Torque - Definition, Unit, Examples, Types, Applications, FAQs

Definition of Torque

Definition of Torque
Torque is the turning effect of a force about a fixed point or axis.
Mathematically, torque is defined as the vector product of the position vector ( $\mathbf{r}$ ) and the force vector (F).

$
\vec{\tau}=\vec{r} \times \vec{F}
$

Magnitude of Torque

$
\tau=r F \sin \theta
$

Where:

• $\mathbf{r}=$ Position vector (distance from pivot point)
• $\mathbf{F}=$ Applied force
• $\boldsymbol{\theta}=$ Angle between $\mathbf{r}$ and $\mathbf{F}$

Direction of Torque

The direction of torque is perpendicular to the plane containing $\mathbf{r}$ and $\mathbf{F}$ and is determined using the right-hand rule.

Example

When we apply a force $F$ on a hinged door, it rotates about its hinge. This turning effect of force is called torque (or moment of force).

• SI Unit- Newton-metre
• Dimension- $M L^2 T^{-2}$

If a body is acted upon by more than one force, then we get the resultant torque by doing the vector sum of each torque.
$
\tau=\tau_1+\tau_2+\tau_3 \ldots \ldots
$
Just like force is the cause of translatory motion similarly Torque is the cause of rotatory motion.

Types of Torque

1. Clockwise Torque

When a force causes an object to rotate in the clockwise direction, it produces clockwise torque.

Example:
Turning a screw using a screwdriver in the tightening direction.

2. Anticlockwise (Counterclockwise) Torque

When a force causes rotation in the anticlockwise direction, it produces anticlockwise torque.

Example:
Opening a door by pushing it in the opposite direction.

3. Net Torque

If multiple torques act on an object, the net torque is the algebraic sum of all clockwise and anticlockwise torques.

• If net torque $=0 \rightarrow$ Rotational equilibrium
• If net torque $\neq 0 \rightarrow$ Object rotates

4. Couple Torque

When two equal and opposite forces act on a body at different points and cause rotation without translation, it is called a couple.

Applications of Torque

1. Torque is the turning effect of a force that makes an object rotate about a fixed point or axis.
2. A door opens more easily when pushed at the handle because the handle is farther from the hinges, producing greater torque.
3. A long spanner or wrench helps to loosen a tight nut more easily as it produces more torque than a short one.
4. A large steering wheel allows a driver to turn the vehicle with less effort because it increases the turning effect.
5. A seesaw balances when the torque on one side is equal to the torque on the other side.
6. When we pedal a bicycle, torque is produced which rotates the wheels.
7. Screwdrivers and drills use torque to rotate screws and machine parts.

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Solved Examples Based on Torque

Example 1: Let $\vec{F}$ be the force acting on a particle having a position vector $\vec{r}$ and $\vec{T}$ be the torque of this force about the origin. Then

1) $\vec{r} \cdot \vec{T}=0$ and $\vec{F} \cdot \vec{T} \neq 0$
2) $\vec{r} \cdot \vec{T} \neq 0$ and $\vec{F} \cdot \vec{T}=0$
3) $\vec{r} \cdot \vec{T} \neq 0$ and $\vec{F} \cdot \vec{T} \neq 0$
4) $\vec{r} \cdot \vec{T}=0$ and $\vec{F} \cdot \vec{T}=0$

Solution:

We know that $\vec{\tau}=\vec{r} \times \vec{f}$
The angle between $\mathrm{\tau}$ and $\mathrm{F}$ is $90^{\circ}$ and between $\mathrm{\tau}$ and $\mathrm{r}$ is 900 for example, if we take the dot product between two vectors as $\theta=90^{\circ} \Longrightarrow \cos \theta=\cos 90^{\circ}=0$
$
\begin{aligned}
& \vec{r} \cdot \vec{\tau}=\vec{r} \cdot(\vec{r} \times \vec{f})=0 \\
& \therefore \vec{f} \cdot \vec{\tau}=\vec{f} \cdot(\vec{r} \times \vec{f})=0
\end{aligned}
$

Hence, the answer is option (4).

Example 2: A force of $-F \hat{k}$ acts on $\mathrm{O}$, the origin of the coordinate system. The torque about the point $(1,-1)$ is :
1) $-F(\hat{i}-\hat{j})$
2) $F(\hat{i}-\hat{j})$
3) $-F(\hat{i}+\hat{j})$
4) $F(\hat{i}+\hat{j})$

Solution:

Torque -

$\vec{\tau}=\vec{r} \times \vec{F}$
wherein
This can be calculated by using either $\tau=r_1 F$ or $\tau=r \cdot F_1$ $r_1=$ perpendicular distance from the origin to the line of force.
$F_1=$ component of force perpendicular to line joining force.
given $\vec{\tau}=\vec{r} \times \vec{F}$
$
\begin{aligned}
& \vec{F}=-F \hat{k}, \vec{r}=\hat{i}-\hat{j} \\
& \therefore \quad \vec{r} \times \vec{F}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
1 & -1 & 0 \\
0 & 0 & -F
\end{array}\right| \\
& =\hat{i} F-\hat{j}(-F)=F(\hat{i}+\hat{j})
\end{aligned}
$

Hence, the answer is option (4).

Example 3: A force of 40 N acts on a point B at the end of an L-shaped object, as shown in the figure. The angle θ that will produce the maximum moment of the force about point A is given by :

1) $\tan \theta=\frac{1}{2}$
2) $\tan \theta=2$
3) $\tan \theta=4$
4) $\tan \theta=\frac{1}{4}$

Solution:

Torque

$
\underset{\tau}{\vec{r}}=\underset{F}{\rightarrow}
$
wherein
This can be calculated by using either $\tau=r_1 F$ or $\tau=r \cdot F_1$
$r_1=$ perpendicular distance from the origin to the line of force.
$F_1$ = component of force perpendicular to line joining force.
torque about $\mathrm{A}=\tau_A=(F \cos \theta) \cdot 4+(F \sin \theta) \cdot 2$
for maximum torque
$
\frac{d \tau_A}{d \Theta}=0
$
or $\quad 4 f(-\sin \theta)+2 f \cos \theta=0$
$
2 \sin \theta=\cos \theta \text { or } \tan \theta=\frac{1}{2}
$

Hence, the answer is option (1).

Example 4: A uniform rod AB is suspended from a point X, at a variable distance x from A, as shown. To make the rod horizontal, a mass m is suspended from its end A. A set of (m, x) values is recorded. The appropriate variables that give a straight line, when plotted, are :

1) $m, x$
2) $m, \frac{1}{x}$
3) $m, \frac{1}{x^2}$
4) $m, x^2$

Solution:

Torque -

$
\vec{\tau}=\underset{r}{\rightarrow} \times \underset{F}{\vec{F}}
$
wherein
This can be calculated by using either $\tau=r_1 F$ or $\tau=r \cdot F_1$
$r_1=$ perpendicular distance from the origin to the line of force.
$F_1$ = component of force perpendicular to line joining force.
Balancing Torque w.r.t point of suspension
$
\begin{aligned}
& m g x=M g\left(\frac{l}{2}-x\right) \\
& \Rightarrow m x=M \frac{l}{2}-M x \Rightarrow m=\frac{M l}{2} \cdot \frac{1}{x}-M
\end{aligned}
$

This represents a straight line

Hence, the answer is option (2).

Example 5: The magnitude of torque on a particle of mass 1 kg is 2.5 Nm about the origin. If the force acting on it is 1N, and the distance of the particle from the origin is 5 m, the angle between the force and the position vector is (in radians) :

1) 0.52

2) 30

3) 10

4) 1.04

Solution

$\begin{aligned}
& T=\vec{r} \times \vec{F} \\
& T=|\vec{r}| \cdot|\vec{F}| \cdot \sin \theta \ldots-(1) \\
& T=2.5 \mathrm{Nm} \\
& |\vec{r}|=5 m \\
& |\vec{F}|=1 N \\
& \text {-put in (1) } \\
& T=2.5=1 \times 5 \times \sin \theta \\
& \sin \theta=0.5=\frac{1}{2} \\
& \theta=\frac{\pi}{6} \\
& \theta=0.52 \text { radian } \\
&
\end{aligned}$

Hence, the answer is option (1).

Summary

In short, torque can be conceptualised as an object's twist around a certain axis, similar to how a linear force is a push or a pull. Another definition of torque is the product of the force's magnitude and the force's line of action's perpendicular distance from the axis of rotation. It is possible to find the torque immediately given the length of the moment arm.