A body of weight 50N is pulled along a rough horizontal plane by force of 18N acting at 14° with the horizontal .Find the coefficient of friction .
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Answer:
F1=18N
\text{let }\vec F \text{ resultant force}let F resultant force
\vec F= m \vec aF=ma
\vec a = 0 \text{ since the body is hauled}a=0 since the body is hauled
\vec F=0F=0
\vec F = \vec F_1+\vec F_{fr}+\vec NF=F1+Ffr+N
\text{for vertical axis Y:}for vertical axis Y:
\vec N-mg+\vec F_1\sin14\degree= 0N−mg+F1sin14°=0
\vec N =mg - \vec F_1\sin14\degree= 50-18*\sin14\degree=45.66N=mg−F1sin14°=50−18∗sin14°=45.66
\text{for horizontal axis X:}for horizontal axis X:
\vec F_1 \cos14\degree-F_{fr}=0F1cos14°−Ffr=0
F_{fr}=\vec F_1 \cos14\degree=18*\cos14\degree=17.47Ffr=F1cos14°=18∗cos14°=17.47
μ=NFfr=45.6617.47=0.383
\text{Answer: } \mu= 0.383Answer: μ=0.383
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