Concept Card: Equilibrium of Rigid Bodies & Couple
1. The Physical Framework of a Rigid Body:
A rigid body is an idealized mechanical system of particles in which the mutual distance between any pair of particles remains invariant under the application of arbitrary external forces. Consequently, a rigid body does not deform, and its general motion in three-dimensional space can be resolved into a combination of:
- Translation of its centre of mass ($CM$), possessing 3 translational degrees of freedom.
- Rotation about an instantaneous axis passing through the $CM$, possessing 3 rotational degrees of freedom.
In planar (two-dimensional) mechanics, a rigid body possesses strictly 3 degrees of freedom: two independent translational coordinates in the $xy$-plane and one rotational coordinate about the perpendicular $z$-axis.
2. The Necessary and Sufficient Conditions for Mechanical Equilibrium
For an extended rigid body to remain in complete mechanical equilibrium, two independent conditions must be simultaneously satisfied:
Condition 1: Translational Equilibrium (First Condition of Equilibrium)
The vector sum of all external forces acting on the rigid body must be identically zero:
$$\sum \vec{F}_{\text{ext}} = \vec{0}$$
In Cartesian components:
$$\sum F_x = 0, \quad \sum F_y = 0, \quad \sum F_z = 0$$
Physical Consequence: By Newton’s second law for a system of particles, $\vec{F}_{\text{ext}} = M \vec{a}_{cm} = \vec{0}$. Hence, the linear acceleration of the centre of mass is zero ($\vec{a}_{cm} = \vec{0}$), ensuring either static rest ($\vec{v}_{cm} = \vec{0}$) or uniform rectilinear translation ($\vec{v}_{cm} = \text{constant}$).
Condition 2: Rotational Equilibrium (Second Condition of Equilibrium)
The vector sum of the external torques of all forces acting on the rigid body about any reference point must be identically zero:
$$\sum \vec{\tau}_{\text{ext}} = \vec{0}$$
In Cartesian components:
$$\sum \tau_x = 0, \quad \sum \tau_y = 0, \quad \sum \tau_z = 0$$
Physical Consequence: By the rotational analogue of Newton’s second law, $\sum \vec{\tau}_{cm} = \frac{d\vec{L}_{cm}}{dt} = I_{cm} \vec{\alpha} = \vec{0}$. Hence, the angular acceleration of the body about its centre of mass is zero ($\vec{\alpha} = \vec{0}$), ensuring either no rotation ($\vec{\omega} = \vec{0}$) or uniform rotation with constant angular velocity ($\vec{\omega} = \text{constant}$).
3. Origin-Independence Theorem of Torque in Translational Equilibrium
A fundamental theorem in rigid body statics governs the choice of pivot point for calculating torques:
Theorem: If a system of forces satisfies translational equilibrium ($\sum \vec{F}_i = \vec{0}$), then the net torque of the forces is completely independent of the choice of reference point.
Proof:
Let the net torque about an origin $O$ be:
$$\vec{\tau}_O = \sum_{i} \vec{r}_i \times \vec{F}_i$$
Now consider an alternative arbitrary origin $O’$ with position vector $\vec{r}_{O’}$ relative to $O$. The position vector of the point of application of force $\vec{F}_i$ with respect to $O’$ is $\vec{r}_i’ = \vec{r}_i – \vec{r}_{O’}$. The net torque about $O’$ is:
$$\vec{\tau}_{O’} = \sum_{i} \vec{r}_i’ \times \vec{F}_i = \sum_{i} (\vec{r}_i – \vec{r}_{O’}) \times \vec{F}_i = \sum_{i} \vec{r}_i \times \vec{F}_i – \vec{r}_{O’} \times \left( \sum_{i} \vec{F}_i \right)$$
Since $\sum \vec{F}_i = \vec{0}$ (translational equilibrium):
$$\vec{\tau}_{O’} = \vec{\tau}_O – \vec{r}_{O’} \times \vec{0} = \mathbf{\vec{\tau}_O} \quad (\text{Q.E.D.})$$
Strategic Problem-Solving Rule: In equilibrium problems, always choose the reference pivot point at the intersection of unknown forces (such as hinge reactions, normal forces, or friction). Since the moment arm of these forces becomes zero, they are eliminated from the torque equation, drastically simplifying the algebraic solution.
4. Theory of a Couple and Moment of a Couple
A couple is defined as a pair of equal and opposite parallel forces whose lines of action do not coincide.
- Resultant Force of a Couple:
$$\vec{F}_{\text{net}} = \vec{F} + (-\vec{F}) = \mathbf{\vec{0}}$$
A couple produces zero linear acceleration ($\vec{a}_{cm} = \vec{0}$). It cannot translate a body. - Moment (Torque) of a Couple:
Let force $\vec{F}_1 = \vec{F}$ act at position $\vec{r}_1$ and force $\vec{F}_2 = -\vec{F}$ act at position $\vec{r}_2$ relative to origin $O$. The net torque is:
$$\vec{\tau}_{\text{couple}} = \vec{r}_1 \times \vec{F} + \vec{r}_2 \times (-\vec{F}) = (\vec{r}_1 – \vec{r}_2) \times \vec{F}$$
Let $\vec{r}_{12} = \vec{r}_1 – \vec{r}_2$ be the relative position vector between the two application points. The magnitude of the torque is:
$$\tau = |\vec{r}_{12}| |\vec{F}| \sin\theta = \mathbf{F \cdot d}$$
where $d = |\vec{r}_{12}| \sin\theta$ is the perpendicular distance between the lines of action of the two forces, termed the arm of the couple. - Free Vector Property: Notice that the origin $\vec{r}_O$ does not appear anywhere in the expression $\vec{\tau} = (\vec{r}_1 – \vec{r}_2) \times \vec{F}$. Therefore, the torque of a couple is a free vector: its magnitude, direction, and sense are identical about every point in the universe.
- Irreducibility: A couple cannot be balanced by any single force. It can only be equilibrated by another couple of equal magnitude and opposite rotational sense.
5. The Principle of Moments & Center of Gravity
- The Principle of Moments (Lever Law): For a body in rotational equilibrium under parallel forces:
$$\sum \tau_{\text{anticlockwise}} = \sum \tau_{\text{clockwise}}$$
For a simple lever pivoted at a fulcrum:
$$\text{Load} \times \text{Load Arm} = \text{Effort} \times \text{Effort Arm} \implies \text{Mechanical Advantage (MA)} = \frac{\text{Load}}{\text{Effort}} = \frac{\text{Effort Arm}}{\text{Load Arm}}$$ - Center of Gravity ($CG$) vs. Center of Mass ($CM$):
The Center of Gravity is the point about which the total gravitational torque on the body is zero:
$$\sum (\vec{r}_i – \vec{r}_{cg}) \times m_i \vec{g} = \vec{0}$$
If the gravitational field $\vec{g}$ is uniform across the entire extent of the body, $\vec{g}$ factors out of the sum, and $\vec{r}_{cg} \equiv \vec{r}_{cm}$. However, for extremely large bodies (e.g., a mountain or tall skyscraper in a divergent planetary gravitational field), $\vec{g}$ decreases with height, placing the $CG$ strictly below the $CM$.
6. Common JEE Traps & Advanced Pitfalls
- Trap 1 (Three Coplanar Forces in Equilibrium): If three non-parallel coplanar forces hold a rigid body in equilibrium, their lines of action must intersect at a single concurrent point. If they did not intersect at one point, the torque about the intersection of any two forces would be non-zero due to the third force, violating rotational equilibrium.
- Trap 2 (Partial Equilibrium Misconception): A body can be in translational equilibrium without being in rotational equilibrium (e.g., when subjected to a couple), or in rotational equilibrium without translational equilibrium (e.g., a rotating wheel with an accelerating axle). Equilibrium requires BOTH conditions.
- Trap 3 (Toppling vs. Sliding): When a horizontal force is applied to a block on a rough surface, the normal reaction shifts away from the centre toward the tipping edge to counteract the overturning torque. Toppling occurs precisely when the normal reaction reaches the extreme edge ($x = b/2$) and can shift no further.
Solved Examples
Example 1 (Direct Conceptual Application – Support Reactions on a Non-Uniformly Loaded Beam):
A uniform horizontal steel rod $AB$ of mass $M = 2.0\text{ kg}$ and length $L = 1.0\text{ m}$ is supported horizontally on two sharp knife-edge supports. Support 1 is located at distance $d_1 = 0.10\text{ m}$ from end $A$, and Support 2 is located at distance $d_2 = 0.20\text{ m}$ from end $B$. A concentrated load of mass $m = 3.0\text{ kg}$ is suspended from the rod at a distance $x = 0.30\text{ m}$ from end $A$. Taking $g = 9.8\text{ m/s}^2$:
(a) Set up the equations for translational and rotational equilibrium.
(b) Calculate the normal reaction forces $N_1$ and $N_2$ exerted by Support 1 and Support 2 respectively.
Solution:
Let end $A$ be the coordinate origin ($x = 0$).
– Position of Support 1: $x_1 = 0.10\text{ m}$, upward normal reaction $N_1$.
– Position of Support 2: $x_2 = L – d_2 = 1.0 – 0.20 = 0.80\text{ m}$, upward normal reaction $N_2$.
– Position of suspended load: $x_m = 0.30\text{ m}$, downward force $W_m = m g = (3.0)(9.8) = 29.4\text{ N}$.
– Position of rod’s centre of mass: $x_c = L/2 = 0.50\text{ m}$, downward weight $W_{\text{rod}} = M g = (2.0)(9.8) = 19.6\text{ N}$.
(a) Translational Equilibrium ($\sum F_y = 0$):
$$N_1 + N_2 – W_m – W_{\text{rod}} = 0$$
$$N_1 + N_2 = (3.0 + 2.0)(9.8) = (5.0)(9.8) = \mathbf{49.0\text{ N}} \quad \text{— (Equation 1)}$$
(b) Rotational Equilibrium ($\sum \tau = 0$ about Support 1 at $x_1 = 0.10\text{ m}$):
Taking torque about $x_1 = 0.10\text{ m}$ completely eliminates $N_1$ from the equation:
$$\sum \tau_{x_1} = 0$$
Counterclockwise torque from $N_2$ must balance clockwise torques from the weights:
$$N_2 \cdot (x_2 – x_1) – W_m \cdot (x_m – x_1) – W_{\text{rod}} \cdot (x_c – x_1) = 0$$
Substitute known distances:
$$x_2 – x_1 = 0.80 – 0.10 = 0.70\text{ m}$$
$$x_m – x_1 = 0.30 – 0.10 = 0.20\text{ m}$$
$$x_c – x_1 = 0.50 – 0.10 = 0.40\text{ m}$$
$$N_2 (0.70) = (29.4)(0.20) + (19.6)(0.40)$$
$$N_2 (0.70) = 5.88 + 7.84 = 13.72\text{ N}\cdot\text{m}$$
$$N_2 = \frac{13.72}{0.70} = \mathbf{19.6\text{ N}}$$
Substituting $N_2 = 19.6\text{ N}$ into Equation 1:
$$N_1 = 49.0 – 19.6 = \mathbf{29.4\text{ N}}$$
Takeaway: Choosing the pivot at one of the knife-edge supports decouples the two equations, solving for the second reaction directly without solving simultaneous algebraic equations.
Example 2 (Mathematical Manipulation – Classical Ladder Statics & Slipping Limits):
A uniform ladder of mass $M = 20.0\text{ kg}$ and length $L = 5.0\text{ m}$ rests with its upper end against a smooth (frictionless) vertical wall and its lower end on a rough horizontal floor. The ladder is inclined at an angle $\theta = 53^\circ$ to the horizontal ($\cos 53^\circ = 0.60, \sin 53^\circ = 0.80$, take $g = 9.8\text{ m/s}^2$).
(a) Determine the normal force from the wall ($N_w$), the normal force from the floor ($N_f$), and the static friction force ($f_s$) at the floor.
(b) Calculate the minimum coefficient of static friction $\mu_{\min}$ required to prevent the ladder from slipping.
(c) If a person of mass $m = 60.0\text{ kg}$ climbs the ladder when the actual coefficient of friction is $\mu_s = 0.50$, find the maximum distance $x$ along the ladder the person can climb before it begins to slip.
Solution:
(a) Forces Acting on the Ladder:
– At top (wall): Horizontal normal reaction $N_w$ pointing away from the wall (smooth wall $\implies f_{\text{wall}} = 0$).
– At bottom (floor): Vertical normal force $N_f$ upward, static friction $f_s$ horizontally toward the wall.
– At $CM$ (distance $L/2$ along ladder): Gravity $W = M g = (20)(9.8) = 196\text{ N}$ vertically downward.
From translational equilibrium:
$$\sum F_y = 0 \implies N_f = M g = (20)(9.8) = \mathbf{196\text{ N}}$$
$$\sum F_x = 0 \implies f_s = N_w$$
From rotational equilibrium about the base contact point on the floor:
$$\sum \tau_{\text{base}} = 0$$
Clockwise torque of gravity balances counterclockwise torque of $N_w$:
$$N_w (L \sin\theta) – M g \left(\frac{L}{2} \cos\theta\right) = 0$$
$$N_w = \frac{M g \cos\theta}{2 \sin\theta} = \frac{1}{2} M g \cot\theta$$
Substitute numerical values:
$$N_w = \frac{1}{2} (196) \left(\frac{0.60}{0.80}\right) = (98)(0.75) = \mathbf{73.5\text{ N}}$$
$$f_s = N_w = \mathbf{73.5\text{ N}}$$
(b) Minimum Coefficient of Friction:
To prevent slipping, static friction must not exceed its maximum available limit:
$$f_s \le \mu_s N_f \implies \mu_{\min} = \frac{f_s}{N_f} = \frac{\frac{1}{2} M g \cot\theta}{M g} = \frac{1}{2} \cot\theta$$
$$\mu_{\min} = \frac{1}{2} \left(\frac{0.60}{0.80}\right) = \frac{0.75}{2} = \mathbf{0.375}$$
(c) Person Climbing the Ladder:
With the person of mass $m = 60\text{ kg}$ at distance $x$ from the base:
$$\sum F_y = 0 \implies N_f’ = (M + m) g = (20 + 60)(9.8) = 80 \times 9.8 = 784\text{ N}$$
The maximum static friction available at the floor is:
$$f_{s,\max} = \mu_s N_f’ = (0.50)(784\text{ N}) = 392\text{ N}$$
At the verge of slipping, $N_w’ = f_{s,\max} = 392\text{ N}$.
Taking torque about the base contact point:
$$N_w’ (L \sin\theta) = M g \left(\frac{L}{2} \cos\theta\right) + m g (x \cos\theta)$$
Divide both sides by $\cos\theta$:
$$N_w’ L \tan\theta = M g \frac{L}{2} + m g x$$
Substitute numerical values ($L = 5.0\text{ m}$, $\tan 53^\circ = \frac{0.80}{0.60} = \frac{4}{3}$):
$$(392)(5.0) \left(\frac{4}{3}\right) = (196)(2.5) + (60 \times 9.8) x$$
$$\frac{7840}{3} = 490 + 588 x$$
$$2613.33 = 490 + 588 x \implies 588 x = 2123.33 \implies x = \frac{2123.33}{588} \approx \mathbf{3.61\text{ m}}$$
Takeaway: As a person ascends the ladder, the overturning torque about the base increases, demanding a progressively higher normal reaction from the wall and thus larger frictional resistance at the floor.
Example 3 (Multi-Concept Linkage – Advanced JEE Scenario: Toppling Versus Sliding of a Block):
A uniform solid block of mass $M$, rectangular cross-section of width $b$ and height $h$ rests on a rough horizontal surface with coefficient of static friction $\mu$. A horizontal force $F$ is applied to the block at height $y$ above the surface.
(a) Determine the location of the normal reaction force $N$ as a function of $F, M, g, y$.
(b) Derive the critical force $F_{\text{slide}}$ required to cause sliding and the critical force $F_{\text{topple}}$ required to initiate toppling.
(c) Deduce the universal criterion for the block to topple before sliding.
(d) If a uniform cube of side $a = 0.50\text{ m}$ and mass $M = 10\text{ kg}$ is pulled at its top edge ($y = a$) on a floor with $\mu = 0.60$, determine whether it slides or topples first, and find the threshold force.
Solution:
(a) Shift of the Normal Reaction:
In static equilibrium before any motion:
$$\sum F_y = 0 \implies N = M g$$
$$\sum F_x = 0 \implies f_s = F$$
Let the normal force act at a forward horizontal distance $x$ from the vertical centerline passing through the $CM$.
Taking torque about the centre of mass ($CM$, located at height $h/2$):
– The applied force $F$ exerts a clockwise torque: $\tau_F = F \left(y – \frac{h}{2}\right)$.
– The friction force $f_s = F$ at the bottom exerts a clockwise torque: $\tau_f = F \left(\frac{h}{2}\right)$.
– Total clockwise overturning torque:
$$\tau_{\text{overturn}} = F \left(y – \frac{h}{2}\right) + F\left(\frac{h}{2}\right) = F \cdot y$$
– The normal reaction $N = M g$ at distance $x$ forward exerts a restoring counterclockwise torque: $\tau_N = N \cdot x = M g x$.
For rotational equilibrium:
$$M g x = F y \implies \mathbf{x = \frac{F y}{M g}}$$
(b) Critical Forces:
1. Sliding Condition: Sliding begins when $F$ reaches the maximum static friction:
$$\mathbf{F_{\text{slide}} = \mu N = \mu M g}$$
2. Toppling Condition: The restoring torque has a physical upper bound because the normal force cannot act beyond the front edge of the base: $x_{\max} = \frac{b}{2}$.
Setting $x = \frac{b}{2}$:
$$\frac{b}{2} = \frac{F y}{M g} \implies \mathbf{F_{\text{topple}} = M g \frac{b}{2y}}$$
(c) Condition for Toppling Before Sliding:
The block will topple before it slides if and only if $F_{\text{topple}} < F_{\text{slide}}$:
$$M g \frac{b}{2y} < \mu M g \implies \mathbf{\mu > \frac{b}{2y}}$$
For a horizontal force applied at the very top edge ($y = h$):
$$\mathbf{\mu > \frac{b}{2h}}$$
(d) Analysis of the Cube ($b = h = a = 0.50\text{ m}$, $y = a$, $\mu = 0.60$):
The critical friction threshold for a cube pulled at its top edge is:
$$\mu_{\text{crit}} = \frac{a}{2a} = 0.50$$
Since $\mu = 0.60 > 0.50$, the cube will topple before sliding!
The threshold force required to initiate toppling is:
$$F_{\text{topple}} = M g \frac{a}{2a} = \frac{M g}{2} = \frac{(10\text{ kg})(9.8\text{ m/s}^2)}{2} = \mathbf{49.0\text{ N}}$$
(Whereas sliding would require $F_{\text{slide}} = \mu M g = 0.60 \times 98 = 58.8\text{ N} > 49.0\text{ N}$).
Takeaway: In toppling problems, the normal force does not stay at the center; it shifts to provide a counter-torque. Equilibrium breaks down when the required point of action leaves the physical boundary of the base.
Example 4 (Edge Case – Hinge Reactions Under External Loads & Applied Couples):
A uniform rigid beam $AB$ of mass $M = 4.0\text{ kg}$ and length $L = 2.0\text{ m}$ is freely hinged to a vertical wall at end $A$. The beam is maintained in a horizontal position by a light cable attached to end $B$, inclined at an angle $\alpha = 30^\circ$ to the beam. In addition, an external counterclockwise couple of moment $\Gamma = 20.0\text{ N}\cdot\text{m}$ is applied to the beam at its midpoint (take $g = 10.0\text{ m/s}^2$).
(a) Determine the tension $T$ in the supporting cable.
(b) Calculate the horizontal and vertical components of the reaction force at hinge $A$.
(c) Calculate the magnitude and direction of the resultant hinge force.
Solution:
(a) Evaluation of Cable Tension $T$:
Forces acting on beam $AB$ ($L = 2.0\text{ m}$):
– Weight $W = M g = (4.0)(10.0) = 40.0\text{ N}$ downward at $x = L/2 = 1.0\text{ m}$.
– Cable tension $T$ at end $B$ ($x = 2.0\text{ m}$):
$$T_x = -T \cos 30^\circ, \quad T_y = T \sin 30^\circ$$
– Hinge reaction at $A$: Horizontal component $R_x$ (to the right) and vertical component $R_y$ (upward).
– Applied couple: $\Gamma = +20.0\text{ N}\cdot\text{m}$ (counterclockwise).
Taking torque about hinge $A$ eliminates $R_x$ and $R_y$:
$$\sum \tau_A = 0$$
$$(T \sin 30^\circ) L + \Gamma – W \left(\frac{L}{2}\right) = 0$$
Note that the couple $\Gamma$ enters the torque equation with its full magnitude regardless of where it is applied!
Substitute values ($L = 2.0\text{ m}$, $\sin 30^\circ = 0.50$):
$$T (0.50)(2.0) + 20.0 – (40.0)(1.0) = 0$$
$$1.0 T + 20.0 – 40.0 = 0 \implies 1.0 T = 20.0 \implies \mathbf{T = 20.0\text{ N}}$$
(b) Components of the Hinge Reaction:
From translational equilibrium:
$$\sum F_x = 0 \implies R_x – T \cos 30^\circ = 0$$
$$R_x = T \cos 30^\circ = 20.0 \times \frac{\sqrt{3}}{2} = 10\sqrt{3} \approx \mathbf{17.32\text{ N}}$$
$$\sum F_y = 0 \implies R_y + T \sin 30^\circ – W = 0$$
$$R_y = W – T \sin 30^\circ = 40.0 – (20.0)(0.50) = 40.0 – 10.0 = \mathbf{30.0\text{ N}}$$
(c) Resultant Hinge Reaction:
Magnitude:
$$R = \sqrt{R_x^2 + R_y^2} = \sqrt{(17.32)^2 + (30.0)^2} = \sqrt{300 + 900} = \sqrt{1200} = 20\sqrt{3} \approx \mathbf{34.64\text{ N}}$$
Direction with horizontal:
$$\theta = \tan^{-1}\left(\frac{R_y}{R_x}\right) = \tan^{-1}\left(\frac{30.0}{10\sqrt{3}}\right) = \tan^{-1}(\sqrt{3}) = \mathbf{60^\circ}$$
Takeaway: A pure couple contributes zero net linear force to $\sum \vec{F} = \vec{0}$, but it directly alters the tension in supporting members and the reaction forces at constraints through the rotational equilibrium condition $\sum \vec{\tau} = \vec{0}$.
Worksheet: 10 Practice Problems
Problem 1 (JEE Main – Single Correct):
A couple acting on a rigid body produces:
(A) Pure translational acceleration without angular acceleration
(B) Pure rotational acceleration without linear acceleration
(C) Both linear and angular acceleration simultaneously
(D) Neither linear nor angular acceleration
Problem 2 (JEE Main – Single Correct):
If a rigid body is in translational equilibrium under the action of several coplanar forces ($\sum \vec{F} = \vec{0}$), which of the following statements about the total torque $\vec{\tau}$ acting on the body is strictly true?
(A) The total torque is identically zero about every point.
(B) The total torque has the same value about any arbitrary reference point.
(C) The total torque is zero only about the centre of mass.
(D) The total torque depends linearly on the coordinates of the chosen origin.
Problem 3 (JEE Main – Single Correct):
A uniform horizontal wooden beam of length $4.0\text{ m}$ and weight $600\text{ N}$ rests on two simple supports placed at its extreme ends. A concentrated downward load of $300\text{ N}$ is placed on the beam at a distance of $1.0\text{ m}$ from the left support. The vertical reaction force at the left support is:
(A) $375\text{ N}$
(B) $450\text{ N}$
(C) $525\text{ N}$
(D) $600\text{ N}$
Problem 4 (JEE Main – Single Correct):
A uniform rod of length $L$ and mass $M$ is pivoted smoothly at its lower end to a horizontal floor. A horizontal force $F$ is applied at its upper free end so that the rod remains in static equilibrium inclined at an angle $\theta$ to the vertical. The magnitude of the force $F$ is:
(A) $M g \tan\theta$
(B) $\frac{1}{2} M g \tan\theta$
(C) $\frac{1}{2} M g \sin\theta$
(D) $M g \cot\theta$
Problem 5 (JEE Main – Single Correct):
A uniform ladder rests in limiting equilibrium with its top end leaning against a smooth vertical wall and its base resting on a rough horizontal floor with static friction coefficient $\mu$. The minimum angle $\theta$ that the ladder can make with the horizontal without slipping satisfies:
(A) $\tan\theta = \frac{1}{2\mu}$
(B) $\tan\theta = \frac{1}{\mu}$
(C) $\tan\theta = 2\mu$
(D) $\tan\theta = \frac{\mu}{2}$
Problem 6 (JEE Advanced – One or More Correct):
Which of the following statements regarding the equilibrium of a rigid body is/are correct?
(A) If $\sum \vec{F} = \vec{0}$, the resultant torque of the forces is independent of the choice of reference origin.
(B) If $\sum \vec{\tau} = \vec{0}$ about one specific point, it is guaranteed to be zero about all points only if $\sum \vec{F} = \vec{0}$.
(C) Three non-parallel coplanar forces acting on a rigid body in equilibrium must be concurrent.
(D) A pure mechanical couple can be completely balanced by a single force of equal magnitude applied through the centre of mass.
Problem 7 (JEE Advanced – One or More Correct):
A uniform rectangular block of mass $M$, width $b$, and height $h$ rests on a rough horizontal surface with coefficient of friction $\mu$. A horizontal force $F$ is slowly increased at the top edge of the block ($y = h$). Which of the following statements is/are correct?
(A) The block slides before toppling if $\mu < \frac{b}{2h}$.
(B) The block topples before sliding if $\mu > \frac{b}{2h}$.
(C) At the verge of toppling, the normal reaction force acts precisely through the front tipping edge.
(D) If $\mu = \frac{b}{2h}$, the block reaches the threshold of sliding and toppling simultaneously.
Problem 8 (JEE Advanced – One or More Correct):
A uniform thin horizontal rod of mass $M$ and length $L$ is supported symmetrically by two identical vertical inextensible strings attached at its ends $A$ and $B$. If the string at end $B$ is suddenly cut at $t = 0$, then at that exact instant:
(A) The tension in the string at $A$ immediately becomes $T = \frac{1}{4} M g$.
(B) The angular acceleration of the rod about hinge point $A$ is $\alpha = \frac{3g}{2L}$.
(C) The downward linear acceleration of the centre of mass is $a_{cm} = \frac{3g}{4}$.
(D) The tension in the string at $A$ remains $\frac{1}{2} M g$ because the rod has non-zero rotational inertia.
Problem 9 (JEE Main / Advanced – Numerical Value Type):
A uniform horizontal steel rod of mass $M = 6.0\text{ kg}$ and length $L = 3.0\text{ m}$ is supported horizontally by two vertical suspension wires placed at distances of $0.50\text{ m}$ and $2.50\text{ m}$ from its left end. A point mass $m = 4.0\text{ kg}$ is attached to the rod at a distance of $1.0\text{ m}$ from the left end. Taking $g = 10.0\text{ m/s}^2$, calculate the tension (in Newtons) in the left suspension wire.
Problem 10 (JEE Main / Advanced – Numerical Value Type):
A uniform solid cube of mass $M = 10.0\text{ kg}$ and side length $a = 0.60\text{ m}$ rests on a rough horizontal plane with coefficient of static friction $\mu = 0.80$. A horizontal pulling force $F$ is applied to the cube at a height $h = 0.50\text{ m}$ above the floor. Taking $g = 10.0\text{ m/s}^2$, determine the maximum force $F$ (in Newtons) that can be applied without causing the cube to topple.
Solutions & Explanations
Answer Key Summary:
1. (B) | 2. (B) | 3. (C) | 4. (B) | 5. (A) | 6. (A, B, C) | 7. (A, B, C, D) | 8. (A, B, C) | 9. 60 | 10. 60
Solution 1:
A couple consists of two forces $\vec{F}$ and $-\vec{F}$. The resultant force is $\vec{F}_{\text{net}} = \vec{F} + (-\vec{F}) = \vec{0}$, so $\vec{a}_{cm} = \vec{0}$ (no linear acceleration). However, the torque $\vec{\tau} = \vec{r}_{12} \times \vec{F} \neq \vec{0}$, which generates angular acceleration $\vec{\alpha} = \vec{\tau}/I \neq \vec{0}$. Thus, a couple produces pure rotation.
Correct Option: (B)
Solution 2:
By the origin-independence theorem of torque:
$$\vec{\tau}_{O’} = \vec{\tau}_O – \vec{r}_{O’} \times \left(\sum \vec{F}\right)$$
When $\sum \vec{F} = \vec{0}$, $\vec{\tau}_{O’} = \vec{\tau}_O$. Thus, the total torque has the same value about any arbitrary reference point.
Correct Option: (B)
Solution 3:
Length of beam $L = 4.0\text{ m}$.
– Downward weight $W = 600\text{ N}$ acts at center ($x = 2.0\text{ m}$).
– Downward load $P = 300\text{ N}$ acts at $x = 1.0\text{ m}$.
Let $R_L$ and $R_R$ be the reaction forces at left ($x = 0$) and right ($x = 4.0\text{ m}$) supports.
Taking torque about the right support ($x = 4.0\text{ m}$):
$$\sum \tau_{\text{right}} = 0$$
$$R_L (4.0) – P (4.0 – 1.0) – W (4.0 – 2.0) = 0$$
$$4.0 R_L = 300(3.0) + 600(2.0) = 900 + 1200 = 2100\text{ N}\cdot\text{m}$$
$$R_L = \frac{2100}{4.0} = \mathbf{525\text{ N}}$$
Correct Option: (C)
Solution 4:
The rod of length $L$ is inclined at $\theta$ to the vertical.
– The gravity force $M g$ acts at the midpoint (distance $L/2$ from pivot), with perpendicular lever arm $\frac{L}{2} \sin\theta$.
– The horizontal force $F$ acts at the top (distance $L$ from pivot), with perpendicular vertical lever arm $L \cos\theta$.
Taking torque about the pivot:
$$\sum \tau_{\text{pivot}} = 0$$
$$F (L \cos\theta) = M g \left(\frac{L}{2} \sin\theta\right)$$
$$F = \frac{1}{2} M g \frac{\sin\theta}{\cos\theta} = \mathbf{\frac{1}{2} M g \tan\theta}$$
Correct Option: (B)
Solution 5:
Let the ladder have length $L$ and mass $M$.
– Translational equilibrium: $N_f = M g$ and $f_s = N_w$.
– Rotational equilibrium about base:
$$N_w (L \sin\theta) = M g \left(\frac{L}{2} \cos\theta\right) \implies N_w = \frac{1}{2} M g \cot\theta$$
– Limiting friction condition:
$$f_s \le \mu N_f \implies \frac{1}{2} M g \cot\theta \le \mu M g \implies \cot\theta \le 2\mu \implies \tan\theta \ge \frac{1}{2\mu}$$
The minimum angle for equilibrium is $\tan\theta = \frac{1}{2\mu}$.
Correct Option: (A)
Solution 6:
– (A) is correct: Proved via $\vec{\tau}_{O’} = \vec{\tau}_O – \vec{r}_{O’} \times \sum \vec{F} = \vec{\tau}_O$.
– (B) is correct: If $\sum \vec{\tau} = 0$ at one point, it is zero at all points if and only if $\sum \vec{F} = 0$.
– (C) is correct: If three non-parallel forces are in equilibrium, the torque about the intersection of any two must be zero, meaning the line of action of the third must pass through that same point (concurrency).
– (D) is incorrect: A couple cannot be balanced by a single force; a single force creates a non-zero $\sum \vec{F} \neq \vec{0}$, violating translational equilibrium.
Correct Options: (A, B, C)
Solution 7:
For a block of width $b$ and height $h$ pulled at the top ($y = h$):
– Maximum static friction: $F_{\text{slide}} = \mu M g$.
– Toppling threshold: $F_{\text{topple}} h = M g \left(\frac{b}{2}\right) \implies F_{\text{topple}} = M g \frac{b}{2h}$.
– If $\mu < \frac{b}{2h}$, $F_{\text{slide}} < F_{\text{topple}}$, so the block slides first (A is correct).
– If $\mu > \frac{b}{2h}$, $F_{\text{topple}} < F_{\text{slide}}$, so the block topples first (B is correct).
– At toppling, the normal force shifts to the extreme front edge $x = b/2$ (C is correct).
– If $\mu = \frac{b}{2h}$, $F_{\text{slide}} = F_{\text{topple}}$ (D is correct).
Correct Options: (A, B, C, D)
Solution 8:
When string $B$ is cut, point $A$ acts as an instantaneous pivot.
Moment of inertia about $A$: $I_A = \frac{1}{3} M L^2$.
Torque of gravity about $A$: $\tau_A = M g \left(\frac{L}{2}\right)$.
– Angular acceleration:
$$\alpha = \frac{\tau_A}{I_A} = \frac{M g L / 2}{M L^2 / 3} = \mathbf{\frac{3g}{2L}} \quad \text{(B is correct)}$$
– Downward acceleration of the $CM$ (at distance $L/2$ from $A$):
$$a_{cm} = \alpha \left(\frac{L}{2}\right) = \left(\frac{3g}{2L}\right) \left(\frac{L}{2}\right) = \mathbf{\frac{3g}{4}} \quad \text{(C is correct)}$$
– Writing Newton’s second law for the vertical motion of the $CM$:
$$M g – T = M a_{cm} = M \left(\frac{3g}{4}\right) \implies T = M g – \frac{3}{4} M g = \mathbf{\frac{1}{4} M g} \quad \text{(A is correct)}$$
Option (D) is incorrect.
Correct Options: (A, B, C)
Solution 9:
Rod mass $M = 6.0\text{ kg}$, $W_{\text{rod}} = 60\text{ N}$ at center ($x = 1.5\text{ m}$).
Left wire at $x_1 = 0.50\text{ m}$ (tension $T_1$).
Right wire at $x_2 = 2.50\text{ m}$ (tension $T_2$).
Load $m = 4.0\text{ kg}$, $W_{\text{load}} = 40\text{ N}$ at $x = 1.0\text{ m}$.
Taking torque about the right wire attachment point ($x_2 = 2.50\text{ m}$):
$$\sum \tau_{x_2} = 0$$
$$T_1 (2.50 – 0.50) = W_{\text{load}} (2.50 – 1.0) + W_{\text{rod}} (2.50 – 1.50)$$
$$T_1 (2.0) = (40)(1.50) + (60)(1.0) = 60 + 60 = 120\text{ N}\cdot\text{m}$$
$$T_1 = \frac{120}{2.0} = \mathbf{60\text{ N}}$$
Correct Answer: 60
Solution 10:
Cube of side $a = 0.60\text{ m}$, mass $M = 10.0\text{ kg}$, weight $W = 100\text{ N}$.
Force $F$ applied at height $h = 0.50\text{ m}$.
At the verge of toppling, the normal force acts through the front bottom corner (distance $a/2 = 0.30\text{ m}$ from center).
Taking torque about the front bottom edge:
$$F \cdot h = M g \left(\frac{a}{2}\right)$$
$$F (0.50) = (10.0)(10.0)(0.30) = 30.0\text{ N}\cdot\text{m}$$
$$F = \frac{30.0}{0.50} = \mathbf{60\text{ N}}$$
Check sliding threshold:
$$F_{\text{slide}} = \mu M g = (0.80)(100) = 80\text{ N}$$
Since $F_{\text{topple}} = 60\text{ N} < F_{\text{slide}} = 80\text{ N}$, the cube will topple when $F$ reaches $60\text{ N}$.
Correct Answer: 60