A) Mg
B) Ma
C) \[\frac{Ma}{2}\]
D) \[\frac{Ma}{3}\]
Correct Answer: D
Solution :
Force of friction at point P, friction \[{{\text{F}}_{\text{friction}}}=\frac{1}{3}Ma\,\sin \theta \] \[=\frac{1}{3}Ma\,\sin {{90}^{o}}\] \[[\text{here}\,\theta ={{90}^{o}}]\] \[=\frac{Ma}{3}\]You need to login to perform this action.
You will be redirected in
3 sec