Question

A body of mass m is placed on a rough surface with a coefficient of friction μ inclined at angle θ. If the mass is in limiting equilibrium, then

• A. θ = tan^-1 (m/μ)
• B. θ = tan^-1 (μ)
• C. θ = tan^-1 (μ/m)
• D. θ = tan^-1 (1/μ)

Answer 1

The correct option is B

θ = tan^-1 (μ)

Step 1: Given data

1. The mass of the body is m.
2. The coefficient of friction of the rough surface is $\mu$.
3. The angle of the inclined plane = $\theta$.

Step 2: Frictional force

1. The frictional force is defined as the force that prevents the motion of the body at the point of contact.
2. Frictional force, $F=\mu N$, where $\mu$ and $N$ are the coefficient of friction of the rough surface and normal force on the body.

Step 3: Diagram

Step 4: Finding the angle of the inclined plane

According to the question, the block is in the equilibrium stage on the inclined plane. From the above figure,

Normal force,

$N=mg\cos \theta$ ……………..(1)

and frictional force,

$f=mg\sin \theta$ ………………(2)

Again, we know, that, the frictional force

$f=\mu N$

or $f=\mu mg\cos \theta$ ………………(3)

From equations (2) and (3) we get,

$\frac{mg\sin \theta }{\mu mg\cos \theta }=1$

$\frac{\tan \theta }{\mu }=1$

$\tan \theta =\mu$

or $\theta ={\tan }^{-1}\left(\mu \right)$

Hence, option (B) is correct.