Question

28. A body of mass 50 kg is rest on an inclined plane of angle 30°
with the horizontal. The coefficient of friction between the
body and the plane is​

Answer 1

Answer:

..............

Explanation:

its the answer

Answer 2

Answer:

The coefficient of friction between the body and the plane is​ $\frac{1}{\sqrt{3}}$.

Explanation:

After breaking the weight of the body into different components we get that,

$N=mgcos\theta$       (1)

Where,

N=normal force acting on the body

m=mass of the body

g=acceleration due to gravity

θ=angle the inclined plane makes with the ground

And frictional force is given as,

$f=mgsin\theta$           (2)

f=frictional force

The frictional force is also given as,

$f=\mu N$             (3)

μ=coefficient of friction

By using equations (1) and (2) in equation (3) we get;

$mgsin\theta=\mu\times mgcos\theta$

$\mu=\frac{sin\theta}{cos\theta}=tan\theta$

$\mu=tan 30\textdegree=\frac{1}{\sqrt{3}}$       (∵θ=30° given in question)

Hence, the coefficient of friction between the body and the plane is​ $\frac{1}{\sqrt{3}}$.