A box of mass 8 kg is placed on a rough inclined plane on inclination theta. its downwrd motion can be prevented by applying an upwrd pullf. and it can be mde to slide ipwrds by applying a force 2f . the coffecient of friction between the box and inclind plane is
Answers
N = normal force by the incline on the block
f' = frictional force on the block
m = mass of the block = 8 kg
θ = angle of incline
μ = Coefficient of friction
consider the force diagram to prevent the motion of block in down direction :
force equation perpendicular to incline is given as
N = mg Cosθ
frictional force on the block is given as
f' = μ N
f' = μ mg Cosθ
parallel to incline , the force equation is given as
f + f' = mg Sinθ
f + μ mg Cosθ = mg Sinθ
f = mg Sinθ - μ mg Cosθ eq-1
consider the force diagram to make the block slide upward:
force equation perpendicular to incline is given as
N = mg Cosθ
frictional force on the block is given as
f' = μ N
f' = μ mg Cosθ
parallel to incline , the force equation is given as
2 f = mg Sinθ + f'
2 f = mg Sinθ + μ mg Cosθ
using eq-1
2 (mg Sinθ - μ mg Cosθ) = mg Sinθ + μ mg Cosθ
2 mg Sinθ - 2 μ mg Cosθ = mg Sinθ + μ mg Cosθ
mg Sinθ = 3 μ mg Cosθ
tanθ = 3 μ
μ = tanθ/3
The answer is :
tan theta / 3
