# 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 respectivelya} mg/2,mg/2b} $\sqrt{3}$mg/2,mg/2c} mg/2,$\sqrt{3}$mg/2d}mg,mg

souvik das
33 Points
11 years ago

(a) is the correct ans

36 Points
11 years ago

Ans: a

In first case when u resolve the forces and find the frictional force value(f) = µN.

N=mg*cos(60)=mg/2.

f=mg/2          (µ=1)

but the component of force acting parallel to surface is mg*sin(60)=$\sqrt{3}$mg/2

But if u observe the parallel force component is less than frictional force value.

Here the point to be observed is that frictional force will always develop as opposite to the acting force .And µN is the maximum frictional force that can develop but it doesnt mean that always that much frictional force will develop.So the least among above two forces has to be considered.

For the other case its just straight forward µN=1(mg sin(30))=mg/2.

OCEAN GODRE
40 Points
11 years ago

θthe force is acting upwards so the frictional force will act downwards

and its magnitude will be =coeeff of friction *normal reaction=μ*n

but here the normal reaction is balanced by mgcosΘ

therefore n=mgcosθ

therefore frictional force=μmgcosθ

for θ =30 ,force of friction=root3*mg/2 and for θ =60, force =mg/2

hence b is the correct answer