A mass of 200 Kg is placed on a rough inclined plane of angle 30 degree. If coefficient of limiting friction is (1/√3), find the greatest and the least forces in Newton acting parallel to the plane to keep the massin equilibrium.
Answers
The least force required to keep the mass in equilibrium is 0 N.
The weight of the force which will act in downward direction parallel to the plane
w=mgsinθ
The friction force which will act in the opposite direction of the movement of the block is given as
f_friction=μmgcosθ
Therefore the net force on the body
F=mgsinθ-μmgcosθ
Plugging the values in the above equation
F=200*9.81*sin30-(1/√3)*200*9.81*cos30
=0 N.
This means that the inclination of the plane is not enough to move the block, and block itself will be in equilibrium, there is no need to apply external force.
Answer:
Greatest Force - 1960N
Least Foce - 0N
Explanation:
R=mgcosθ
fmax= μR = μMG COSθ
So the greatest force requires to keep the body in equilibrium
Fmax = mgsinθ + fmax = mgsinθ +μmgcosθ
= 200 x 9.8 {1/2 +1/√3 x √3/2} = 1960N
Fmin = mgsinθ -fmax
= mgsinθ - μmgcosθ
mg{sinθ - μcosθ}
=200 x 9.8{1/2 - 1/√3 x √3/2}
=200 x 9.8 x 0
=0N