A) mg
B) \[mg\sin {{74}^{o}}\]
C) \[mg\cos {{74}^{o}}\]
D) none of the above
Correct Answer: A
Solution :
[a] Using Lami's theorem \[\frac{mg}{\sin ({{180}^{\circ }}-{{37}^{\circ }})}=\frac{{{N}_{2}}}{\sin ({{180}^{\circ }}-{{37}^{\circ }})}\] \[\therefore \] \[{{N}_{2}}=mg\]You need to login to perform this action.
You will be redirected in
3 sec