block of mass m is kept on a plank. The
coefficient of friction between the plank and the
block is 1. The plank is slowly raised from one end
so that it makes angle O with horizontal. The force
of friction acting on the plank when 0 = 30° is​

Answer 1

your complete question -> A block of mass m is kept on a plank. The coefficient of friction b/w plank and block is 1.The plank is slowly raised from one end so that it makes angle Θ with horizontal. The forces of friction acting on the planck,when Θ=30° and Θ=60° are respectively

• a} mg/2,mg/2

• b} $\sqrt{3}mg/2$,mg/2

• c} mg/2,$\sqrt{3}$mg/2

• d} mg,mg

answer : option (a)

solution : first find angle of repose, α = tan^-1(u) , where u is coefficient of friction.

= tan^-1(1) = 45°

hence, if inclination angle less than angle of repose then, body won't move downward. at this condition

friction = weight of body along plane

here in first case Θ = 30° < 45°

so, friction = mgsin30°

= mg/2

now in second case , Θ = 60° > 45° here angle of repose is less than angle of inclination, so body will move downward automatically.

then, friction force = coefficient of friction × normal reaction acting between block and plane of surface.

i.e., friction = umgcos60° = 1 × mg × 1/2 = mg/2

hence, in both cases we get friction = mg/2. so, option (a) is correct.

Answer 2

Answer:

Mg/2^1/2

Explanation:

30° < tan–1