Question

A block of mass 20 kg is acted upon by a force F= 30Nat an angle 53 with the horizontal in downward direction as shown .The coefficient of friction between and the horizontal surface is 0.2 .The friction force acting on the block by the ground is
(a) 40 N (b) 30 N (c) 18 (d) 44.8

Answer 1

by taking f =u.N
the friction force coming is 44.8 N

by taking f = u. F cos 53°
friction force coming is 18 N
since... block is in rest therefore all forces should be balanced by each other

and friction force is adjustable...so ans. is 18 N

Answer 2

Answer:

The frictional force acting on the block by the ground is 18 N.

(c) is correct option.

Explanation:

Given that,

Mass = 20 kg

Force = 30 N

Angle = 53°

Coefficient of friction = 0.2

We need to calculate the maximum friction force

Using maximum frictional force

$F_{max}=\mu N$

$F_{max}=\mu(mg+F\sin\theta)$

Put the value into the formula

$F_{max}=0.2(20\times9.8+30\sin53)$

$F_{max}=43.99\ N$

We need to calculate the frictional force

Using formula of frictional

$f_{\mu}=F\cos\theta$

Put the value into the formula

$f_{\mu}=30\cos53$

$f_{\mu}=18\ N$

Hence, The frictional force acting on the block by the ground is 18 N.