A body of weight 30 N rests on a horizontal floor. A gradually increas...
The ratio of the limiting force of friction (F) to the normal reaction (R or N) between two bodies is known as co-efficient of friction (μ).
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A body of weight 30 N rests on a horizontal floor. A gradually increas...
Problem:
A body of weight 30 N rests on a horizontal floor. A gradually increasing horizontal force is applied to the body which just starts moving when the force is 9 N. The coefficient of friction between the body and the floor will be
a) 10/3
b) 3/10
c) 1/3
d) 1/9
Solution:
The force required to just start moving an object is known as the static friction force. We are given that a gradually increasing horizontal force is applied to the body until it just starts moving when the force is 9 N.
Let's assume that the coefficient of friction between the body and the floor is 'μ'. The weight of the body is 30 N, and the force required to just start moving the body is 9 N.
Normal Force:
The normal force is the force exerted by a surface to support the weight of an object resting on it. In this case, the normal force is equal to the weight of the body, which is 30 N.
Friction Force:
The friction force can be calculated using the formula: Friction force = coefficient of friction * normal force.
The friction force required to just start moving the body is equal to the force applied, which is 9 N.
Therefore, we can write the equation as:
9 N = μ * 30 N
Solving for μ:
To solve for μ, divide both sides of the equation by 30 N:
(9 N) / (30 N) = μ
Simplifying the equation:
0.3 = μ
Therefore, the coefficient of friction between the body and the floor is 0.3.
Answer:
The correct answer is option b) 3/10.