A rod of length L and mass M is acted on by two unequal forces F_1 and F_2 ( lt F) as shown in the figure. The tension in the rod at a distance y from the end A is given by

A rod of length L and mass M is acted on by two unequal forces F_1 and

1.	A rod of length L and mass M is acted on by two unequal forces F_1 and F_2 ( lt F) as shown in the figure. The tension in the rod at a distance y from the end A is given by
Answer» `F_1 (1- (y)/(L))+F_2 ((y)/(L))` `F_2 (1-(y)/(L ) )+ F_1((Y)/(L))` `(F_1 -F_2) (y)/(L)` None of these Solution :NET forceon therod`= F_1-F_2` Asmassof therodis M ,so ACCELERATIONOF therodis ` a=((F_1 -F_2))/(M)` Now consideringthe motionofthe rod[ whosemassisequalto `(m//L)`y]then theequationof MOTIONIS ` F_1-T= (M )/(L )y xx a` whereT isthetensionin therodat B . `orF_1 -T=(M )/(L)y xx ((F_1-F_2)/(M))` `orT= F_1 (1- (y)/(L ) ) +F_2((y)/(L ))` secondmethod: tocalculatetensionat Bwecan alsoconsiderthe motionof theotherof the rodi.e.,BC. thentheequationof motionwill be ` T- F_2=(M )/(L )(l-y) xx a` ` orT =F_2(M )/(L )(L-y) xx ((F_1 -F_2))/(M)` ( USING(i)) `orT=F_1 ( 1- (y)/(L ))+F_2((y)/(L ))`