# A block of mass 2kg rests on a rough inclined plane making an angle of 30degree with the horizontal.the coefficient of ststic friction between the block and the plane is 0.7.What is the magnitude of the friction force on the block?(g=908meter/second square)

2
by avinash01lover

2014-09-01T22:03:22+05:30
Given, m = 2 kg
θ = 30 degree
μ = 0.7

As it is clear for figure :
R = m g cos ​θ
Therefore, force of friction, F = ​μ R = ​μ m g cos ​θ = 0.7 X 2 X 9.8 X cos 30 = 11.86 N
2014-09-02T06:11:18+05:30

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See the diagram.
If the block is resting with out slipping on the inclined surface, it means that the frictional force is greater than the gravitational pull down the plane. we calculate the minimum value of static friction at the surface of the block.

Friction force Ff = coeff of static friction * Normal force
Ff           = Mu * N
Normal force N = mg cos Ф  as the block is in equilibrium. and net force in the  direction of perpendicular to plane is zero.  Ф is the angle of inclination of plane.

Ff = Mu * m g Cos Ф = 0.7 * 2 kg * 9.8m/s² * cos 30 = 11.88 Newtons

Suppose we calculate the component of weight along the inclined plane:
= m g Sin Ф = 2 kg * 9.8m/s² * sin 30 = 9.8 Newtons

So we see that the static frictional force is higher than the gravity along the inclined plane, so the block rests.

If Ф is increased, cos Ф reduces and sin Ф increases, and then there will be a point when the block starts to slide.