A) \[mg{{\cos }^{2}}\theta \]
B) \[mg{{\sin }^{2}}\theta \]
C) \[mg\cos \theta \]
D) \[mg\tan \theta \]
Correct Answer: D
Solution :
By applying Lamis theorem at P, we have \[\frac{R}{\sin {{90}^{o}}}=\frac{F}{\sin ({{180}^{o}}-\theta )}=\frac{mg}{\sin ({{90}^{o}}+\theta )}\] \[\Rightarrow \]\[\frac{R}{1}=\frac{F}{\sin \theta }=\frac{mg}{\cos \theta }\] \[\Rightarrow \]\[F=mg\tan \theta \]You need to login to perform this action.
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