Question

A body of weight 50 N is hauled along a rough horizontal plane by a pull of 18 N acting at an angle of 14° with the horizontal. Find the coefficient of friction.

Answer 1

Frictional force f=28.2cos45
o
=28.2
2
​

1
​
=20N
Normal reaction R=50−28.2sin45
o
=50−20=30N

Answer 2

Answer:

The coefficient of friction is 0.2493

Explanation:

We know that

Normal reaction force = N = F cos $\theta$

where $\theta$ is the angle between the horizontal and the pull

Therefore N = 50 cos 14°

We also know that

Frictional force = F = F sin $\theta$

where $\theta$ is the angle between the horizontal and the pull

Therefore F = 50 sin 14°

We also know that F = N μ

where F is the frictional force

          N is the normal reaction force

          μ is the coefficient of friction

therefore, on substitution,

=> 50 sin 14° = 50 cos 14° * μ

=> tan 14° = μ

=>  μ = 0.2493

Therefore, the coefficient of friction is 0.2493