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Two forces F(1) " and " F(2) are acting...

Two forces ` F_(1) " and " F_(2)` are acting on a rod abc as shown in figure.

A

`sqrt(31)N`

B

`sqrt(26)N`

C

`sqrt(41)N`

D

`sqrt(36)N`

Text Solution

Verified by Experts

The correct Answer is:
c
Net force `F = sqrt(F_(1)^(2) +F_(2)^(2))= sqrt(41)N`
`f_(1) = 0.4 xx 10g = 40 N f_(k) = 0.3 xx 10g = 30 N`
net force is less than `f_(0)` hence
Required friction force = applined force` = sqrt(41)N`
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