A) \[\frac{F}{3}\] rightward
B) \[\frac{F}{3}\]leftward
C) \[\frac{2F}{3}\]rightward
D) \[\frac{2F}{3}\]leftward
Correct Answer: B
Solution :
[b] Newton's second law is in the horizontal direction, \[F+{{f}_{F}}=m{{a}_{G}}\] ... (1) For angular acceleration a, \[FR-{{F}_{f}}R={{l}_{G}}\alpha \] \[\therefore \,\,\,\,\,\,\,F-{{F}_{1}}=\frac{mR\alpha }{2}=\frac{m{{a}_{G}}}{2}\] ? (2) Solving for \[{{F}_{f}}\] from equation (1) and (2), \[{{F}_{f}}=\frac{F}{3}\,\,\,\,\,\,\therefore \,\,\,{{F}_{f}}=\frac{F}{3}\] rightward \[\therefore \] The force of friction exerted on the surface \[=\frac{F}{3}\]leftwardYou need to login to perform this action.
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