A body of mass $m$ is kept on a rough horizontal surface (coefficient of friction = $\mu$). A horizontal force

Question

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:

Accepted Answer

1. **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].
2. **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.
   $$ |\vec{F}| = \sqrt{N^2 + f_s^2} $$
3. **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 ($f_s = \mu mg$): $|\vec{F}|_{max} = \sqrt{(mg)^2 + (\mu mg)^2} = mg\sqrt{1+\mu^2}$.
4. **Conclusion:** The magnitude of the resultant force satisfies the condition:
   $$ |\vec{F}| \le mg\sqrt{1+\mu^2} $$

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