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Coefficients of Friction

What will be ...

Question

What will be the force of friction, if $190 N$ force is being applied over a block $40 k g$ on the rough surface? (Coefficient of static friction $μ_{s} = 0.5, g = 10 {m s}^{- 2})$

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Solution

Limiting friction= $μ m g = 0.5 \times 40 \times 10 = 200 N$
As applied force is less than limiting friction,
therefore frictional force=applied force=190N

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