Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

A body of mass $m$ is kept on a rough horizontal surface (coefficient of friction = $\mu$ ). A horizontal force is applied to the body, but it does not move. The resultant of normal reaction and the frictional force acting on the object is given by $\vec{F}$ , where:

$|\vec{F}| = mg + \mu mg$

$|\vec{F}| = \mu mg$

$|\vec{F}| \le mg\sqrt{1+\mu^2}$

$|\vec{F}| = mg$

Step-by-Step Solution

Identify Forces: The forces exerted by the surface on the body are the Normal Reaction ( $N$ ) and the Static Friction ( $f_s$ ).

Since the body is on a horizontal surface and there is no vertical acceleration, $N = mg$ [NCERT Class 11, Physics Part I, Chapter 5, Section 5.9].
The static friction $f_s$ balances the applied horizontal force to keep the body at rest. Its magnitude adjusts such that $0 \le f_s \le f_{s,max}$ , where the limiting friction $f_{s,max} = \mu N = \mu mg$ [NCERT Class 11, Physics Part I, Eq. 4.14].

Resultant Force (Contact Force): The resultant force $\vec{F}$ (often called the total contact force) is the vector sum of $\vec{N}$ and $\vec{f_s}$ . Since $\vec{N}$ is vertical and $\vec{f_s}$ is horizontal, they are perpendicular. $|\vec{F}| = \sqrt{N^2 + f_s^2}$
Apply Limits: Substituting $N = mg$ and the inequality for friction:

Minimum $|\vec{F}|$ occurs when $f_s = 0$ (no applied force): $|\vec{F}|_{min} = \sqrt{(mg)^2 + 0} = mg$ .
Maximum $|\vec{F}|$ occurs when friction is limiting ( $f_s = \mu mg$ ): $|\vec{F}|_{max} = \sqrt{(mg)^2 + (\mu mg)^2} = mg\sqrt{1+\mu^2}$ .

Conclusion: The magnitude of the resultant force satisfies the condition: $|\vec{F}| \le mg\sqrt{1+\mu^2}$

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