Question 15: Laws of Motion

Question

A block of mass 10 kg rests on a rough inclined plane making an angle of 30$^\circ$ with the horizontal. The coefficient of static friction between the block and the plane is 0.1. The friction force on the block is

Accepted Answer

To find the friction force on the block, we first need to analyze the forces acting on the block on an inclined plane.
The forces acting on the block include:
Gravitational force (weight) acting vertically downward, $mg$
Normal force, $N$, exerted by the plane perpendicular to its surface
Frictional force, which acts parallel to the inclined surface and opposite to the direction of possible motion
The gravitational force can be resolved into two components:
Perpendicular to the inclined plane: $mg \cos 	heta$
Parallel to the inclined plane: $mg \sin 	heta$
In this problem:
Mass of the block, $m$ = 10 kg
Angle of the incline, $	heta$ = 30$^\circ$
Coefficient of static friction, $\mu_s$ = 0.1
First, we calculate the perpendicular component of the gravitational force to find the normal force:
$N = mg \cos 	heta$
Substituting the values:
$N = (10 \, 	ext{kg}) (9.8 \, 	ext{m/s}^2) \cos 30^\circ$
$\cos 30^\circ = \frac{\sqrt{3}}{2}$
$N = 10 	imes 9.8 	imes \frac{\sqrt{3}}{2}$
$N = 49 \sqrt{3} \, 	ext{N}$
Next, we find the maximum static friction force:
$f_{	ext{max}} = \mu_s N$
$f_{	ext{max}} = 0.1 	imes 49 \sqrt{3} \, 	ext{N}$
$f_{	ext{max}} = 4.9 \sqrt{3} \, 	ext{N}$
We now compare this with the component of the gravitational force parallel to the inclined plane:
$mg \sin 	heta$
Substituting the values:
$mg \sin 30^\circ = 10 	imes 9.8 	imes \frac{1}{2}$
$mg \sin 30^\circ = 49 \, 	ext{N}$
The friction force will be equal to the component of the gravitational force parallel to the inclined plane but within the limit of the maximum static friction force. Here, we recognize that the maximum static friction force (4.9 $\sqrt{3}$ N) is indeed greater than 4.9 N (the component of the gravitational force parallel to the incline). Hence, the actual friction force counteracting the motion will be:
$49 \, 	ext{N}$
Therefore, the correct answer is Option C:
Option C: 49 N

Answer

4.9 N

Answer

49$\sqrt3$ N

Answer

49 N

Answer

0.1 $	imes$ 49$\sqrt3$ N

Laws of Motion

A block of mass 10 kg rests on a rough inclined plane making an angle of 30^\circ with the horizontal. The coefficient of static friction between the block and the plane is 0.1. The friction force on the block is

Choose the correct answer: