A) \[2mg\]
B) \[mg\]
C) \[\frac{3mg}{2}\]
D) \[4mg\]
Correct Answer: A
Solution :
[a] \[F=2T\sin \theta \] or \[T\sin \theta =\frac{F}{2}\] \[T\sin \theta +N=mg\] \[\Rightarrow N=mg-T\sin \theta \] For no loss of contact: N>0 \[\Rightarrow mg-T\sin \theta >0\] \[\Rightarrow mg-\frac{F}{2}>0\] \[\Rightarrow F<2mg\] So \[{{F}_{\max }}=2mg\]You need to login to perform this action.
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