Question

A force of 100N is applied on a block of mass 3kg as shown in figure the coefficient of friction between the surface of the block is 1/4 the frictional force acting on the body is

Answer 1

Answer: The frictional force is approximate 20 N downward.

Explanation:

Given that,

Force F = 100 N

Mass = 3 kg

Angle $\theta = 30^{0}$

Now, the frictional force  

$F_{\mu} = \mu N$

Where, N = normal force

$N = F cos\theta$

$N = 100\times cos 30^{0}$

$N =  86.6 N$

Now, the frictional force

$F_{\mu} = \dfrac{1}{4}\times86.6$

$F_{\mu} = 21.6 N$

Hence, the frictional force is approximate 20 N downward.

 

Answer 2

"Given that,

Force F = 100 N

Mass = 3 kg

$Angle \theta \quad =\quad 30^{\circ}$

Now, the frictional force  F= μN

Where, N = normal force

and $N\quad =\quad F\cos { \theta}$

$N\quad =\quad 100\times \cos {30} ^ {\circ}$

Now, the frictional force

${ F }_{ \mu}\quad =\quad \frac { 1 }{ 4 } \times 86.6$

${ F }_{ \mu}\quad =\quad 21.6N$

Hence, frictional force is approximate 20 N downward."