a block of mass 10 kg is placed on an inclined plane when the angle of inclination is 30°the block just begin to slide down the plane the force of friction is.
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Answer:
Given
Angle of inclination of θ=30
o
Normal reaction, N=mgcos30
o
=10×10×
2
3
=50
3
Limiting Friction force, F
=μN=μ50
3
N
Hence, during motion limiting friction force is =0.5×50
3
=25
Explanation:
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