Question

for a body of mass 5kg
on a plane
at a limiing Static friction of 30°
what is the force of Friction?​

Answer 1

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Given,

Normal reaction, N=mgcosθ

Limiting friction force, Fr

​  

=μN=μmgcosθ

The angle of inclination is 30°

Mass of lock, m=5kg

mgsinθ−μN=ma

Sliding constant velocity mean acceleration, a=0

mgsin30°−μmgcos30°=0

μ = tan30° = $\large\bold{\frac{1}{\sqrt{3} } }$.

Hence, the coefficient of friction is   $\large\bold{\frac{1}{\sqrt{3} } }$.

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Answer 2

Answer:

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Explanation:

Given,

Normal reaction, N=mgcosθ

Limiting friction force, Fr =μN=μmgcosθ

The angle of inclination is 30°

Mass of lock, m=5kg

mgsinθ−μN=ma

Sliding constant velocity mean acceleration, a=0

mgsin30°−μmgcos30°=0

μ = tan30° = 1/√3

Hence, the coefficient of friction is 1/√3