Question

The magnitude of friction force acting on the block if the block is in equilibrium ($g = 10m/{s^{ - 2}}$ )

A.$\dfrac{{20}}{3}N $
B.$20N$
C.$10\sqrt{3} N$
D.$10N$

Answer 1

Hint: Force is the push and the pull on an object. Since its magnitude and direction both are important it is the vector quantity. Force is the product of mass into acceleration and is measured in kilogram into metre per Second Square and its SI (system international) unit is Newton. Here we are asked to find the magnitude of the force when the block is placed in the equilibrium, so use formula $F = mg\cos \theta $and place the given values and simplify.

Complete step by step answer:
Given that Mass, $m = 2kg$
$g = 10m/{s^{ - 2}}$
The magnitude of the force means the size of the force.
Now, the block in the equilibrium makes an angle. By using vertically opposite angles are equal and sum of the two angles at the perpendicular is $ = 90^\circ $
Therefore, $\theta = 90^\circ - 30^\circ = 60^\circ $
Now, by using the formula –
$F = mg\cos \theta $
Place the given values in the above equation
$F = 2 \times 10 \times \cos 60^\circ $
Now, we know that $\cos 60^\circ = \dfrac{1}{2}$
$ \Rightarrow F = 2 \times 10 \times \dfrac{1}{2}$
Simplify the above equation
$ \Rightarrow F = 10N$
Therefore, the required answer – the magnitude of the given force is $10N$.

So, the correct answer is “Option D”.

Note:
The magnitude of the force is the sum of all the forces acting on the body. If two forces are acting in the same direction, the magnitude of the force increases and is the sum of both the forces. When the two forces are acting in the opposite direction, the resultant magnitude of the force will be the difference of both the forces.