A body of mass 5kg is placed on rough horizontal surface. If coefficient of friction is 1/√3.Find how much pulling force should act on the body at an angle 30° from horizontal so that the body just begins to move.
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Answer:
50N.
Explanation:
The block will have its own weight in downward direction as mg and will have two components as mgsin30 as horizontal and mgcos30 as vertically downward. The mgsin30 will be in the direction of the frictional force f.
So, we know that the f=μN or f= μ*mgcos30.
Since, normal reaction force is opposite to the vertical component of the block. So, N=mgcos30.
Now, f=(1/√3)*5*10*√3/2 which on solving we will get the frictional force as 25N.
We also know that the force experienced by the block will be F=f+mgsin30.
F=25+5*10sin30.
F=25 + 50*(1/2) or F=25+25 = 50N of the force is required for pulling the body at an angle of 30 degree.
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