Block A is kept on frictionless surface of a truck, moving with acceleration 10ms^2.if mass of block is 5 kg, find Normal force on it.
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Answer:
Mass of the block=5kg
Coeffecient of friction=0.2
external applied force, F=40N
The angle at which the force is applied=30degree
So the horizontal component of force=Fcos30=40×
2
3
=20
3
N
While the uertical component of the force acting in upward direction=Fsin30=40×
2
1
=20N
The normal reaction from the surface (N)=mg−Fsin30=50−20=30N
So the ualue of limiting friction=μN=0.2×30=6N
Hence the net horizontal force on the block=Fcos30=μN=20
3
N−6N=28.64N
The horizontal acceleration of the block=
m
Fcos30−μN
=
5
28.64
=5.73m/s
2
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Block A is kept on frictionless surface of a truck, moving with acceleration 10ms^2.if mass of block is 5 kg, find Normal force on it.
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