Question

A block of mass 3 kg is placed on a floor. The coefficient of friction between block and floor is 0.4. A horizontal force of 4 N is applied on the block. Calculate the friction force acting between the block and the floor.​

Answer 1

Answer : 4N

From the F.B.D

From the F.B.DHere,

From the F.B.DHere, Normal reaction by the floor is balanced by the weight of object.

$→Mg = N \: ....(i)$

Now, Limiting value of friction :

$→f = uN \: ....(from \: i) \\→f = 0.4 \times 3 \times 10 \\ →f = 12N$

Since, The applied force is less than the limiting value.

→ Frictional force acting between the

block and floor = 4N

(in opposite direction)

.........................................

Explanation :

The static friction is self adjustable force which can oppose the applied force to the extent of limiting value.

Moreover, If the applied force is greater than the limiting value of statistic friction, then linetic friction comes into role which is constant for a given Mass.