A box of mass 50 kg is placed on an inclined plane. When the angle of the plane is increased to 30º, the box begins to slide downwards. Calculate the coefficient of static friction between the plane and the box. Draw the free body diagram.
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if inclination of plane is increased to 30° degree then , box start to sliding . it means 30° is angle of repose for this plane .
at equilibrium conditions ,
along horizontal force balanced by frictional force
mgsin30° = fr
we know, fr = u× N
where u is coefficient of friction and N is normal reaction between plane and block .
but Normal reaction balanced with perpendicular component of weight
e.g N = mgcos30°
so, mgsin30 = umgcos30
u = tan30°
u = 1/√3
at equilibrium conditions ,
along horizontal force balanced by frictional force
mgsin30° = fr
we know, fr = u× N
where u is coefficient of friction and N is normal reaction between plane and block .
but Normal reaction balanced with perpendicular component of weight
e.g N = mgcos30°
so, mgsin30 = umgcos30
u = tan30°
u = 1/√3
Answered by
2
Mass = 50 kg
Angle = 30°
Frictional force = Μ mgcos∅
Downward force = mgsin∅
Since the box begins to slide down, both the forces will be equal ;
M mgcos 30° = mg sin30°
M × (√3/2) = (1/2)
M = (1/√3)
M = √3/3
________________________
Hence the coefficient of static Friction will be = (1/√3) = (√3/3)
Angle = 30°
Frictional force = Μ mgcos∅
Downward force = mgsin∅
Since the box begins to slide down, both the forces will be equal ;
M mgcos 30° = mg sin30°
M × (√3/2) = (1/2)
M = (1/√3)
M = √3/3
________________________
Hence the coefficient of static Friction will be = (1/√3) = (√3/3)
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