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PhysicsRotational Motion
A uniform rod of mass 20 kg and length 5 m leans against a smooth vertical wall making an angle of $60°$ with it. The foot of the rod rests on a rough floor. What is the friction force on the rod from the floor?
Options
1
$50\sqrt{3}$ N
2
$100$ N
3
$100\sqrt{3}$ N
4
$200$ N
Correct Answer
$100\sqrt{3}$ N
Solution
1

Rod at $60°$ to wall = $30°$ to floor. Weight $W = 200$ N at midpoint.

Torque about foot of rod ($\sum\tau = 0$):

$$N_{wall} \cdot L\sin30° = W \cdot \frac{L}{2}\cos30°$$$$N_{wall} = \frac{200 \times \cos30°}{2\sin30°} = \frac{200 \times \frac{\sqrt{3}}{2}}{2 \times \frac{1}{2}} = 100\sqrt{3} \text{ N}$$
2

Horizontal equilibrium: Friction $f = N_{wall} = \boxed{100\sqrt{3} \text{ N}}$

Torques about foot → $N_{wall} = 100\sqrt{3}$ N
Horizontal equilibrium → Friction $= N_{wall} = 100\sqrt{3}$ N
Theory: Rotational Motion
1. Static Equilibrium of Rigid Bodies

Three conditions for complete static equilibrium: (1) $\sum F_x = 0$ (horizontal forces balance). (2) $\sum F_y = 0$ (vertical forces balance). (3) $\sum \tau = 0$ (torques about any point balance). Choose the torque pivot wisely — often at a point where unknown forces act (eliminates those unknowns from torque equation). For a ladder/rod leaning against wall: three unknown forces (floor normal, friction, wall normal). Three equations exactly solve the problem.

2. Torque Calculation

Torque $\tau = r \times F = rF\sin\theta$ (angle between $\vec{r}$ and $\vec{F}$). Or: $\tau = F \times d$ where $d$ = perpendicular distance from pivot to line of action of force. For rod leaning at angle: torque of weight about foot $= W \times (L/2)\cos\theta$ (horizontal distance from foot to midpoint of rod). Torque of wall normal about foot $= N_{wall} \times L\sin\theta$ (vertical distance from foot to top of rod). Sign convention: counterclockwise positive.

3. Ladder Problem — General Approach

Setup: uniform rod/ladder of mass $m$, length $L$, against smooth wall at angle $\alpha$ to floor. Forces: (1) Normal from smooth wall $N_w$ (horizontal). (2) Weight $mg$ at midpoint (vertical). (3) Normal from floor $N_f$ (vertical). (4) Friction from floor $f$ (horizontal). Equations: $\sum F_x = 0$: $f = N_w$. $\sum F_y = 0$: $N_f = mg$. Torque about foot: $N_w \cdot L\sin\alpha = mg \cdot (L/2)\cos\alpha$. Gives: $N_w = mg\cos\alpha/(2\sin\alpha) = mg/(2\tan\alpha)$.

4. Minimum Friction for Equilibrium

From ladder analysis: $f = N_w = mg/(2\tan\alpha)$. Also $N_f = mg$. Coefficient of friction needed: $\mu \geq f/N_f = 1/(2\tan\alpha)$. For the rod to stay without slipping: $\mu \geq 1/(2\tan\alpha)$. At $\alpha=45°$: $\mu_{min} = 1/2 = 0.5$. At smaller $\alpha$ (more horizontal): more friction needed (rod more likely to slip). At $\alpha=90°$ (vertical): $\mu_{min} = 0$ (no friction needed). This is why it gets harder to lean a ladder near-horizontal.

5. Centre of Mass and Centre of Gravity

CM: $\vec{r}_{cm} = \sum m_i\vec{r}_i/M$. For uniform rod: CM at geometric centre. CG: point where resultant of gravitational forces acts. For uniform $g$: CM $=$ CG. For non-uniform $g$ (very large objects): CG $\neq$ CM (CG is closer to Earth). In problems: weight acts at CM (or CG for rigid body in gravity). For composite bodies: treat each part separately, find overall CM.

6. Parallel Forces and Resultant

For parallel forces: resultant $= $ algebraic sum. Point of action: $x = \sum F_i x_i / \sum F_i$ (weighted average). For a couple (two equal and opposite parallel forces): net force $= 0$, but torque $= F \times d$ (perpendicular distance). Couple does not depend on the choice of pivot — pure rotation. Example: steering wheel, screwdriver. Any force system can be replaced by a single force plus a couple.

7. Mechanical Advantage

Levers: first class (fulcrum between effort and load — seesaw, scissors), second class (load between effort and fulcrum — wheelbarrow, nutcracker), third class (effort between fulcrum and load — tongs, fishing rod). MA $= $ Load/Effort $=$ effort arm/load arm. Pulley systems: MA $=$ number of rope segments supporting load. Hydraulic press: MA $= A_2/A_1$ (area ratio). Inclined plane: MA $= 1/\sin\theta$ (ignoring friction). Screw jack: MA $= 2\pi r/p$ ($r$ = handle radius, $p$ = pitch).

8. Bending Moment and Beam Theory

For a horizontal beam supported at two ends with load $W$ at centre: Reaction at each support $= W/2$. Maximum bending moment at centre $= WL/4$. Bending moment $M = EI/R$ where $E$ = Young modulus, $I$ = second moment of area, $R$ = radius of curvature. I-beam cross section: maximises $I$ for given material → maximum resistance to bending with minimum material. Cantilever (one fixed end): maximum bending moment at fixed end $= WL$. Beams fail at point of maximum bending moment.

Frequently Asked Questions
1. Why is torque taken about the foot of the rod? ⌄
Taking torques about the foot eliminates the unknown friction force and the unknown floor normal from the equation (both pass through the foot, so their moment arms = 0 and they contribute zero torque). This gives one equation with only one unknown ($N_{wall}$), which can be solved directly. If we had taken torques about any other point, we would have 2+ unknowns in the torque equation and would need to solve simultaneous equations. Choosing the pivot wisely is the key skill in equilibrium problems.
2. How does the angle affect the friction required? ⌄
Friction $f = N_{wall} = mg/(2\tan\alpha)$ where $\alpha$ = angle with floor. For $\alpha = 30°$ (like this problem, rod at 60° to wall): $f = mg/(2\tan30°) = mg/(2/\sqrt{3}) = mg\sqrt{3}/2 = 200\sqrt{3}/2 = 100\sqrt{3}$ N. ✓ As $\alpha$ decreases (more horizontal rod): $\tan\alpha$ decreases → friction increases. At $\alpha \to 0$: $f \to \infty$ (impossible — rod would slip). This is why nearly-horizontal ladders slip.
3. What would happen if the wall were also rough? ⌄
If wall has friction $f_w$ (upward, if rod tends to slide down): (1) $\sum F_x = 0$: $f_{floor} = N_{wall}$. (2) $\sum F_y = 0$: $N_{floor} + f_{wall} = mg$. (3) Torque equation: $N_{wall}\cdot L\sin\alpha - f_{wall}\cdot L\cos\alpha = mg(L/2)\cos\alpha$. Now three equations but four unknowns ($N_w$, $f_w$, $N_f$, $f_f$) — statically indeterminate! Need additional constraint (like slipping condition for one surface) to solve.
4. What is the minimum angle for the rod not to slip? ⌄
For rod not to slip: $f \leq \mu_s N_{floor}$. $mg/(2\tan\alpha) \leq \mu_s \times mg$. $1/(2\tan\alpha) \leq \mu_s$. $\tan\alpha \geq 1/(2\mu_s)$. $\alpha_{min} = \tan^{-1}(1/(2\mu_s))$. For $\mu_s = 0.5$: $\alpha_{min} = \tan^{-1}(1) = 45°$. For $\mu_s = 1/\sqrt{3} \approx 0.577$: $\alpha_{min} = 30°$. For $\mu_s = \sqrt{3}/2$: $\alpha_{min} = \tan^{-1}(1/\sqrt{3}) = 30°$.
5. What is the reaction from the floor in this problem? ⌄
Vertical equilibrium: $N_{floor} = W = mg = 20\times10 = 200$ N. The floor reaction is purely vertical (providing the normal force balancing weight). Horizontal equilibrium: friction $= N_{wall} = 100\sqrt{3}$ N. Total floor reaction (resultant of normal and friction): $R_{floor} = \sqrt{N_{floor}^2 + f^2} = \sqrt{200^2 + (100\sqrt{3})^2} = \sqrt{40000 + 30000} = \sqrt{70000} = 100\sqrt{7} \approx 265$ N. Direction: $\tan^{-1}(f/N) = \tan^{-1}(100\sqrt{3}/200) = \tan^{-1}(\sqrt{3}/2) \approx 40.9°$ from vertical.
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