Answer to Question #208052 in Mechanics | Relativity for Harshal Ahire

Question

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

Answer 1

$F_1= 18N$

$\text{let }\vec F \text{ resultant force}$

$\vec F= m \vec a$

$\vec a = 0 \text{ since the body is hauled}$

$\vec F=0$

$\vec F = \vec F_1+\vec F_{fr}+\vec N$

$\text{for vertical axis Y:}$

$\vec N-mg+\vec F_1\sin14\degree= 0$

$\vec N =mg - \vec F_1\sin14\degree= 50-18*\sin14\degree=45.66$

$\text{for horizontal axis X:}$

$\vec F_1 \cos14\degree-F_{fr}=0$

$F_{fr}=\vec F_1 \cos14\degree=18*\cos14\degree=17.47$

$\mu= \frac{F_{fr}}{N}=\frac{17.47}{45.66}=0.383$

$\text{Answer: } \mu= 0.383$