Informative line

Learn torque definition and cross product vectors for calculation of magnitude and direction of a?×b. Practice torque problems with solutions.

# Cross Product

• If two vectors  $$\vec a$$ and $$\vec b$$ having magnitudes $$|\vec a |$$ and $$|\vec b |$$ respectively, making an angle $$\theta$$ with each other, then the cross product of $$\vec a$$ and $$\vec b$$ produces third vector $$\vec c$$, as

$$\vec c = \vec a × \vec b$$

### Magnitude of cross product $$\vec c$$

Magnitude of $$\vec c$$

$$|\vec c| = |\vec a|\,|\vec b| \,sin \theta$$

### Direction of cross product $$\vec c$$

• The direction can be found by Right Hand Thumb Rule.
• Place vectors $$\vec a$$ and $$\vec b$$ tail to tail without altering their orientations, imagine a line that is perpendicular to their plane where they meet.
• Place right hand around  $$\vec a$$ , $$\vec b$$ that fingers sweep $$\vec a$$ towards $$\vec b$$.
• The direction of thumb gives the direction of $$\vec c$$.

Important Note

• Cross product of two non - zero vectors having same direction is, zero.
• $$(\vec a× \vec b) = - (\vec b × \vec a)$$

Here, magnitudes are same but directions are opposite.

### Symbols of Directions

Calculation of magnitude and direction of $$\vec a × \vec b$$ :

• To find direction, point fingers of yours right hand towards $$\vec a$$.
• Curl them to bend towards $$\vec b$$.
• The direction of thumb gives the direction $$\vec c$$ i.e. inside the screen.
• For magnitude, calculate

$$|\vec c| = |\vec a|\,|\vec b| \,sin \theta$$

### Unit vector notation

• If vectors are given in unit vector notation;

$$\vec a = a_x \hat i +a_y \hat j +a_z \hat k$$

$$\vec b = b_x \hat i +b_y \hat j +b_z \hat k$$

then; $$\vec c = \vec a × \vec b$$

$$= (a_x \hat i +a_y \hat j +a_z \hat k) × ( b_x \hat i +b_y \hat j +b_z \hat k)$$

### Use of distributive law

• Cross product of each component of first vector with each component of second vector

$$\vec c = c_x \hat i +c_y \hat j +c_z \hat k$$

$$=a_x b_x \hat i × \hat i + a_x b_y \hat i × \hat j + a_x b_z \hat i × \hat k +a_y b_x \hat j × \hat i+a_y b_y \hat j × \hat j+a_y b_z \,\hat j × \hat k+a_zb_x \hat k× \hat i + a_z b_y \hat k × \hat j + a_z b_z \hat k × \hat k$$

Apply vector product in terms of unit vector

$$\hat i ×\hat i =\hat j ×\hat j = \hat k ×\hat k =0$$

$$\hat i ×\hat j = \hat k$$;  $$\hat j ×\hat i = -\hat k$$

$$\hat j ×\hat k = \hat i$$;  $$\hat k ×\hat j = - \hat i$$

$$\hat k ×\hat i = \hat j$$ ;  $$\hat i ×\hat k = - \hat j$$

$$\vec c = (a_y b_z - b_y a_z)\hat i + (a_z b_x - b_z a_x)\hat j + (a_x b_y - b_x a_y)\hat k$$

This can also be represented in the form of determinants:

$$\left[\begin{matrix} c_x \\ c_y \\ c_z \end{matrix}\right] = \left[\begin{matrix} \hat i & \hat j & \hat k \\ a_x & a_y & a_z \\ b_x & b_y & b_z \end{matrix}\right]$$

### Opening  the determinant

• For any $$1 \leq i$$$$j \leq n$$ ; let $$B_{i j }$$ be the matrix of order (n– 1) × (n– 1) obtained from A, by deleting the $$i^{th}$$ row and $$j^{th}$$ column.

The determinant A is given by,

$$|A| = \sum\limits^n_{j=1} (-1)^{1+j}$$ $$a_{ij}$$ determinant $$B_{ij}$$

Also, cofactor of $$a_{ij} =A_{ij} = (-1) ^{i+j} (det \, B_{ij})$$

det $$A= |A| = \sum\limits^n_{j=1} a_{ij} A_{ij}$$  for any $$1\leq i\leq n$$

$$= \sum\limits^n_{i=1} a_{ij} A_{ij}$$   for any $$1\leq j\leq n$$

That is,

$$[A] = a_{11} A_{11} + a_{12} A_{12} + ..... a_{1n} A_{1n}$$ for 3×3 matrix

$$[A] = \left[\begin{matrix} \hat i & \hat j & \hat k \\ a_x & a_y & a_z \\ b_x & b_y & b_z \end{matrix}\right]$$

$$[A] = \left|\begin{matrix} a_y & a_z \\ b_y & b_z \end{matrix}\right| \hat i -\left|\begin{matrix} a_x & a_z \\ b_x & b_z \end{matrix}\right| \hat j + \left|\begin{matrix} a_x & a_y \\ b_x & b_y \end{matrix}\right| \hat k$$

$$=( a_y b_z - a_z b_y) \hat i - ( a_x b_z - a_z b_x) \hat j + ( a_x b_y - a_y b_x) \hat k$$

Rearrange it,

$$\vec c=( a_y b_z - a_z b_y) \hat i +( a_z b_x- a_x b_z) \hat j + ( a_x b_y - a_y b_x) \hat k$$

#### Find the vector $$\vec c = \vec a × \vec b$$. Given: $$\vec a = 2\,\hat i + 3\,\hat j$$ $$\vec b = \,\hat i + 2\,\hat j$$

A $$\hat k$$

B $$-2\,\hat k$$

C $$3\,\hat i$$

D $$4\,\hat j$$

×

Given: $$\vec a = 2\,\hat i + 3\,\hat j$$

$$\vec b = \,\hat i + 2\,\hat j$$

Vector $$\vec c$$ is given as,

$$\vec c = \vec a × \vec b = (2\, \hat i + 3\, \hat j) × ( \hat i + 2\, \hat j)$$

$$\vec c= (2× 1 ) \hat i × \hat i + (2× 2 ) \hat i × \hat j+(3× 1 ) \hat j × \hat i +(3× 2 ) \hat j × \hat j$$

$$\because$$ vector product in terms of unit vector

$$\begin {pmatrix}{{\hat i × \hat i = \hat j × \hat j = \hat k × \hat k = 0\, \\ \hat i × \hat j = \hat k\,,\, \hat j × \hat i =- \hat k\,\\ \hat j × \hat k = \hat i\,,\, \hat k × \hat j = -\hat i\,\\ \hat k× \hat i = \hat j\,,\, \hat i × \hat k = - \hat j}}\end {pmatrix}$$

$$= 2 × 0 + 4 \,\hat k + 3 (-\hat k) + 6× 0$$

$$= 4\,\hat k - 3\,\hat k$$

$$\vec c = \hat k$$

### Find the vector $$\vec c = \vec a × \vec b$$. Given: $$\vec a = 2\,\hat i + 3\,\hat j$$ $$\vec b = \,\hat i + 2\,\hat j$$

A

$$\hat k$$

.

B

$$-2\,\hat k$$

C

$$3\,\hat i$$

D

$$4\,\hat j$$

Option A is Correct

# Calculation of Perpendicular Distance from a Point to a Line of Action of Force

• Let a force $$\vec F$$ is acting upon point P.
• To calculate perpendicular distance from point O to line of action of force, extend the line of action of force from both the sides, as shown in figure.

• Now, $$\vec F$$ is extended both sides and choose the perpendicular distance accordingly as shown in Action-1 and Action-2.

### Action -2

• If position vector of P with respect to O is  $$\vec r$$ and angle between  $$\vec r$$  and  $$\vec F$$  is  $$\theta$$ (by joining tail to tail),

then,  $$sin (\pi - \theta) = r_ \bot$$

$$r_\bot = r \, sin (\pi - \theta)$$

$$|r_\bot| = r \,sin \,\theta$$

#### Find the perpendicular distance from point O on the line of action of force as shown in figure.

A $$\ell \, cos \theta$$

B $$\ell$$

C $$\ell \, sin \theta$$

D $$\ell \, tan \theta$$

×

Extend line of action of force on both sides.

Draw a perpendicular on line of action of force from point O.

From $$\Delta \,$$OPN,

$$ON =O\,P \,sin \theta$$

$$ON = r_{\bot}$$ , $$OP = \ell$$

$$r_{\bot } = \ell \,sin \theta$$

### Find the perpendicular distance from point O on the line of action of force as shown in figure.

A

$$\ell \, cos \theta$$

.

B

$$\ell$$

C

$$\ell \, sin \theta$$

D

$$\ell \, tan \theta$$

Option C is Correct

# Torque as Vector Product

• If a force $$\vec F$$ acts on a particle P and the position of particle P, relative to the origin O, is given by position vector $$\vec r$$, then $$\vec \tau$$ acting on the particle, relative to fixed point O is a vector quantity, which is defined as:

$$\vec \tau = \vec r × \vec F$$

• $$\vec r$$ is the position vector to point of application of force $$\vec F$$, with respect to point, about which $$\vec \tau$$ is to be calculated.

### Magnitude of $$\vec c$$

As  $$\vec c =\vec a× \vec b$$

$$|\vec c| = a\,b \, sin \,\theta$$

Similarly, $$\vec \tau =\vec r× \vec F$$

$$|\vec \tau| = r\,F \, sin \,\phi$$

where $$\phi$$ is a smaller angle between the directions of $$\vec r$$ and $$\vec F$$.

### Direction of $$\vec \tau$$

• To determine the direction of $$\vec \tau$$, slide $$\vec F$$ (without changing its direction) until its tail is at origin, as shown in figure.

• Use of right-hand thumb rule, for vector product by sweeping the fingers of the right hand from $$\vec r$$ (the first vector in the product) towards $$\vec F$$ (the second).
• Outstretched thumb gives the direction of $$\vec\tau$$, as shown in figure.

### Unit vector notation of Torque $$\vec \tau$$

As $$\vec \tau = \vec r × \vec F$$

$$\vec \tau= (r_x \hat i + r_y \hat j + r_z \hat k) × (F_x \hat i + F_y \hat j + F_z \hat k)$$

Expanding, using distributive law

$$\vec \tau= (r_y F_z - r_z F_y)\hat i + (r_z F_x - r_xF_z)\hat j + (r_x F_y - r_yF_x)\hat k$$

#### Find the torque of force $$\vec F$$ about O. $$Given:tan \,37° = \dfrac{3}{4},\,sin \,37° =\dfrac{3}{5}$$

A $$- 9\,\hat k$$

B $$6\, \hat k$$

C $$3\, \hat k$$

D $$2\, \hat k$$

×

$$(\vec r = 4 \,\hat i + 3 \,\hat j)$$

$$\vec r$$ is the position vector of point P.

$$\vec r$$, in unit vector notation,

$$\vec r = r_x \,\hat i + r_y \,\hat j + r_z \,\hat k$$  as  $$r_x= 4,$$ $$r_y = 3,$$  $$r_z = 0$$

$$\vec r = 4 \, \hat i + 3\,\hat j$$

Given: $$\vec F = 3 \, \hat i$$

Torque on force $$\vec F$$ about O,

As $$\vec \tau = \vec r × \vec F$$

$$\because$$ vector product in terms of unit vector

$$\begin {pmatrix}{{\hat i × \hat i = \hat j × \hat j = \hat k × \hat k = 0\, \\ \hat i × \hat j = \hat k\,,\, \hat j × \hat i =- \hat k\,\\ \hat j × \hat k = \hat i\,,\, \hat k × \hat j = -\hat i\,\\ \hat k× \hat i = \hat j\,,\, \hat i × \hat k = - \hat j}}\end {pmatrix}$$

$$= (4 \, \hat i +3\,\hat j) × (3\, \hat i)$$

$$\vec \tau = - 9 \, \hat k$$

# Alternate Method

From figure,

$$tan \theta = \dfrac{3}{4}$$

$$\theta = 37°$$

Torque at force F about O is given as

$$|\vec \tau |=|\vec r |\,|\vec F | \,sin \theta$$

Magnitude of position vector $$(\vec r)$$

$$|\vec r | = \sqrt{(4)^2+(3)^2}$$

= 5 units

Torque,

$$|\vec \tau | = 5 × 3 × sin \,37°$$

$$= 15 × \dfrac{3}{5}$$

$$|\vec \tau | = 9 \, unit$$

For direction, right hand thumb rule is used.

$$\vec \tau$$ is inside the plane of screen, according to right hand thumb rule.

### Find the torque of force $$\vec F$$ about O. $$Given:tan \,37° = \dfrac{3}{4},\,sin \,37° =\dfrac{3}{5}$$

A

$$- 9\,\hat k$$

.

B

$$6\, \hat k$$

C

$$3\, \hat k$$

D

$$2\, \hat k$$

Option A is Correct

# Calculation of Torque due to more than Two Forces

• To calculate torque on a body due to two or more forces about the same axis, the vector sum of all individual torque is done.

$$\vec \tau = \vec \tau_1+\vec \tau_2+\vec \tau_3+........$$

Important note

Take one of the sense of rotation, either clockwise or anticlockwise rotation, as +ve and the other one as – ve.

$$\tau =\tau_1+ \tau_2+\tau_3+\tau_4........$$

Put all values with proper sign.

#### A rod of length 2 m is hinged at one end 'O', as shown in figure. Find the torque due to F1 and F2 about O.

A 15 Nm (clockwise)

B 25 Nm (clockwise)

C 25 Nm (Anticlockwise)

D 30 Nm (Anticlockwise)

×

Torque, $$\tau_\bot$$due to $$\vec F_1$$

Extend the line of action of force $$\vec F_1$$ on both sides and draw a perpendicular on it from O.

Perpendicular distance of F1 from O

here $$(r_\bot)_1 = 2\,sin 60°$$

$$= \sqrt 3 \,m$$

Torque $$(\tau_1)$$ due to force F1

$$\tau_1 = (r_\bot)_1 \, F_1$$

$$= \sqrt 3 × 10 \sqrt 3$$

= 30 Nm

The sense of rotation due to $$\vec F_1$$ is anticlockwise.

Torque, $$\tau_2$$ due to $$\vec F_2$$

Extend the line of action of force, $$\vec F_2$$ on both sides and draw a perpendicular on it from O.

Perpendicular distance of F2 from O

$$(r_\bot)_2 = 1 \,sin 45°$$

$$= \dfrac{1}{\sqrt 2}\,m$$

Torque $$(\tau_2)$$ due to F2

$$\tau_2 = (r_\bot )_2 \, F_2$$

$$= \dfrac{1}{\sqrt 2} × 5\sqrt 2$$

= 5 Nm

The sense of rotation due to F2 is clockwise.

Let anticlockwise as positive direction.

Total torque

$$\tau = \tau _1 + \tau _2$$

$$= 30 + (-5)$$

$$= 25 \,Nm$$  (Anticlockwise)

### A rod of length 2 m is hinged at one end 'O', as shown in figure. Find the torque due to F1 and F2 about O.

A

15 Nm (clockwise)

.

B

25 Nm (clockwise)

C

25 Nm (Anticlockwise)

D

30 Nm (Anticlockwise)

Option C is Correct

# Direction of Torque

• Torque, $$\tau$$ is the rotational equivalent of force.

A rod, is free to rotate about hinge H. A force, $$\vec F_1$$ is applied on it as shown in figure.

• Torque measures the effectiveness of a force, causing  the rod to rotate about a hinge.
• The tendency of force F1 to rotate the rod is in clockwise direction as shown in figure.

→ Now $$\vec F_2$$ is applied as shown in figure.

• The tendency of force F2 to rotate the rod is in anticlockwise direction.

→ $$\vec F _3$$ is applied as shown in figure.

• The tendency of force $$\vec F_3$$ to rotate the rod is in clockwise direction.

• F4 has tendency to rotate the rod in anticlockwise direction.

• F5 and F6 have no tendency to rotate the rod.

Important note:

Even if, the hinge is not given and direction of torque of a force about a particular point O, is to be determined, then the given point O can be assumed as hinge.

For example:

• If direction of torque about the point O is to be determined, then the direction of torque $$\tau$$, would be the same, even if, the point O is not mentioned as hinge.

#### Which of the following forces give a clockwise torque about O?

A $$\vec F_1$$

B $$\vec F_2$$

C $$\vec F_3$$

D $$\vec F_4$$

×

For force $$\vec F_4 ,\, \vec F _1$$

No tendency to rotate the rod.

For force, $$\vec F_2$$

$$\vec F_2$$ has tendency to rotate it anticlockwise.

For force, $$\vec F_3$$

$$\vec F_3$$ has tendency to rotate it clockwise.

So, option (C) is correct.

### Which of the following forces give a clockwise torque about O?

A

$$\vec F_1$$

.

B

$$\vec F_2$$

C

$$\vec F_3$$

D

$$\vec F_4$$

Option C is Correct

# Torque

• Torque is the rotational equivalent of force.
• The ability of a force $$\vec F$$ to cause rotation depends on three factors:
1. The magnitude $$\vec F$$ of the force.
2. The distance $$r$$ from the point of application to the pivot.
3. The angle at which the force is applied.

• Figure shows the cross - section of a body, free to rotate about an axis passing  through O and perpendicular to the cross-section.
• A force $$\vec F$$ is applied at point P, whose position relative to O is defined as position vector $$\vec r$$.

• To determine force $$\vec F$$, which causes the rotation of body about the axis of rotation, force is resolved into two components, as shown in figure:

(A) The radial component of $$\vec F$$$$\vec F_r$$

(B) The tangential Component of $$\vec F$$$$\vec F_t$$

• The radial component $$F_r$$, does not cause rotation of body. As the door cannot be opened by pulling it parallel to the plane of the door.
• The tangential component $$F_t$$, causes the rotation.

### Magnitude of $$F_t$$

• The tangential component $$F_t$$, which is perpendicular to position vector $$\vec r$$ is given by;

$$|\vec\tau| = r\,F\,sin \phi$$

### Two equivalent ways of computing  the torque

1. $$|\vec \tau| = r (F\, sin\phi)$$

$$|\vec \tau| = r \, F_\bot$$

where $$F_\bot$$ is tangential component of force.

2. $$|\vec \tau| = (r \,sin \phi) F$$

$$= r_\bot \, F$$

where $$r_\bot\,$$ also called as moment arm of $$\vec F,$$ is perpendicular distance between the rotation axis at O and line of action of $$\vec F$$, as shown in figure.

#### The figure shows six equal forces of magnitude 10 N acting on a rod of length  L = 2 m, hinged at one end. Find the torque due to  $$F_1 ,F_2 ,F_3 ,F_4 ,F_5 ,F_6$$ separately.

A $$20\,Nm , 0\,Nm, 10\sqrt 2\,Nm, 10\,Nm ,5\sqrt 3\,Nm, 0 \,\,Nm$$

B $$0\,Nm , 20\,Nm, 0\,Nm,10\,Nm,10\sqrt 2\,Nm, 10\sqrt 3\,Nm$$

C $$20\,Nm , 0\,Nm, 10\,Nm,10\sqrt 2\,Nm, 5\sqrt 3\,Nm,0\,Nm,0 \,Nm$$

D $$20\,Nm, 10\,Nm,0\,Nm,0\,Nm,0\,Nm,10\sqrt 2\,Nm$$

×

Torque due to Force F1

$$\tau_1 = r \, sin \,90º × F_1$$

$$= 2 × 1 × 10$$

$$= 20 \, Nm$$   (clockwise)

Torque due to force F2

$$\tau_2 = r \,sin \,0º × F_2$$

= 0

Torque due to F3

$$\tau_3 = r\,sin \,45° × F_3$$

$$= 2× \dfrac{1}{\sqrt2} × 10$$

$$= 10 \,\sqrt 2 \, Nm$$     (Anticlockwise)

Torque due to force F4

$$\tau_4 = r \,sin \, 90° × F_4$$

$$= 1× 10$$

$$= 10 \,Nm$$    (Anticlockwise)

Torque due to force F

$$\tau_5 = r \,sin \, 60° × F_5$$

$$= 1× 10 × \dfrac{\sqrt 3 }{2}$$

$$= 5\sqrt 3 \,Nm$$    (Clockwise)

Torque due to force F

$$\tau_6 = r \,sin \, 0° × F_6$$

$$= 0 ×10×1$$

$$= 0 \,Nm$$

### The figure shows six equal forces of magnitude 10 N acting on a rod of length  L = 2 m, hinged at one end. Find the torque due to  $$F_1 ,F_2 ,F_3 ,F_4 ,F_5 ,F_6$$ separately.

A

$$20\,Nm , 0\,Nm, 10\sqrt 2\,Nm, 10\,Nm ,5\sqrt 3\,Nm, 0 \,\,Nm$$

.

B

$$0\,Nm , 20\,Nm, 0\,Nm,10\,Nm,10\sqrt 2\,Nm, 10\sqrt 3\,Nm$$

C

$$20\,Nm , 0\,Nm, 10\,Nm,10\sqrt 2\,Nm, 5\sqrt 3\,Nm,0\,Nm,0 \,Nm$$

D

$$20\,Nm, 10\,Nm,0\,Nm,0\,Nm,0\,Nm,10\sqrt 2\,Nm$$

Option A is Correct

# Torque of Three Parallel or Anti-parallel Forces about Different Points

• To calculate torque on a body due to two or more forces about the same axis, the vector sum of all individual torque is done.

$$\vec \tau = \vec \tau_1+\vec \tau_2+\vec \tau_3+........$$

Important note

Take one of the sense of rotation, either clockwise or anticlockwise rotation, as +ve and the other one as – ve.

$$\tau =\tau_1+ \tau_2+\tau_3+\tau_4........$$

Put all values with proper sign.

#### Three forces $$\vec F_1, \,\vec F_2\,and\, \vec F_3,$$ acting on a rod are kept on a smooth surface, as shown in figure. Find the torque on the rod about point A and point B.

A $$- 10 \, Nm , 10 \, Nm$$

B $$- 20 \, Nm ,- 30 \, Nm$$

C $$- 10 \, Nm , 0 \, Nm$$

D $$10 \, Nm , 10 \, Nm$$

×

Let point A is assumed to be hinge.

Torque $$(\tau _1)_ A$$ due to $$F_1$$

Here, force  $$F_1$$ is not causing the rotation because, $$r_\bot=0$$.

$$F_1$$ itself is acting on point A.

Torque at A due to $$F_1$$

$$(\tau _1)_ A = (r_\bot ) _{1/A} × F_1$$

= 0 ×10

= 0 Nm

Torque $$(\tau_2)_A$$ due to $$\vec F_2$$

Extend line of action of force $$F_2$$ on both sides and draw a perpendicular from point A.

Torque at A due to F2

$$(\tau_2)_A = (r_\bot )_{2/A} . F_2$$

= 1 × 50

= 50 Nm

and sense of rotation of F2 is clockwise.

Torque $$(\tau_3) _A$$ due to force $$\vec F_3$$

Extend line of action of force $$F_3$$ on both sides and draw a perpendicular from point A.

Torque at A due to F3

$$(\tau_3)_A = (r_\bot)_ {3/A} . F_3$$

= 2 × 20

= 40 Nm

and sense of rotation of Fis anticlockwise.

Take Anticlockwise to be positive

Total torque at A

$$(\vec \tau)_A = (\vec\tau_1)_A + (\vec\tau_2)_A+(\vec\tau_3)_A$$

= 0 – 50 + 40

= – 10 Nm

Negative sign indicates the direction of torque is opposite to the direction that is considered as positive.

$$\therefore$$ The $$\tau$$ about point A is 10 Nm clockwise.

Let point B is assumed as hinge.

Torque $$(\tau_1)_B$$ due to $$\vec F_1$$

Extend line of action of force $$F_1$$ on both sides and draw a perpendicular from point B.

Torque at point B due to F1

$$(\vec\tau_1)_B = (r_\bot)_{1/B} × \vec F_1$$

= 1 × 10

= 10 Nm

and sense of rotation of $$\vec F_1$$ is clockwise.

Torque $$(\tau_2)_B$$ due to $$F_2$$

Force  $$F_2$$ is not causing the rotation because, $$r_\bot=0$$.

$$F_2$$ itself is acting on point B.

Torque at B due to $$F_2$$

$$(\tau _2)_ B = (r_\bot ) _{2/B} × F_2$$

= 0× 50

= 0 Nm

Torque $$(\tau_3)_B$$ due to $$\vec F_3$$

Extend line of action of force $$F_3$$ on both sides and draw a perpendicular from point B.

Torque at B due to $$\vec F_3$$

$$(\tau_3)_B = (r_\bot)_{3/B} × F_3$$

= 1 × 20 Nm

= 20 Nm

and sense of rotation is direction is anticlockwise.

Total torque at B

$$\tau _B =\tau_{1/B}+ \tau_{2/B} +\tau_{3/B}$$

=  – 10 + 0 + 20

=   10 Nm

Torque about point B is 10 Nm in anticlockwise direction.

Here, $$\vec \tau_A \neq \vec \tau_B$$

### Three forces $$\vec F_1, \,\vec F_2\,and\, \vec F_3,$$ acting on a rod are kept on a smooth surface, as shown in figure. Find the torque on the rod about point A and point B.

A

$$- 10 \, Nm , 10 \, Nm$$

.

B

$$- 20 \, Nm ,- 30 \, Nm$$

C

$$- 10 \, Nm , 0 \, Nm$$

D

$$10 \, Nm , 10 \, Nm$$

Option A is Correct

# Torque about any Two Points on a Body which is in Translational Equilibrium

• To calculate torque about P, the position vector of all points of application of force are to be written with respect to P.

• In translational equilibrium $$\sum \vec F = 0$$

$$\Rightarrow \vec r_{i/P} = \vec r_i '$$

$$= \vec r _{Q/P} + \vec r _{i/Q}$$

$$= \vec r _{Q/P} + \vec r _{i}$$

$$\vec \tau _P = \sum (\vec r_{Q/P} + \vec r _i) × \vec F _i$$

$$= \sum \vec r_{Q/P} × \vec F_i +\sum \vec r_i × \vec F_i$$

$$= \vec r _{Q/P} \sum \vec F_i + \sum \vec r _i × \vec F_i$$

as $$\sum \vec F_i = 0$$

$$\vec\tau_P = 0 + \sum \vec r_i × \vec F _i$$

as  $$\vec \tau_Q = \sum \vec r_i × \vec F_i$$

$$\therefore \, \vec \tau_P = \vec \tau_Q$$

#### Three forces $$\vec F_1,\vec F_2$$ and $$\vec F_3$$ acting on a rod of length $$3 \, \ell$$ , is kept on a smooth surface, as shown in figure. Find the torque on the rod about points A, B and C. Given : - $$|\vec F_1| = |\vec F_2| =F$$; $$|\vec F_3| = 2 F$$

A $$1\, \ell F, 2\, \ell F, 3\, \ell F$$

B $$-3\, \ell F, -3\, \ell F, -3\, \ell F$$

C $$2\, \ell F, 3\, \ell F, 4\, \ell F$$

D $$2\, \ell F, 2\, \ell F, 2\, \ell F$$

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Torque about point A due to $$\vec F _1$$

$$(\tau_1)_A = (r_\bot)_{1/A} \,F_1$$

$$= \ell F_1$$

$$=\ell\, F$$    as   $$|\vec F_1| = F$$

Sense of rotation is anticlockwise.

Taking anticlockwise as positive.

$$(\tau_1)_A = +\ell \,F$$

Torque about point A due to $$\vec F_2$$

$$(\tau_2) _A = (r_\bot)_{2/A} . F_2$$

$$= 2\, \ell .F_2$$   as  $$|\vec F_2| = F$$

$$= + 2 \, \ell\, F$$

As sense of rotation is anticlockwise.

Torque about point A due to $$\vec F_3$$

$$(\tau_3) _A = (r_\bot)_{3/A} . F_3$$

$$= 3\, \ell .F_3$$

$$= 3\, \ell\,(2 F)$$

$$6 \,\ell\,F$$

As sense of rotation is clockwise.

$$(\tau_3)_A = - 6 \,\ell F$$

$$\tau_A =\tau_1+\tau_2+\tau_3$$

$$= \ell F+2\,\ell F-6\,\ell F$$

$$= - 3\,\ell F$$

Torque about point B due to $$\vec F_1$$

$$(\tau)_{1/B} = (r_\bot) _{1/B} . F_1$$

$$=\ell .F$$

Sense of rotation is clockwise.

$$(\tau)_{1/B} = - \ell F$$

Torque about point B due to $$\vec F_2$$

$$\vec F_2$$ itself is acting on point B.

$$\therefore (r_\bot) _{2/B} = 0$$

$$\tau_{2/B} = (r_\bot)_{2/B} .F_2$$

= 0.F2

= 0

Torque about point B due to $$\vec F_3$$

$$\tau_{3/B} = (r_\bot)_{3/B} .F_3$$

$$= \ell \, .2F$$

$$=2 \ell F$$

As sense of rotation is clockwise.

$$(\tau_3)_B = - 2\,\ell F$$

$$\tau_B = (\tau_1)_B + (\tau_2)_B + (\tau_3)_B$$

$$= - \ell F + 0 - 2\,\ell F$$

$$= - 3 \, \ell F$$

Torque about point C, due to $$\vec F_1$$

$$(\tau_1)_C = (r_\bot)_{1/C} . F_1$$

$$= 2 \,\ell .F$$

Sense of rotation is clockwise.

$$(\tau_1)_C = - 2\,\ell F$$

Torque about point C, due to $$\vec F_2$$

$$(\tau_2)_C = (r_\bot)_{2/C} . F_2$$

$$= \,\ell .F$$

$$(\tau_2)_C = - \,\ell F$$     (clockwise)

Torque about point C, due to $$\vec F_3$$

As force  $$\vec F_3$$ is acting on point C itself.

$$\therefore \,(r_\bot)_{3/C} = 0$$

$$(\tau)_{3/C} = (r_\bot) _{3/C} . F_3$$

= 0.2F

= 0

$$\tau_C = (\tau_1)_C + (\tau_2)_C +(\tau_3)_C$$

$$= - 2\,\ell F - \ell F + 0$$

$$= - 3\,\ell F$$

As $$\tau_A = \tau_B = \tau_C$$

Torque about all the points is same if the body is in translational equilibrium.

$$F_{net} = 0$$

### Three forces $$\vec F_1,\vec F_2$$ and $$\vec F_3$$ acting on a rod of length $$3 \, \ell$$ , is kept on a smooth surface, as shown in figure. Find the torque on the rod about points A, B and C. Given : - $$|\vec F_1| = |\vec F_2| =F$$; $$|\vec F_3| = 2 F$$

A

$$1\, \ell F, 2\, \ell F, 3\, \ell F$$

.

B

$$-3\, \ell F, -3\, \ell F, -3\, \ell F$$

C

$$2\, \ell F, 3\, \ell F, 4\, \ell F$$

D

$$2\, \ell F, 2\, \ell F, 2\, \ell F$$

Option B is Correct