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

Torque is something that everyone understands. People may find it difficult to express, but they are aware of it. Still, it's simply understandable if you picture a window or a door. A door must be pushed away from the hinge in order to be opened. This makes it possible to open the door with relatively little effort. More force is needed the closer you go to the hinge.

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This Story also Contains

1. Definition of Torque
2. Solved Examples Based on Torque
3. Summary

In this article, we will cover the concept of Torque. This topic falls under the broader category of Rotational Motion which is a crucial chapter in Class 11 physics. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination (JEE Main), National Eligibility Entrance Test (NEET), and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE and others. Over the last ten years of the JEE Main exam (from 2013 to 2023), more than nineteen questions have been asked on this concept. And for NEET five questions were asked from this concept.

Let's read this entire article to gain an in-depth understanding of Torque.

Definition of Torque

The vector product of the Force vector and position vector is known as torque.

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

• Its direction is always perpendicular to the plane containing vector r and F and with the help of the right-hand screw rule, we can find it.

• The magnitude of torque is calculated by using either

$\tau=r_1 F \text { or } \tau=r \cdot F_1$

$r_1=$ perpendicular distance from the origin to the line of force.
$F_1$ = component of force perpendicular to line joining force.
- $\tau=r . F . \sin \phi$

Where $\phi$ is the angle between vector $\mathrm{r}$ and $\mathrm{F}$
- $\tau_{\max }=r . F\left(\right.$ when $\left.\phi=90^{\circ}\right)$

Solved Examples Based on Work Done By a Constant Force

• $\tau_{\min }=0\left(\text { when } \phi=0^0\right)$

• If a pivoted, hinged body tends to rotate due to an applied force, then it is said that torque is acted on the body by force.

Example

In the rotation of a hinged door when we apply torque with the help of force $F$.
- 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.

Recommended Topic Video

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 -

$
\underset{\tau}{\vec{\tau}}=\underset{r}{\vec{F}} \times \underset{\vec{F}}{\overrightarrow{ }}
$
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.