Informative line

Torque About Point

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,

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

Illustration Questions

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 -1

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,  \(sin (\pi - \theta) = r_ \bot \)

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

         \(|r_\bot| = r \,sin \,\theta\)

Illustration Questions

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.

image image

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

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From \(\Delta \, \)OPN,

 \(ON =O\,P \,sin \theta\)

\(ON = r_{\bot} \) , \(OP = \ell\)

\(r_{\bot } = \ell \,sin \theta\)

image

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

image
A

\(\ell \, cos \theta\)

.

B

\(\ell\)

C

\(\ell \, sin \theta\)

D

\(\ell \, tan \theta\)

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.

Illustration Questions

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.

image image

For force, \(\vec F_2\)

\(\vec F_2\) has tendency to rotate it anticlockwise.

image image

For force, \(\vec F_3\)

\(\vec F_3\) has tendency to rotate it clockwise.

 

image image

So, option (C) is correct.

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

image
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.

In addition, 

  • 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.

Illustration Questions

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) 

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Torque due to force F2

\(\tau_2 = r \,sin \,0º × F_2\)

 = 0

image image

Torque due to F3

  \(\tau_3 = r\,sin \,45° × F_3 \)

  \(= 2× \dfrac{1}{\sqrt2} × 10\)

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

image image

Torque due to force F4  

\(\tau_4 = r \,sin \, 90° × F_4 \)

\(= 1× 10\)

\( = 10 \,Nm\)    (Anticlockwise)

image image

Torque due to force F

\(\tau_5 = r \,sin \, 60° × F_5\)

\(= 1× 10 × \dfrac{\sqrt 3 }{2}\)

\( = 5\sqrt 3 \,Nm\)    (Clockwise)

image image

Torque due to force F

\(\tau_6 = r \,sin \, 0° × F_6 \)

\(= 0 ×10×1\)

\( = 0 \,Nm\)    

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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.

image
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 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

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

Illustration Questions

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.

image

\(\vec r\), in unit vector notation, 

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

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

image

Given: \(\vec F = 3 \, \hat i\)

image

Torque on force \(\vec F\) about O, 

As \(\vec \tau = \vec r × \vec F\)

\(\because\) vector product in terms of unit vector

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

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

    \(\vec \tau = - 9 \, \hat k\)

image

Alternate Method 

From figure, 

\(tan \theta = \dfrac{3}{4}\)

\(\theta = 37°\)

image

Torque at force F about O is given as 

\(|\vec \tau |=|\vec r |\,|\vec F | \,sin \theta \)

image

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

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

= 5 units  

image

Torque,

\(|\vec \tau | = 5 × 3 × sin \,37° \)

\(= 15 × \dfrac{3}{5}\)

\(|\vec \tau | = 9 \, unit\)

image

For direction, right hand thumb rule is used.

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

image image

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

image
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.

Illustration Questions

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.

image image

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

image

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

image

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.

image image

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

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The sense of rotation due to F2 is clockwise.

image

Let anticlockwise as positive direction.

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Total torque 

\(\tau = \tau _1 + \tau _2\)

  \(= 30 + (-5)\)

\(= 25 \,Nm\)  (Anticlockwise)

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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. 

image
A

15 Nm (clockwise)

.

B

25 Nm (clockwise)

C

25 Nm (Anticlockwise)

D

30 Nm (Anticlockwise)

Option C 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.

Illustration Questions

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 \(10 \, Nm , 10 \, Nm\)

×

Torque about point A

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

image image

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.

image image

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.

image image

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.

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Torque about point B 

Let point B is assumed as hinge.

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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.

image image

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

image

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.

image image

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, \(\vec \tau_A \neq \vec \tau_B\)

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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. 

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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  \(\vec \tau_Q = \sum \vec r_i × \vec F_i\)

    \(\therefore \, \vec \tau_P = \vec \tau_Q\)

Illustration Questions

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 \(2\, \ell F, 2\, \ell F, 2\, \ell F\)

×

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.

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

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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.

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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.

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

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Total torque about point A 

\(\tau_A =\tau_1+\tau_2+\tau_3 \)

\(= \ell F+2\,\ell F-6\,\ell F\)

\(= - 3\,\ell F\)

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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.

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

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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

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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.

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

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Total torque about point B

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

\(= - \ell F + 0 - 2\,\ell F \)

 \(= - 3 \, \ell F\)

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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.

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

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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)

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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

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Total torque about point C

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

 \(= - 2\,\ell F - \ell F + 0\)

 \(= - 3\,\ell F\)

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As \(\tau_A = \tau_B = \tau_C\)

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

\(F_{net} = 0\)

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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 : - \(|\vec F_1| = |\vec F_2| =F\); \(|\vec F_3| = 2 F\)

image
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

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