Torque Explained: Definition, Formula, and Solved Problems

By Pankaj Kumar|Last Updated: July 12, 2026|

Key Takeaways

Torque is the turning effect of a force about a fixed point, measured in N-m.
Anticlockwise torque is positive; clockwise torque is negative by convention.
A couple produces rotation without translation; its moment equals force times perpendicular distance.
A rigid body in mechanical equilibrium has zero net force and zero net torque.
Rotational equilibrium is independent of the origin chosen for calculating torques.

What is Torque?

Definition

Torque is the turning effect of force about a fixed point. It is also called the Moment of force. If you are working through rotational mechanics and need structured support, a economics tutor is not the right fit — but for physics and engineering topics like this one, one-on-one guidance can make a real difference.

Unit of Torque

N-m

Dimension of Torque

$[M{L^2}{T^{ – 2}}]$

Usual Sign Convention for Torque

A torque that produces anticlockwise rotation is taken positive and clockwise rotation is taken negative.

Formula for Torque

The magnitude of the torque is the product of the magnitude of the force and the perpendicular distance of the line of action of the force from the fixed point.

Diagram showing torque of force F about point O with perpendicular distance

Right angle triangle OBP used in torque derivation

Torque of force F about the point O

${Gamma _0} = OP times F$

${Gamma _0} = left| {OP} right|,left| F right|sin theta ,hat n$ …(i)

Now, from right angle triangle OBP,

$sin theta = frac{{OB}}{{OP}}quad Rightarrow quad OB = (OP)sin theta $

$or,quad OB = left| {OP} right|sin theta $

Now, from (i),

${Gamma _0} = (OB)(F)hat n$

${Gamma _0} = {r_1} times F$

Some Quick Facts on Torque

A force of a given magnitude applied at right angles to the door at its outer edge is most effective in producing rotation.

The moment of a force vanishes if either the magnitude of the force is zero or if the line of action of the force passes through the fixed point.

If the direction of force F is reversed, the direction of the moment of force is also reversed. If the direction of both r and F is reversed, the direction of the moment of force remains the same.

Students who find these concepts challenging in broader engineering contexts may also benefit from exploring how torque principles apply in practice — for example, those studying with an aerospace engineering tutor regularly encounter rotational mechanics in structural and propulsion analysis.

Solved Problems on Torque

Ex-1: A simple pendulum of length l is pulled aside to make an angle q with the vertical. Find the magnitude of the torque of the weight W of the bob about the point of suspension.

Ans. From figure, torque of weight W about point O,

Simple pendulum diagram showing torque of weight W about point of suspension

${Gamma _0} = {r_1} times F$

$ = (OPsin theta )(mg)( – hat k)$

$ = (ell ,sin theta )(mg)( – hat k)$

Therefore, the magnitude of torque of weight W about the point of suspension is

$left| {{Gamma _0}} right| = mgell sin theta = Well sin theta $

Ex-2: A cubical block of mass m and edge a slides down a rough inclined plane of inclination q with a uniform speed. Find the torque of the normal force acting on the block about its center.

Ans. Since block is sliding with constant speed $mgsin theta = {f_k}$

Cubical block sliding down a rough inclined plane with forces labeled

Again, cubical block has no any angular acceleration $(alpha = 0)$

$sum {{Gamma _{cm}}} = 0quad Rightarrow quad {Gamma _{N,,cm}} + {Gamma _{{f_k},,cm}} = 0$

$ Rightarrow quad {Gamma _{N,,cm}} = left( { – frac{a}{2}{f_k}sin {{90}^o}hat k} right)$

$ Rightarrow quad left| {{Gamma _{N,,cm}}} right| = frac{a}{2} cdot mgsin theta $

Understanding how forces and torques interact in mechanical systems is also central to aircraft maintenance — students pursuing an aircraft maintenance tutor will recognise these equilibrium principles in real-world inspection and repair contexts. For a broader look at how online learning resources compare when working through problems like these, see this comparison of Khan Academy vs paid homework sites.

Ex-3: If $F = (2N)hat i – (3N)hat j$ and $r = (3m)hat i + (2m)hat j$, then find torque of force F about the origin.

Ans. As we know, ${Gamma _0} = r times F$

$r times F = left| {begin{array}{*{20}{c}} {hat i}&{hat j}&{hat k} 3&2&0 2&{ – 3}&0 end{array}} right| = hat i(0 – 0) – hat j(0 – 0) + hat k( – 9 – 4) = ( – 13,N{rm{ – m}})hat k$

So, torque of force F about the origin $ = (13,N{rm{ – m}})( – hat k)$.

Moment of Couple

A pair of equal and opposite forces with different lines of action is known as a couple. A couple produces rotation without translation. If an object is not on a pivot, a couple causes it to rotate about its center of mass.

Diagram illustrating a couple with two equal and opposite forces separated by distance d

This couple can produce a turning effect or torque on the body. The moment of a couple is a measure of the turning effect (t).

$Gamma $ = moment of couple = Magnitude of either force X perpendicular distance between the forces.

$Gamma = Fd$

The principles of couples and moments also appear in economic modelling of equilibrium — students working with a microeconomics tutor will encounter analogous balancing conditions in market equilibrium analysis. Risk and balance concepts similarly underpin fields like construction, as explored in this guide to risk management in the construction industry.

Mechanical Equilibrium of a Rigid Body

A rigid body is said to be in mechanical equilibrium if its linear and angular momentum do not change with time. Equivalently, the body has neither linear nor angular acceleration.

Condition for Translational Equilibrium: No Linear Acceleration

The vector sum of the forces on a rigid body is zero.

${F_1} + {F_2} + …… + {F_n} = sumlimits_{i = 1}^n {{F_i}} = 0$

If the total force on the body is zero, then the total linear momentum of the body does not change with time;

$sum P = {rm{constant}}$

Condition for Rotational Equilibrium: No Angular Acceleration

The vector sum of all the torques on the rigid body is zero,

${tau _1} + {tau _2} + …… + {tau _n} = sumlimits_{i = 1}^n {{tau _i}} = 0$

If the total torque on the rigid body is zero, the total angular momentum of the body does not change with time. The rotational equilibrium condition is independent of the location of the origin about which the torques are taken.

Equilibrium analysis is a concept that extends well beyond physics — macroeconomic models of supply and demand rest on similar balancing principles, which is why students working with a macroeconomics tutor often find cross-disciplinary thinking valuable. Quality control in engineering also depends on understanding system balance, as discussed in this piece on why mastering quality control matters for your career.

Sample Problem: Mechanical Equilibrium of a Rigid Body

Ex.: An uniform meter stick of 200g is suspended from the ceiling through two identical strings of equal length fixed at the ends. A small object of mass of 20 grams is placed on the stick at a distance of 70 cm from the left end. Find tensions in two strings.

Ans. Since the system is in equilibrium, so for translational equilibrium

$sum {F_x} = 0,,,sum {F_y} = 0$

${T_1} + {T_2} = (M + m)g$ …(i)

Meter stick suspended by two strings with small mass placed at 70 cm from left end

M = 200 gm, AB = 100 cm, AD = 70 cm, m = 20 gm.

Again, the system has no angular acceleration;

$sum Gamma = 0$

${Gamma _{{T_1},,A}} + {Gamma _{Mg,,A}} + {Gamma _{mg,,A}} = 0$

$(AB){T_2}hat k + (AC)Mg( – hat k) + (AD)mg( – hat k) = 0$

$(AB){T_2} = (AC)Mg + (AD)mg$

$(100,cm){T_2} = (50,cm)Mg + (70,cm)mg$

$ Rightarrow quad 10{T_2} = 5Mg + 7mg$

Solving for ${T_2}$; ${T_2} = 1.12,N$. Now, putting this value of ${T_2}$ into equation (i), we obtain ${T_1} = 1.04,N$.

For students who want to deepen their understanding of online learning options alongside self-study resources like this, the article on the benefits of online tutoring and expert teaching offers useful perspective on structured learning approaches.