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

- 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 \(\vec c\),

\(|\vec c| = |\vec a|\,|\vec b| \,sin \theta\)

- 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.

**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

\(|\vec c| = |\vec a|\,|\vec b| \,sin \theta\)

- 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; \(\vec c = \vec a × \vec b\)

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

- 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:

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

- 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,

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

A \(\hat k\)

B \(-2\,\hat k\)

C \(3\,\hat i\)

D \(4\,\hat j\)

- 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.

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

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

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

A \(\ell \, cos \theta\)

B \(\ell\)

C \(\ell \, sin \theta\)

D \(\ell \, tan \theta\)

- 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.

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 \(\vec F \).

- 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.

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

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

A \(- 9\,\hat k\)

B \(6\, \hat k\)

C \(3\, \hat k\)

D \(2\, \hat k\)

- 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 15 Nm (clockwise)

B 25 Nm (clockwise)

C 25 Nm (Anticlockwise)

D 30 Nm (Anticlockwise)

- 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 F
_{1}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 F
_{2}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.

- F
_{4}has tendency to rotate the rod in anticlockwise direction.

- F
_{5}and F_{6}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.

A \(\vec F_1\)

B \(\vec F_2\)

C \(\vec F_3\)

D \(\vec F_4\)

- Torque is the rotational equivalent of force.
- The ability of a force \(\vec F\) to cause rotation depends on three factors:

- The magnitude \(\vec F\) of the force.
- The distance \(r\) from the point of application to the pivot.
- 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.

- The tangential component \(F_t\), which is perpendicular to position vector \(\vec r\) is given by;

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

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.

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 \(20\,Nm, 10\,Nm,0\,Nm,0\,Nm,0\,Nm,10\sqrt 2\,Nm\)

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

B \(- 20 \, Nm ,- 30 \, Nm\)

C \(- 10 \, Nm , 0 \, Nm\)

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

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

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

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