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Torque and Equilibrium
Torque: The Turning Effect
Torque (τ), or the moment of a force, is the measure of the force's tendency to produce rotation about a pivot point.
$$\tau = r F \sin(\theta)$$
Torque is determined by force magnitude, distance from pivot, and the angle of application.
$\tau$=Torque magnitude(Nm)
$r$=Position vector (distance from pivot)(m)
$F$=Applied force(N)
$\theta$=Angle between force and position vector(degrees)
$\theta = 90^\circ$
→$\tau = rF$ (Maximum torque)
$\theta = 0^\circ$
→$\tau = 0$ (Force passes through pivot)
Moment Arm: The perpendicular distance ($l = r \sin\theta$) from the pivot to the line of action of the force.
Vector Form: Torque as a cross product: $\vec{\tau} = \vec{r} \times \vec{F}$. Direction given by the right-hand rule, perpendicular to the plane of $\vec{r}$ and $\vec{F}$.
Proportionality: $\tau \propto r$ and $\tau \propto \sin\theta$ — doubling the lever arm doubles the torque; maximum torque occurs at $\theta = 90°$.
Dimensional Check: $[\tau] = [r][F][\sin\theta] = \text{m} \times \text{N} = \text{Nm}$. Torque has the same dimensions as energy (J), but they are physically distinct — torque is never expressed in joules.
Two Equivalent Views: Either decompose $\vec{F}$ into $F\sin\theta$ perpendicular to $\vec{r}$, or decompose $\vec{r}$ into $r\sin\theta$ perpendicular to $\vec{F}$. Both give $\tau = rF\sin\theta$.
Conditions of Equilibrium
Total static equilibrium requires that a body has neither linear nor angular acceleration, satisfying two distinct mathematical conditions.
$$\sum \vec{F} = 0, \quad \sum \vec{\tau} = 0$$
The sum of all external forces and all external torques must be zero.
$\sum \vec{F}$=Net external force(N)
$\sum \vec{\tau}$=Net external torque(Nm)
$\sum F = 0$ only
→Translational equilibrium; object may still rotate.
$\sum \tau = 0$ only
→Rotational equilibrium; object may still translate.
Principle of Moments: For a body in rotational equilibrium, the sum of clockwise torques equals the sum of counter-clockwise torques.
Axis Independence: If an object is in equilibrium, the net torque is zero about any arbitrary axis chosen for calculation.
Static vs Dynamic: A body at rest is in static equilibrium; a body moving with constant velocity is in dynamic equilibrium. Both satisfy the same conditions.
Choosing a smart pivot point simplifies torque equations by eliminating unknown forces whose lines of action pass through that pivot.
Strategy: Place the pivot where the most unknown forces act — their torques become zero, reducing the number of unknowns.
Coplanar Forces: For 2D problems, resolve forces into x and y components. Apply $\sum F_x = 0$, $\sum F_y = 0$, and $\sum \tau = 0$.
Sign Convention: Counter-clockwise torques are positive, clockwise torques are negative. Choose one convention and stick to it.
Negative Reactions: If solving yields a negative force, the assumed direction was wrong — the force actually acts opposite to your assumption. Don't discard it; flip the direction.
Center of Gravity and Stability
The Center of Gravity (CG) is the point where the entire weight of a body appears to act, determining its stability limit.
$$x_{cg} = \frac{\sum w_i x_i}{\sum w_i}$$
The CG position is the weighted average of the positions of all constituent masses.
$x_{cg}$=Position of CG(m)
$w_i$=Weight of individual component $i$(N)
$x_i$=Position of component $i$(m)
Stable Equilibrium: A small displacement raises the CG, creating a restoring torque that returns the object to its original position.
Unstable Equilibrium: A small displacement lowers the CG, and the object topples further away from its original position (e.g., a ball on top of a hill).
Neutral Equilibrium: Displacement neither raises nor lowers the CG, so the object stays in its new position (e.g., a ball on a flat surface).
Uniform Bodies: For a uniform object, the CG is at the geometric center (e.g., midpoint of a uniform beam, center of a uniform disc).
Base of Support: An object remains stable as long as the vertical line from its CG falls within its base.
Couples and Pure Rotation
A couple consists of two equal, opposite, and parallel forces whose lines of action do not coincide, producing pure rotation.
$$\tau_{couple} = F \times d$$
The torque of a couple is the product of one force and the perpendicular distance between them.
$F$=Magnitude of one of the forces(N)
$d$=Arm of the couple (perpendicular distance)(m)
Zero Net Force: A couple always results in $\sum F = 0$, meaning it cannot produce translational motion.
Constant Torque: Unlike a single force, the torque of a couple is the same about any point in the plane.
Proportionality: $\tau_{couple} \propto d$ — widening the separation between forces linearly increases the torque while keeping net force zero.
Application Strategy
Worked Examples
Problem Statement
Calculate the torque produced by a 15N force applied perpendicular to a lever 0.5m from the pivot.
Definition of TorquePerpendicular Force
Strategic Toolkit
$\tau = rF \sin\theta$
Step 1
Identify Variables
We are given a force $F = 15$ N and a distance (lever arm) $r = 0.5$ m.
$F = 15$ N, $r = 0.5$ m
Step 2
Determine Angle
The force is applied perpendicularly, which means the angle $\theta = 90^\circ$. Note that $\sin(90^\circ) = 1$.
$\sin(90^\circ) = 1$
Step 3
Calculate Torque
Substitute values into the general torque formula.
$\tau = 0.5 \times 15 \times 1 = 7.5$ Nm
Problem Statement
A 100g uniform meter rule is balanced at the 40cm mark when a mass $m$ is placed at the 10cm mark. Find $m$.
Principle of MomentsCenter of Gravity
Strategic Toolkit
$\sum \tau_{cw} = \sum \tau_{ccw}$
Step 1
Locate Weights
A uniform meter rule has its CG at the 50cm mark. Its weight is $0.1 \times g$. The unknown mass $m$ is at 10cm.
$W_{rule} = 0.1g$ at 50cm, $W_m = mg$ at 10cm
Step 2
Calculate Distances from Pivot
The pivot is at 40cm. Calculate distances for both weights.
$d_{rule} = 50 - 40 = 10$ cm, $d_m = 40 - 10 = 30$ cm
Step 3
Apply Equilibrium Condition
Clockwise torque (rule's CG) must balance counter-clockwise torque (mass $m$).
$mg \times 30 = 0.1g \times 10$
Step 4
Solve for m
Cancel $g$ and solve the linear equation.
$m = \frac{0.1 \times 10}{30} = 0.0333$ kg = 33.3 g
Problem Statement
A boy (300 N) stands at the edge of a 4.0 m uniform board (200 N). The board rests on two pedestals: one at the left end (A) and one 1.0 m from A (D). Find the reaction forces.
Both conditions of equilibriumSmart pivot choice
Strategic Toolkit
$\sum F_y = 0, \quad \sum \tau = 0$
Step 1
Draw Force Diagram
Identify forces: $R_1$ (up at A), $R_2$ (up at D), board weight 200 N at center (2.0 m), boy 300 N at far end (4.0 m).
Forces: $R_1$, $R_2$, $200$ N, $300$ N
Step 2
Choose Pivot at D
Taking torques about D eliminates $R_2$ from the equation. Distances: A is 1.0 m left, board CG is 1.0 m right, boy is 3.0 m right.
Distances from D: $d_A = 1$ m, $d_{cg} = 1$ m, $d_{boy} = 3$ m
Step 3
Apply Torque Equation
Set net torque about D to zero. CW torques (boy + board CG) equal CCW torques ($R_1$).
$R_1(1) = -300(3) - 200(1)$, so $R_1 = -1100$ N
Step 4
Find R₂ from Force Balance
The negative sign means $R_1$ actually pushes downward. Use $\sum F_y = 0$ to find $R_2$.
$R_2 = 300 + 200 + 1100 = 1600$ N
Problem Statement
A force $\vec{F} = 2\hat{i} - 3\hat{j} + 4\hat{k}$ N acts at point P ($\hat{i} - 2\hat{j} + \hat{k}$ m). Find the torque about point A ($2\hat{i} + \hat{j} + \hat{k}$ m).
Vector cross productTorque about an arbitrary point
Strategic Toolkit
$\vec{\tau} = \vec{r} \times \vec{F}$
Step 1
Find Position Vector
The torque is about A, so we need $\vec{r} = \vec{AP} = \vec{r}_P - \vec{r}_A$.
$\vec{r} = (\hat{i} - 2\hat{j} + \hat{k}) - (2\hat{i} + \hat{j} + \hat{k}) = -\hat{i} - 3\hat{j}$
Step 2
Set Up Cross Product
Compute $\vec{\tau} = \vec{r} \times \vec{F}$ using the determinant method with $\vec{r} = (-1, -3, 0)$ and $\vec{F} = (2, -3, 4)$.
$\vec{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & -3 & 0 \\ 2 & -3 & 4 \end{vmatrix}$
Step 3
Expand Determinant
Expand along the first row: $\hat{i}[(-3)(4) - (0)(-3)] - \hat{j}[(-1)(4) - (0)(2)] + \hat{k}[(-1)(-3) - (-3)(2)]$.
$\vec{\tau} = \hat{i}(-12) - \hat{j}(-4) + \hat{k}(3 + 6) = -12\hat{i} + 4\hat{j} + 9\hat{k}$ Nm
Problem Statement
A driver applies equal and opposite 25 N forces to opposite sides of a steering wheel of diameter 40 cm. Calculate the torque of the couple.
CouplePure rotation
Strategic Toolkit
$\tau = F \times d$
Step 1
Identify Parameters
Force $F = 25$ N, diameter = 40 cm, so the perpendicular distance between the forces $d = 0.40$ m.
$F = 25$ N, $d = 0.40$ m
Step 2
Apply Couple Formula
The torque of a couple equals one force times the full perpendicular separation.
$\tau = 25 \times 0.40 = 10$ Nm
Step 3
Key Insight
This torque is the same regardless of which point you choose as the pivot — a defining property of couples. The net force is zero ($25 - 25 = 0$), so no translation occurs.
Memorize formulas
General Torque Formula
Force × Distance × sine of the angle.
Formula
$\tau = r F \sin(\theta)$
Rotational Equilibrium
Sum of all turning effects must cancel out.
Formula
$\sum \tau_{cw} = \sum \tau_{ccw}$
Torque of a Couple
One force magnitude times the total perpendicular separation.
Formula
$\tau = F \times d_{\perp}$
Center of Gravity Position
The 'average' location of weight in the system.
Formula
$x_{cg} = \frac{\sum w_i x_i}{\sum w_i}$
Torque Vector (Cross Product)
Torque as a vector cross product of position and force.
Formula
$\vec{\tau} = \vec{r} \times \vec{F}$
Review
1
A uniform beam of length 4m and mass 10kg is supported at its center. A 5kg mass is placed 1m from the left end. Where must a 2kg mass be placed to maintain equilibrium?
Ans
It is impossible to balance with only 2kg
Review
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Balance torques about the center
The beam's own weight acts at the center, so it produces no torque about the support. Only the 5kg mass and the 2kg mass need to be matched.
$$5 \times 1 = 2 \times d \Rightarrow d = 2.5\text{ m}$$
Check whether that position exists on the beam
The right side of the beam is only 2m long from the center, so the required 2.5m placement lies beyond the end of the beam.
No valid placement on the beam, so equilibrium is impossible.
2
A system of forces satisfies $\sum F = 0$ but $\sum \tau \neq 0$. The forces must constitute:
Ans
A couple (equal, opposite, non-coincident parallel forces)
Review
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Use the defining property of a couple
A couple consists of two equal and opposite parallel forces separated by a distance. Their net force is zero, but they still produce a turning moment.
$$\sum F = 0, \qquad \sum \tau = Fd \neq 0$$
3
A force of 50N is applied to a door at an angle of 30° to the door's surface, at a distance of 0.8m from the hinges. What is the magnitude of the torque produced?
Ans
20 Nm
Review
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Apply the torque formula
The torque magnitude is the product of the lever arm, the force, and the sine of the angle between them. Here that angle is 30°.
$$\tau = rF\sin\theta = 0.8 \times 50 \times \sin 30^\circ = 20\text{ N m}$$
4
A couple produces a torque of 24 Nm using two forces separated by 0.6m. If the separation is increased to 1.2m while keeping the same torque, the magnitude of each force becomes:
Ans
20 N
Review
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Use the couple relation
For a couple, torque equals one force times the separation between the forces. Keeping torque fixed means force and separation vary inversely.
$$F = \frac{\tau}{d} = \frac{24}{1.2} = 20\text{ N}$$
5
A uniform ladder of weight 300N and length 4m leans against a smooth (frictionless) wall at 45° to the horizontal. A 500N firefighter stands 3m from the base along the ladder. The friction force exerted by the ground on the ladder is:
Ans
525 N
Review
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Take moments about the base
Choosing the base removes the ground reaction from the torque equation. The beam weight acts at its midpoint and the firefighter acts 3m along the ladder.
$$N_w(4) = 300(2) + 500(3)$$
Solve for the wall reaction and friction
The torque balance gives the horizontal wall force, and a smooth wall means the only horizontal ground force must match it in magnitude.
$$N_w = \frac{600 + 1500}{4} = 525\text{ N}, \qquad f = 525\text{ N}$$
6
A 60kg diver stands at the edge of a 3m long uniform diving board (mass 40kg) supported at the left end and 1m from the left end. Calculate the upward force exerted by the support at the 1m mark.
Ans
2400 N
Review
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Balance moments about the left end
The left support is the pivot, so only the board's weight and the diver's weight create clockwise torque, while the support at 1m creates counterclockwise torque.
$$R(1) = (40 \cdot 10 \cdot 1.5) + (60 \cdot 10 \cdot 3)$$
Evaluate the support force
Substituting the numbers gives the upward reaction needed at the 1m support.
$$R = 600 + 1800 = 2400\text{ N}$$
7
The torque of a couple about any point in its plane is:
Ans
The same constant value
Review
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Recall the definition of a couple
A couple produces pure rotation with zero net force, so shifting the reference point does not change the resulting moment.
The torque of a couple is the same constant value about any point.
8
A wrench of length $L$ requires a perpendicular force $F$ to loosen a bolt. If the force is instead applied at 30° to the wrench, the required force magnitude becomes:
Ans
$2F$
Review
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Match the required torque
The bolt needs the same torque in both cases, but the effective perpendicular component of the force is smaller at 30°.
$$FL = F' L \sin 30^\circ = \frac{1}{2}F' L$$
Solve for the new force
Cancelling the common lever arm gives the new force as twice the original perpendicular force.
$$F' = 2F$$
9
A uniform meter rule is balanced on a knife-edge at the 30 cm mark when a 150g mass is hung from the 5 cm mark. What is the mass of the ruler?
Ans
187.5 g
Review
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Take moments about the knife-edge
The pivot is at 30 cm, so the 150g mass acts 25 cm to the left and the ruler's weight acts at its center of gravity, 20 cm to the right.
$$0.150 \times 25 = M \times 20$$
Solve for the ruler mass
Rearranging the moment balance gives the ruler's mass in kilograms, then converting to grams gives the final result.
$$M = \frac{3.75}{20} = 0.1875\text{ kg} = 187.5\text{ g}$$
10
If the net torque on a body is zero but the net force is not zero, the body will:
Ans
Translate without rotating
Review
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Separate rotation from translation
Zero net torque prevents rotation, but a nonzero net force still accelerates the center of mass linearly.
The body translates without rotating.
11
A horizontal beam of length 5m and weight 800N is supported by a hinge at the wall and a cable attached to the free end at 45° to the beam. What is the tension in the cable?
Ans
566 N
Review
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Use torque balance about the hinge
The hinge forces create no torque about the hinge, so only the beam's weight and the cable tension appear in the moment equation.
$$T\sin 45^\circ \cdot 5 = 800 \cdot 2.5$$
Solve the tension
Substitute $\sin 45^\circ = \frac{\sqrt{2}}{2}$ and isolate $T$.
$$T = \frac{2000}{5\sin 45^\circ} \approx 566\text{ N}$$
12
A body is in unstable equilibrium. When slightly displaced, its center of gravity:
Ans
Falls, increasing the displacement further
Review
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Identify the unstable behavior
When the body is nudged, its center of gravity moves to a lower position, so the gravitational torque pushes it farther away from the original equilibrium.
The center of gravity falls, increasing the displacement further.
13
The line of action of force $\vec{F} = \hat{i} - 2\hat{j}$ passes through the point with position vector $-\hat{j} + \hat{k}$. The moment of $\vec{F}$ about the origin is:
Ans
$2\hat{i} + \hat{j} + \hat{k}$
Review
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Form the position and force vectors
The point on the line of action gives $\vec r = -\hat{j} + \hat{k}$ and the force is $\vec F = \hat{i} - 2\hat{j}$.
$$\vec r = (0,-1,1), \qquad \vec F = (1,-2,0)$$
Compute the cross product
Take $\vec r \times \vec F$ component by component to get the moment about the origin.
$$\vec \tau = 2\hat{i} + \hat{j} + \hat{k}\ \text{N m}$$
14
When solving for reaction forces on a beam in equilibrium, a calculation gives $R_1 = -500$ N (assumed upward). This means:
Ans
$R_1$ is actually 500 N directed downward
Review
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Interpret the sign of the reaction
A force written as negative relative to the assumed upward direction simply means the actual force points the other way.
$$R_1 = -500\text{ N} \Rightarrow 500\text{ N downward}$$
15
A horizontal beam is hinged at a wall and supported by a cable. Three unknown forces act on the beam: hinge horizontal reaction $H_x$, hinge vertical reaction $H_y$, and cable tension $T$. To find $T$ most efficiently, you should take torques about:
Ans
The hinge point (eliminates $H_x$ and $H_y$)
Review
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Choose the hinge as the pivot
Any force passing through the hinge has zero moment arm, so both hinge reaction components disappear from the torque equation.
Only the cable tension remains as the unknown in the moment balance.
Question 1 of 15
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1
A uniform beam of length 4m and mass 10kg is supported at its center. A 5kg mass is placed 1m from the left end. Where must a 2kg mass be placed to maintain equilibrium?