The 4 kg block is kept on a table has friction coefficient 0.8. Find the resultant force applied by table on 4 kg block and value of friction force.
Answers
50N
Normal force on the block, N=mg=4×10=40N
Maximum friction force which can act on block =μN=0.8×40=32N
As the force applied is less than the maximum friction force so friction force on block is 30N (to stop relative motion)
Both friction force and normal force are exerted by ground and their direction is horizontal and vertical respectively (i.e perpendicular to each other). Hence, net force exerted by ground is
30
2
+40
2
=50N
Normal force on the block, N=mg=4×10=40N
Maximum friction force which can act on block =μN=0.8×40=32N
As the force applied is less than the maximum friction force so friction force on block is 30N (to stop relative motion)
Both friction force and normal force are exerted by ground and their direction is horizontal and vertical respectively (i.e perpendicular to each other). Hence, net force exerted by ground is
2 2
30+40=50N
Static friction is a self-adjusting force.
The limiting value of friction up to which the body stays at rest till it starts moving is given by, F
s
=μmg
where μ is the coefficient of static friction.
Below that limit, the frictional force is equal to the force applied.
Here F
s
=0.8×4×10=32N ; g=10m/s
2
Whereas force applied is 19N less than the limiting value of friction.
Hence Frictional force is 19N.