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As , unequal forces F(1) and F(2) (F(2) ...

As , unequal forces `F_(1)` and `F_(2) (F_(2) lt F_(1))` act on a rod of length L. Calculate the tension at a point situated at a distance y from end A.
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Text Solution

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Here `F_(2) le F_(1)` so resultant force on a rod `F= F_(1) - F_(2)`
If its acceleration is a then `a= (F ) /( M )` where
`a= (F_(1) - F_(2))/( M )`
for y distance mass of rod m `=((M)/(L)) y `
Now equation `F_(1) T= ma`
`: T = F_(1) - (F_(1) - F_(2))/( L )`
`T = F_(1) (1- (y )/(L)) + (F_(2)y)/(L)`
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