question_answer

A rod of length L and mass M is acted on by two unequal forces \[{{\operatorname{F}}_{1}} arid {{F}_{2}} \left( < {{F}_{1}} \right)\]as shown in the figure. The tension in the rod at a distance y from the end A is given by:

A) \[{{\operatorname{F}}_{1}}\left[ 1-\frac{y}{L} \right]+{{\operatorname{F}}_{2}}\left[ \frac{y}{L} \right]\]

B) \[{{\operatorname{F}}_{2}}\left[ 1-\frac{y}{L} \right]+{{\operatorname{F}}_{1}}\left[ \frac{y}{L} \right]\]

C) \[\left( {{\operatorname{F}}_{1}}-{{\operatorname{F}}_{2}} \right)\frac{y}{L}\]

D) None of these

Correct Answer: A

Solution :

\[\operatorname{Net} force on the rod = {{F}_{1}}-{{F}_{2}}\] \[\operatorname{Acceleration} of rod =\frac{{{\operatorname{F}}_{1}}-{{\operatorname{F}}_{2}}}{\operatorname{M}}\] Rod of length 'L' has mass = M \[\operatorname{Rod} of length 'y' has mass =\frac{M}{L}y\] \[{{\operatorname{F}}_{1}}-\operatorname{T}=\frac{M}{L}y.a\] \[{{\operatorname{F}}_{1}}-\operatorname{T}=\frac{M}{L}y\frac{{{\operatorname{F}}_{1}}-{{\operatorname{F}}_{2}}}{\operatorname{M}}\] \[\operatorname{T}={{\operatorname{F}}_{1}}\left[ 1-\frac{y}{L} \right]+{{\operatorname{F}}_{2}}\frac{y}{L}\]