Question

A block of mass 50 Kg can slide on a rough horizontal surface. The coefficient of friction between the block and the surface is 0.6. The least force of pull acting at an angle of ${30^0}$ to the upward drawn vertical which causes the block to just slide is:
A. 29.43 N
B. 219.6 N
C. 21.96 N
D. 294.3 N

Answer 1

Hint:Here we first need to find the frictional force that will cause the block to slide. For this first we have to find the horizontal and the vertical component of the force given. Also the angle for the force is also given. Then equate the horizontal component of the force with the frictional force in order to get the required answer.

Formula used:
1. Reaction force, $R = mg - F\cos \theta $
2. Frictional force, $f = \mu R$
Where, $\mu $is the coefficient of frictional force.

Complete step by step solution:
The free body diagram can be drawn as

Let the force applied on the body at the given angle be F.
Given angle, $\theta = {30^0}$.
This force will consist of horizontal and vertical components.
Vertical component of force will be $F\cos \theta = F\cos {30^0}$.
Horizontal component of force will be $F\sin \theta = F\sin {30^0}$.
Reactional force will be:
$R = mg - F\cos \theta $
Putting the values from the question, we get:
$R = 50 \times 10 - F\cos {30^0}$
By solving;
$R = 500 - F\dfrac{{\sqrt 3 }}{2}$
Now, frictional force is given as follows:
$f = \mu R$ (equation 1)
Also f will be equal to horizontal component of force:
$f = F\sin {30^0}$ (equation 2)
Equating equation 1 and 2, we get;
$\dfrac{F}{2} = \left[ {(500 - F\left( {\dfrac{{\sqrt 3 }}{2}} \right)} \right]\mu $
By solving;
$F = (1000 - F\sqrt 3 )0.6$
$\therefore F = 294.3\,N$

Hence, the correct answer is option D.

Note: Here the vertical component of force balances the mg force applied on the body in downward vertical direction and the horizontal component balances the frictional force applied on the body. Reactional force will be resultant of all the vertical forces applied on the given body. Then finally multiplying it with the frictional coefficient we will get the final answer.