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A body of mass $10kg$ slides along a rough horizontal surface. The coefficient of friction is $\dfrac{1}{{\sqrt 3 }}.$ Taking $g = 10\,\,{\text{m/}}{{\text{s}}^2},$ the least force which acts at an angle of $30^\circ $ to the horizontal is
(A) $25N$
(B) $100N$
(C) $50N$
(D) $\dfrac{{50}}{{\sqrt 3 }}N$

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Last updated date: 25th Jun 2024
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Answer
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Hint: We are provided with the mass and acceleration due to gravity and use that to find the value of the normal force. After finding the value of the normal force use that in the frictional force formula to get the least frictional force.

Complete step by step answer
Friction force: The friction force is defined as the force exerted by a surface when a body moves across it or makes an effort to move across it.
\[ \Rightarrow {F_f} = \mu N\]
\[{F_f}\] is the frictional force.
\[\mu \] is the frictional coefficient.
$N$ is the normal force between the surfaces.
Mainly, there are two types of frictional forces.
Kinetic friction force: it is the sliding force when an object moves.
Static friction force: it is the force experienced by an object when it makes an effort to move.
The normal force is defined as the product of mass of the object and the gravity due to acceleration. Its unit is Newton. When the object is at rest then,
Normal force is$N = mg$
Normal force on an incline is
$ \Rightarrow N = mg \times \cos \theta $
Where,
$N$ is the normal force between the surfaces
$m$ is the mass
$g$ is the acceleration due to gravity
Given,
The mass of the body, $m = 10kg$
The acceleration due to gravity, $g = 10\,\,{\text{m/}}{{\text{s}}^2},$
The frictional coefficient, $\mu = \dfrac{1}{{\sqrt 3 .}}$
The least force acts at the surface, Angle of incline $\theta = 30^\circ $
We know that the frictional force is
\[ \Rightarrow {F_f} = \mu N\]
Normal force on an incline is
$ \Rightarrow N = mg \times \cos \theta $
Substituting the values,
$ \Rightarrow N = 10 \times 10 \times \cos 30^\circ $
$ \Rightarrow N = 10 \times 10 \times \cos 30^\circ $
$ \Rightarrow N = 86.7N$
Now, the frictional force is
\[ \Rightarrow {F_f} = \mu N\]
Substituting the values,
\[ \Rightarrow {F_f} = \dfrac{1}{{\sqrt 3 }} \times 86.7\]
\[ \Rightarrow {F_f} = 50N\]

Note: The formula for calculating normal force differs according to the condition of the object, whether the object is at rest or motion or exerting upward force or downward force.