A) \[Zero\]
B) \[F\]
C) \[\frac{F\,(L-Y)}{L}\]
D) \[\frac{F\,(L-Y)}{M}\]
Correct Answer: C
Solution :
[c] \[a=\frac{F}{M}\] \[F-T=\frac{M}{L}ya\] \[F-T=\frac{M}{L}y\frac{F}{M}\] \[F-T=F\frac{Y}{L}\] \[T=F-\frac{Fy}{L}\] \[T=F\,\left[ 1\frac{y}{L} \right]=T=F\left[ \frac{L-y}{L} \right]\] In Rod \[L\,(m)\to M\] \[I\,(m)\to \frac{M}{L}\] \[(L-y)m\to \frac{M}{L}[L-y]\] \[y\,(m)\to \frac{M}{L}y\]You need to login to perform this action.
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