Question

A block of mass m is placed at rest on an inclined plane of inclination θ to the horizontal. If the coefficient of friction between the block and the plane is μ, then the total force exerted by the inclined plane on the block is

• A. mg
• B. $\mu mg\cos \theta$
• C. $mg\sin \theta$
• D. $\mu mg\tan \theta$

Answer 1

The correct option is A

mg

Step1: Given data

The block of mass m is placed at rest on an inclined plane of inclination $\theta$ to the horizontal and the coefficient of friction between the block and the plane is $\mu$.

Step2: Formula used

Force acting to the horizontal is the normal force due to the inclined plane on the block which is $mg\cos \theta$

Force acting to the vertical is the frictional force which is $mg\sin \theta$

Step3: Calculating total force

On the basis of the given data, analyzing the diagram

The total force is the sum of normal force and frictional force and normal force and frictional force are perpendicular to each other.

$$\begin{gathered} F_{total}=F_{normal}+F_{frictional} \\ {F}_{total}=mg\sqrt{{\sin }^2\theta +{\cos }^2\theta } \\ {F}_{total}=mg[As{\sin }^2\theta +{\cos }^2\theta =1] \end{gathered}$$

Hence, the total force exerted by the inclined plane on the block is mg

Therefore, option A is the correct answer.