Torque and Couple Questions in English

(B) The torque $\vec{\tau}$ is given by the cross product of the position vector $\vec{r}$ and the force vector $\vec{F}$:
$\vec{\tau} = \vec{r} \times \vec{F}$
$\vec{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & 3 \\ 2 & -3 & 4 \end{vmatrix}$
Expanding the determinant:
$\vec{\tau} = \hat{i} [(2)(4) - (3)(-3)] - \hat{j} [(3)(4) - (3)(2)] + \hat{k} [(3)(-3) - (2)(2)]$
$\vec{\tau} = \hat{i} [8 + 9] - \hat{j} [12 - 6] + \hat{k} [-9 - 4]$
$\vec{\tau} = 17\hat{i} - 6\hat{j} - 13\hat{k} \text{ N m}$

(A) The torque $\vec{\tau}$ is given by the cross product of the position vector $\vec{r}$ and the force vector $\vec{F}$:
$\vec{\tau} = \vec{r} \times \vec{F}$
Given $\vec{r} = 7\hat{i} + 3\hat{j} + \hat{k}$ and $\vec{F} = -3\hat{i} + \hat{j} + 5\hat{k}$,we calculate the determinant:
$\vec{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 7 & 3 & 1 \\ -3 & 1 & 5 \end{vmatrix}$
Expanding the determinant:
$\vec{\tau} = \hat{i}(3 \times 5 - 1 \times 1) - \hat{j}(7 \times 5 - 1 \times (-3)) + \hat{k}(7 \times 1 - 3 \times (-3))$
$\vec{\tau} = \hat{i}(15 - 1) - \hat{j}(35 + 3) + \hat{k}(7 + 9)$
$\vec{\tau} = 14\hat{i} - 38\hat{j} + 16\hat{k}$

(A) The torque $\vec{T}$ is defined as the cross product of the position vector $\vec{r}$ and the force vector $\vec{F}$,given by $\vec{T} = \vec{r} \times \vec{F}$.
By the definition of the cross product,the resulting vector $\vec{T}$ is always perpendicular to the plane containing the vectors $\vec{r}$ and $\vec{F}$.
Since $\vec{T}$ is perpendicular to $\vec{r}$,their dot product must be zero: $\vec{r} \cdot \vec{T} = 0$.
Similarly,since $\vec{T}$ is perpendicular to $\vec{F}$,their dot product must be zero: $\vec{F} \cdot \vec{T} = 0$.

(B) couple is defined as a pair of two equal,parallel,and opposite forces acting at different points on a rigid body. Since the net force acting on the body is zero $(F_{net} = F - F = 0)$,there is no linear acceleration or linear motion. However,because the forces act at different points,they create a net torque about any point,which results in purely rotational motion.

(C) The torque $\vec{\tau}$ of a force $\vec{F}$ acting at a position vector $\vec{r}$ relative to the origin is given by the cross product: $\vec{\tau} = \vec{r} \times \vec{F}$.
Given:
$\vec{r} = (3\hat{i} + 2\hat{j} + 3\hat{k}) \text{ m}$
$\vec{F} = (2\hat{i} - 3\hat{j} + 4\hat{k}) \text{ N}$
Calculating the cross product using the determinant method:
$\vec{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & 3 \\ 2 & -3 & 4 \end{vmatrix}$
Expanding the determinant:
$\vec{\tau} = \hat{i}((2)(4) - (-3)(3)) - \hat{j}((3)(4) - (2)(3)) + \hat{k}((3)(-3) - (2)(2))$
$\vec{\tau} = \hat{i}(8 + 9) - \hat{j}(12 - 6) + \hat{k}(-9 - 4)$
$\vec{\tau} = 17\hat{i} - 6\hat{j} - 13\hat{k} \text{ N m}$.

(D) The torque $\tau$ required to loosen the nut is given by $\tau = r F \sin(\theta)$.
In the first case,$r_1 = 8 \, cm$,$F_1 = 6.0 \, N$,and $\theta_1 = 30^{\circ}$.
Therefore,$\tau = 8 \, cm \times 6.0 \, N \times \sin(30^{\circ}) = 8 \times 6.0 \times 0.5 = 24 \, N \cdot cm$.
In the second case,we need to apply a force $F_2$ perpendicularly $(\theta_2 = 90^{\circ})$ at a distance $r_2 = 16 \, cm$.
Since the required torque remains the same,$\tau = r_2 F_2 \sin(90^{\circ})$.
$24 = 16 \times F_2 \times 1$.
$F_2 = \frac{24}{16} = 1.5 \, N$.

(B) The weight $W$ of the book acts downwards at its center of mass,which is at a horizontal distance of $\frac{a}{2}$ from the point of grip.
The torque $\tau$ produced by the weight about the point of grip is given by the product of the force and the perpendicular distance:
$\tau = \text{Force} \times \text{Perpendicular distance}$
$\tau = W \times \frac{a}{2} = W \frac{a}{2}$
Since the weight acts downwards and the pivot point is at the corner,this force creates a rotational tendency in the clockwise direction. To keep the book in equilibrium,the person must apply an equal and opposite torque,which is anticlockwise. Therefore,the person is producing an anticlockwise torque of $W \frac{a}{2}$.

(B) For the cube to begin to tilt about the edge $O$,the normal reaction from the table must act at the edge $O$ itself.
Taking the torque (moment) about the edge $O$:
The torque due to the applied force $F$ is $\tau_F = F \times \frac{3a}{4}$ (tending to rotate clockwise).
The torque due to the weight $mg$ of the cube is $\tau_g = mg \times \frac{a}{2}$ (tending to rotate counter-clockwise,as the weight acts at the centre of mass).
For the cube to be on the verge of tilting,the net torque about $O$ must be zero:
$F \times \frac{3a}{4} = mg \times \frac{a}{2}$
Solving for $F$:
$F = mg \times \frac{a}{2} \times \frac{4}{3a}$
$F = \frac{2}{3} mg$

(B) Given: Force $\overrightarrow{F} = (2\hat{i} - 4\hat{j} + 2\hat{k}) \ N$ and position vector $\overrightarrow{r} = (3\hat{i} + 2\hat{j} - 4\hat{k}) \ m$.
The torque $\overrightarrow{\tau}$ is given by the cross product $\overrightarrow{r} \times \overrightarrow{F}$:
$\overrightarrow{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & -4 \\ 2 & -4 & 2 \end{vmatrix}$
Expanding the determinant:
$\overrightarrow{\tau} = \hat{i}(2 \times 2 - (-4) \times (-4)) - \hat{j}(3 \times 2 - (-4) \times 2) + \hat{k}(3 \times (-4) - 2 \times 2)$
$\overrightarrow{\tau} = \hat{i}(4 - 16) - \hat{j}(6 + 8) + \hat{k}(-12 - 4)$
$\overrightarrow{\tau} = -12\hat{i} - 14\hat{j} - 16\hat{k} \ N-m$
The magnitude of torque is $|\overrightarrow{\tau}| = \sqrt{(-12)^2 + (-14)^2 + (-16)^2}$
$|\overrightarrow{\tau}| = \sqrt{144 + 196 + 256} = \sqrt{596} \approx 24.41 \ N-m$.
Thus,the correct option is $B$.

(C) The resultant force of the system is $F_{res} = 5 \ N + 3 \ N = 8 \ N$,which acts downwards through point $R$.
For the system to be in rotational equilibrium about point $R$,the sum of the moments of the forces about $R$ must be zero.
The force of $5 \ N$ at $P$ creates a clockwise moment about $R$,and the force of $3 \ N$ at $Q$ creates a counter-clockwise moment about $R$.
Taking moments about $R$:
$5 \times PR - 3 \times RQ = 0$
$5 \times PR = 3 \times RQ$
$PR = \frac{3}{5} RQ$

(B) The two forces form a couple. The torque $\tau$ exerted by the couple is given by $\tau = F \times d$, where $F = 0.1\ N$ is the magnitude of each force and $d = 0.05\ m$ is the perpendicular distance between the forces (the diameter of the wheel).
$\tau = 0.1\ N \times 0.05\ m = 0.005\ N\cdot m$.
The work done $W$ when the wheel rotates by an angle $\theta$ is given by $W = \tau \theta$.
For one complete rotation, $\theta = 2\pi\ \text{radians}$.
$W = 0.005\ N\cdot m \times 2\pi\ \text{rad} = 0.01\pi\ J$.
Using $\pi \approx 3.14$, $W = 0.01 \times 3.14 = 0.0314\ J$.
Thus, the work done is about $0.031\ J$.

(D) The torque $\vec{\tau}$ is defined as the cross product of the position vector $\vec{r}$ and the force vector $\vec{F}$,given by $\vec{\tau} = \vec{r} \times \vec{F}$.
By the definition of the cross product,the resulting vector $\vec{\tau}$ is perpendicular to both the position vector $\vec{r}$ and the force vector $\vec{F}$.
Since $\vec{\tau}$ is perpendicular to $\vec{r}$,the dot product $\vec{r} \cdot \vec{\tau} = |\vec{r}| |\vec{\tau}| \cos(90^{\circ}) = 0$.
Similarly,since $\vec{\tau}$ is perpendicular to $\vec{F}$,the dot product $\vec{F} \cdot \vec{\tau} = |\vec{F}| |\vec{\tau}| \cos(90^{\circ}) = 0$.
Therefore,both dot products are equal to zero.

(D) central force is a force that is always directed towards or away from a fixed point (the center).
For a particle moving in a circular path under the influence of a central force,the force vector passes through the center of the circle.
The torque $\vec{\tau}$ is defined as $\vec{\tau} = \vec{r} \times \vec{F}$.
Since the force $\vec{F}$ is directed towards the center,the position vector $\vec{r}$ and the force vector $\vec{F}$ are collinear (the angle between them is $180^{\circ}$ or $0^{\circ}$).
Therefore,the cross product $\vec{r} \times \vec{F} = 0$,which means the torque acting on the particle is zero.
According to the relation $\vec{\tau} = \frac{d\vec{L}}{dt}$,if the external torque is zero,the rate of change of angular momentum is zero,implying that the angular momentum $\vec{L}$ remains constant.

(C) The force is given by $\vec{F} = F\,\hat{k}$.
The position vector of the origin $O(0, 0, 0)$ with respect to the point $P(1, -1, 0)$ is $\vec{r} = (0 - 1)\hat{i} + (0 - (-1))\hat{j} + (0 - 0)\hat{k} = -\hat{i} + \hat{j}$.
The torque $\vec{\tau}$ is given by $\vec{\tau} = \vec{r} \times \vec{F}$.
$\vec{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 1 & 0 \\ 0 & 0 & F \end{vmatrix}$.
Expanding the determinant:
$\vec{\tau} = \hat{i}(1 \cdot F - 0 \cdot 0) - \hat{j}((-1) \cdot F - 0 \cdot 0) + \hat{k}((-1) \cdot 0 - 1 \cdot 0)$.
$\vec{\tau} = F\,\hat{i} + F\,\hat{j} = F\,(\hat{i} + \hat{j})$.

(B) Let $d$ be the perpendicular distance from the center $O$ to any side of the equilateral triangle.
The torque $\tau$ produced by a force is given by the product of the force and the perpendicular distance from the axis of rotation to the line of action of the force.
Taking the counter-clockwise direction as positive,the torques produced by $F_1$,$F_2$,and $F_3$ about point $O$ are:
$\tau_1 = F_1 \times d$ (counter-clockwise)
$\tau_2 = F_2 \times d$ (counter-clockwise)
$\tau_3 = F_3 \times d$ (clockwise)
Given that the total torque about $O$ is zero:
$\sum \tau = \tau_1 + \tau_2 - \tau_3 = 0$
$F_1 d + F_2 d - F_3 d = 0$
Dividing by $d$ (since $d \neq 0$):
$F_1 + F_2 - F_3 = 0$
$F_3 = F_1 + F_2$

(C) The torque of a couple is given by the sum of the torques of the individual forces about the origin.
$\vec{\tau} = (\vec{r_1} \times \vec{F_1}) + (\vec{r_2} \times \vec{F_2})$
Given $\vec{r_1} = 3\hat{i}$,$\vec{F_1} = 1\hat{j}$ and $\vec{r_2} = 3\hat{j}$,$\vec{F_2} = -1\hat{j}$.
$\vec{\tau} = (3\hat{i} \times 1\hat{j}) + (3\hat{j} \times -1\hat{j})$
Since $\hat{i} \times \hat{j} = \hat{k}$ and $\hat{j} \times \hat{j} = 0$,we get:
$\vec{\tau} = 3\hat{k} + 0 = 3\hat{k} \ Nm$.

(D) The torque is defined as $\vec{\tau} = \vec{r} \times \vec{F}$.
By the properties of the cross product,the resulting vector $\vec{\tau}$ is perpendicular to both $\vec{r}$ and $\vec{F}$.
Therefore,the dot product of $\vec{\tau}$ with $\vec{r}$ is $\vec{r} \cdot \vec{\tau} = \vec{r} \cdot (\vec{r} \times \vec{F}) = 0$.
Similarly,the dot product of $\vec{\tau}$ with $\vec{F}$ is $\vec{F} \cdot \vec{\tau} = \vec{F} \cdot (\vec{r} \times \vec{F}) = 0$.
Thus,both dot products are zero.

(D) The position vector of the point where the force is applied is $\vec{r_1} = \hat{i} + 2\hat{j} + 3\hat{k}$.
The position vector of the point about which the torque is to be calculated is $\vec{r_2} = 3\hat{i} - 2\hat{j} - 3\hat{k}$.
The relative position vector $\vec{r}$ is given by $\vec{r} = \vec{r_1} - \vec{r_2} = (\hat{i} + 2\hat{j} + 3\hat{k}) - (3\hat{i} - 2\hat{j} - 3\hat{k}) = -2\hat{i} + 4\hat{j} + 6\hat{k}$.
The torque $\vec{\tau}$ is given by the cross product $\vec{\tau} = \vec{r} \times \vec{F}$.
$\vec{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -2 & 4 & 6 \\ 4 & -5 & 3 \end{vmatrix}$.
$\vec{\tau} = \hat{i}(4 \times 3 - 6 \times (-5)) - \hat{j}((-2) \times 3 - 6 \times 4) + \hat{k}((-2) \times (-5) - 4 \times 4)$.
$\vec{\tau} = \hat{i}(12 + 30) - \hat{j}(-6 - 24) + \hat{k}(10 - 16)$.
$\vec{\tau} = 42\hat{i} + 30\hat{j} - 6\hat{k} \text{ Nm}$.

(A) The torque $\vec{\tau}$ is given by the cross product of the position vector $\vec{r}$ and the force vector $\vec{F}$.
$\vec{\tau} = \vec{r} \times \vec{F} = (3\hat{i} + 2\hat{j} + 3\hat{k}) \times (4\hat{i} - 3\hat{j} + 4\hat{k})$
Using the determinant method for the cross product:
$\vec{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & 3 \\ 4 & -3 & 4 \end{vmatrix}$
$\vec{\tau} = \hat{i} [(2)(4) - (3)(-3)] - \hat{j} [(3)(4) - (3)(4)] + \hat{k} [(3)(-3) - (2)(4)]$
$\vec{\tau} = \hat{i} [8 + 9] - \hat{j} [12 - 12] + \hat{k} [-9 - 8]$
$\vec{\tau} = 17\hat{i} - 0\hat{j} - 17\hat{k}$
$\vec{\tau} = 17(\hat{i} - \hat{k}) \, N \cdot m$

(D) The torque $\overrightarrow \tau$ is given by the cross product of the position vector $\overrightarrow r$ and the force vector $\overrightarrow F$:
$\overrightarrow \tau = \overrightarrow r \times \overrightarrow F$
$\overrightarrow \tau = \begin{vmatrix} \hat i & \hat j & \hat k \\ 7 & 3 & 1 \\ 2 & 1 & 4 \end{vmatrix}$
Expanding the determinant:
$\overrightarrow \tau = \hat i(3 \times 4 - 1 \times 1) - \hat j(7 \times 4 - 1 \times 2) + \hat k(7 \times 1 - 3 \times 2)$
$\overrightarrow \tau = \hat i(12 - 1) - \hat j(28 - 2) + \hat k(7 - 6)$
$\overrightarrow \tau = 11\hat i - 26\hat j + \hat k$

(C) Given: $\vec{r} = 5\hat{i} - 3\hat{j} + 0\hat{k}$ and $\vec{F} = 4\hat{i} - 10\hat{j} + 0\hat{k}$.
The torque is given by the cross product: $\vec{\tau} = \vec{r} \times \vec{F}$.
Using the determinant method:
$\vec{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 5 & -3 & 0 \\ 4 & -10 & 0 \end{vmatrix}$
Expanding the determinant:
$\vec{\tau} = \hat{i}((-3)(0) - (0)(-10)) - \hat{j}((5)(0) - (0)(4)) + \hat{k}((5)(-10) - (-3)(4))$
$\vec{\tau} = \hat{i}(0 - 0) - \hat{j}(0 - 0) + \hat{k}(-50 + 12)$
$\vec{\tau} = -38\hat{k}$.

(D) The torque $\vec{\tau}$ is defined as the cross product of the position vector $\vec{r}$ and the force vector $\vec{F}$,given by $\vec{\tau} = \vec{r} \times \vec{F}$.
By the definition of the cross product,the resulting vector $\vec{\tau}$ is always perpendicular to both the vectors $\vec{r}$ and $\vec{F}$ that form it.
Since the dot product of two perpendicular vectors is always zero,we have $\vec{r} \cdot \vec{\tau} = 0$ and $\vec{F} \cdot \vec{\tau} = 0$.

(C) The moment of force (torque) is given by the cross product of the position vector relative to the point of rotation and the force vector:
$\overrightarrow{\tau} = (\overrightarrow{r} - \overrightarrow{r_0}) \times \overrightarrow{F}$
Here,the point of rotation is $\overrightarrow{r_0} = 2\hat{i} - 2\hat{j} - 2\hat{k}$ and the point of application of force is $\overrightarrow{r} = 2\hat{i} + 0\hat{j} - 3\hat{k}$.
First,calculate the relative position vector:
$\overrightarrow{r} - \overrightarrow{r_0} = (2\hat{i} + 0\hat{j} - 3\hat{k}) - (2\hat{i} - 2\hat{j} - 2\hat{k}) = 0\hat{i} + 2\hat{j} - 1\hat{k}$.
Now,calculate the cross product $\overrightarrow{\tau} = (0\hat{i} + 2\hat{j} - 1\hat{k}) \times (4\hat{i} + 5\hat{j} - 6\hat{k})$:
$\overrightarrow{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 2 & -1 \\ 4 & 5 & -6 \end{vmatrix}$
$\overrightarrow{\tau} = \hat{i}(2(-6) - (-1)(5)) - \hat{j}(0(-6) - (-1)(4)) + \hat{k}(0(5) - 2(4))$
$\overrightarrow{\tau} = \hat{i}(-12 + 5) - \hat{j}(0 + 4) + \hat{k}(0 - 8)$
$\overrightarrow{\tau} = -7\hat{i} - 4\hat{j} - 8\hat{k}$.

(A) The torque $\overrightarrow{\tau}$ is defined as the cross product of the position vector $\overrightarrow{r}$ and the force vector $\overrightarrow{F}$,given by $\overrightarrow{\tau} = \overrightarrow{r} \times \overrightarrow{F}$.
By the definition of the cross product,the resulting vector $\overrightarrow{\tau}$ is always perpendicular to both vectors $\overrightarrow{r}$ and $\overrightarrow{F}$.
Since the dot product of two perpendicular vectors is zero,we have $\overrightarrow{r} \cdot \overrightarrow{\tau} = 0$ and $\overrightarrow{F} \cdot \overrightarrow{\tau} = 0$.

(C) When we apply force on a tap using two fingers,we apply two equal and opposite forces at different points on the handle. This configuration of forces is known as a couple. $A$ couple produces a rotational effect (torque) about the axis of the tap,which makes it easier to turn the tap compared to applying a single force.

(D) Given: Mass $m = 10 \, kg$,Diameter $D = 0.2 \, m$,Radius $r = D/2 = 0.1 \, m$,Acceleration due to gravity $g = 9.8 \, m/s^2$.
Since the body is at rest,the tension $T$ in the rope is equal to the weight of the body: $T = mg = 10 \times 9.8 = 98 \, N$.
The torque $\tau$ applied about the horizontal axis of the cylinder is given by: $\tau = r \times T$.
Substituting the values: $\tau = 0.1 \times 98 = 9.8 \, N-m$.

(C) vector quantity is a physical quantity that has both magnitude and direction.
Work $(W = \vec{F} \cdot \vec{d})$ is a scalar quantity.
Power $(P = \vec{F} \cdot \vec{v})$ is a scalar quantity.
Torque $(\vec{\tau} = \vec{r} \times \vec{F})$ is a vector quantity because it is defined as the cross product of the position vector and the force vector.
Gravitational constant $(G)$ is a scalar quantity.
Therefore, the correct option is $C$.

(B) couple consists of two equal and opposite forces acting at different points. The net force is zero,so there is no linear motion. However,the net torque is non-zero,which causes only rotational motion.

(A) Rotational motion is produced by a torque or a couple.
$A$ couple consists of two equal and opposite forces acting at different points,which produces a pure rotation without any translational motion.
Therefore,the correct option is $A$.

(A) In linear motion,the dynamics is governed by Newton's second law,$F = ma$,where $F$ is force and $a$ is linear acceleration.
In rotational motion,the corresponding equation is $\tau = I\alpha$,where $\tau$ is torque,$I$ is the moment of inertia,and $\alpha$ is angular acceleration.
Comparing the two,force $(F)$ in linear motion is analogous to torque $(\tau)$ in rotational motion.
Therefore,the correct option is $A$.

(A) The torque $\overrightarrow{\tau}$ is given by the cross product of the position vector $\overrightarrow{r}$ and the force vector $\overrightarrow{F}$.
$\overrightarrow{\tau} = \overrightarrow{r} \times \overrightarrow{F}$
$\overrightarrow{\tau} = (7\hat{i} + 3\hat{j} + \hat{k}) \times (-3\hat{i} + \hat{j} + 5\hat{k})$
Using the determinant method for the cross product:
$\overrightarrow{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 7 & 3 & 1 \\ -3 & 1 & 5 \end{vmatrix}$
$\overrightarrow{\tau} = \hat{i}(3 \times 5 - 1 \times 1) - \hat{j}(7 \times 5 - 1 \times (-3)) + \hat{k}(7 \times 1 - 3 \times (-3))$
$\overrightarrow{\tau} = \hat{i}(15 - 1) - \hat{j}(35 + 3) + \hat{k}(7 + 9)$
$\overrightarrow{\tau} = 14\hat{i} - 38\hat{j} + 16\hat{k}$

(D) The position vector of the point where the force is applied is $\overrightarrow{r_1} = \hat{i} + 2\hat{j} + 3\hat{k}$.
To find the torque about point $\overrightarrow{r_2} = 3\hat{i} - 2\hat{j} - 3\hat{k}$,we first calculate the relative position vector $\overrightarrow{r'} = \overrightarrow{r_1} - \overrightarrow{r_2}$.
$\overrightarrow{r'} = (\hat{i} + 2\hat{j} + 3\hat{k}) - (3\hat{i} - 2\hat{j} - 3\hat{k}) = -2\hat{i} + 4\hat{j} + 6\hat{k}$.
The torque $\overrightarrow{\tau}$ is given by $\overrightarrow{\tau} = \overrightarrow{r'} \times \overrightarrow{F}$.
$\overrightarrow{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -2 & 4 & 6 \\ 4 & -5 & 3 \end{vmatrix}$.
$\overrightarrow{\tau} = \hat{i}(4 \times 3 - 6 \times (-5)) - \hat{j}((-2) \times 3 - 6 \times 4) + \hat{k}((-2) \times (-5) - 4 \times 4)$.
$\overrightarrow{\tau} = \hat{i}(12 + 30) - \hat{j}(-6 - 24) + \hat{k}(10 - 16)$.
$\overrightarrow{\tau} = 42\hat{i} + 30\hat{j} - 6\hat{k}$.

(C) The torque $\vec{\tau}$ is defined as the cross product of the position vector $\vec{r}$ and the force vector $\vec{F}$:
$\vec{\tau} = \vec{r} \times \vec{F}$
The magnitude of the torque is given by $\tau = r F \sin \phi$,where $\phi$ is the angle between the position vector $\vec{r}$ and the force vector $\vec{F}$.
In this expression,$F \sin \phi$ represents the component of the force perpendicular to the position vector,which is known as the transverse component.
The radial component $F \cos \phi$ acts along the line of the position vector and does not contribute to the rotational effect (torque).
Therefore,the rotational effect is produced by the transverse component of the force.

(C) The torque is given by $\vec{\tau} = \vec{r} \times \vec{F}$.
For option $A$: $\vec{r} = b \hat{j} - c \hat{k}$ and $\vec{F} = a \hat{k}$.
$\vec{\tau} = (b \hat{j} - c \hat{k}) \times (a \hat{k}) = ab(\hat{j} \times \hat{k}) - ac(\hat{k} \times \hat{k}) = ab \hat{i}$. The $z$-component is $0$.
For option $B$: $\vec{r} = -b \hat{j} - c \hat{k}$ and $\vec{F} = -a \hat{k}$.
$\vec{\tau} = (-b \hat{j} - c \hat{k}) \times (-a \hat{k}) = ab(\hat{j} \times \hat{k}) + ac(\hat{k} \times \hat{k}) = ab \hat{i}$. The $z$-component is $0$.
For option $C$: $\vec{r} = -b \hat{i} - c \hat{k}$ and $\vec{F} = a \hat{j}$.
$\vec{\tau} = (-b \hat{i} - c \hat{k}) \times (a \hat{j}) = -ab(\hat{i} \times \hat{j}) - ac(\hat{k} \times \hat{j}) = -ab \hat{k} - ac(-\hat{i}) = ac \hat{i} - ab \hat{k}$.
The $z$-component is $-ab$,which is negative since $a, b > 0$.

(D) The position vector $\vec{r}$ of the origin $O(0, 0)$ with respect to the point $P(1, -1)$ is given by $\vec{r} = (0 - 1)\hat{i} + (0 - (-1))\hat{j} = -\hat{i} + \hat{j}$.
The force acting at the origin is $\vec{F} = -F\hat{k}$.
The torque $\vec{\tau}$ about the point $P$ is defined as $\vec{\tau} = \vec{r} \times \vec{F}$.
Substituting the values,we get $\vec{\tau} = (-\hat{i} + \hat{j}) \times (-F\hat{k})$.
Using the cross product rules $\hat{i} \times \hat{k} = -\hat{j}$ and $\hat{j} \times \hat{k} = \hat{i}$,we have:
$\vec{\tau} = F(\hat{i} \times \hat{k}) - F(\hat{j} \times \hat{k})$
$\vec{\tau} = F(-\hat{j}) - F(\hat{i})$
$\vec{\tau} = -F(\hat{i} + \hat{j})$.

(B) The torque $\tau$ is given by $\tau = rF \sin \theta$,where $r$ is the distance from the axis of rotation and $\theta$ is the angle between the position vector and the force vector.
$1$. For the force $F$ acting on the rim at an angle of $45^{\circ}$ to the tangent: The perpendicular distance from the center is $R$. The torque is $\tau_1 = F \cdot R \cos(45^{\circ}) = F \cdot R \cdot \frac{1}{\sqrt{2}} \approx 0.707 \ FR$.
$2$. For the force $F$ acting on the axle of radius $R/2$: The force is applied tangentially,so the torque is $\tau_2 = F \cdot (R/2) = 0.5 \ FR$.
$3$. For the force $2F$ acting on the rim of radius $R$: The force is applied tangentially,so the torque is $\tau_3 = 2F \cdot R = 2.0 \ FR$.
All these torques act in the same rotational sense (counter-clockwise).
Net torque $\tau_{net} = \tau_1 + \tau_2 + \tau_3 = 0.707 \ FR + 0.5 \ FR + 2.0 \ FR = 3.207 \ FR$.
Thus,the magnitude of the net torque is nearly $3.2 \ FR$.

(C) The net force acting on the rod is $F_{net} = 6 \ N + 3 \ N = 9 \ N$.
To find the position of the equivalent single force,we calculate the net torque about the center $C$ of the rod.
Taking clockwise torque as positive,the torque due to the $6 \ N$ force is $\tau_1 = 6 \ N \times 2 \ m = 12 \ N \cdot m$.
The torque due to the $3 \ N$ force is $\tau_2 = -3 \ N \times 1 \ m = -3 \ N \cdot m$ (counter-clockwise).
The net torque is $\tau_{net} = 12 - 3 = 9 \ N \cdot m$.
For a single force $F_{net}$ to have the same effect,it must produce the same net torque about the center $C$ at a distance $x$:
$\tau_{net} = F_{net} \times x$
$9 \ N \cdot m = 9 \ N \times x$
$x = 1 \ m$.

(B) Let the forces acting at $20 \, cm$ and $60 \, cm$ be directed downwards $(-F)$ and the forces at $40 \, cm$ and $80 \, cm$ be directed upwards $(+F)$.
Net force $F_{net} = (-F) + (+F) + (-F) + (+F) = 0$.
Since the net force is zero,the rod does not experience any linear motion.
Now,calculate the net torque about the end of the rod $(x = 0)$:
$\tau_{net} = \sum (F_i \times r_i) = (-F \times 20) + (F \times 40) + (-F \times 60) + (F \times 80)$
$\tau_{net} = F(-20 + 40 - 60 + 80) = F(40) = 40F \neq 0$.
Since the net torque is non-zero,the rod experiences a torque and will rotate.

(B) The torque $\vec \tau$ is defined as the cross product of the position vector $\vec r$ and the force vector $\vec F$: $\vec \tau = \vec r \times \vec F$.
By the definition of the cross product,the resulting vector $\vec \tau$ is perpendicular to both $\vec r$ and $\vec F$.
Since the dot product of two perpendicular vectors is zero,we have $\vec \tau \cdot \vec r = 0$ and $\vec \tau \cdot \vec F = 0$.
Therefore,the correct option is $B$.

(B) The torque $\vec{\tau}$ about the origin is given by the cross product of the position vector $\vec{r}$ and the force vector $\vec{F}$:
$\vec{\tau} = \vec{r} \times \vec{F}$
Given $\vec{r} = \hat{i} - 2\hat{j} + \hat{k}$ and $\vec{F} = -2\hat{i} + 2\hat{j} + 3\hat{k}$.
Calculating the cross product using the determinant method:
$\vec{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 1 \\ -2 & 2 & 3 \end{vmatrix}$
Expanding the determinant:
$\vec{\tau} = \hat{i}[(-2)(3) - (1)(2)] - \hat{j}[(1)(3) - (1)(-2)] + \hat{k}[(1)(2) - (-2)(-2)]$
$\vec{\tau} = \hat{i}[-6 - 2] - \hat{j}[3 + 2] + \hat{k}[2 - 4]$
$\vec{\tau} = -8\hat{i} - 5\hat{j} - 2\hat{k}$
Thus,the torque is $-8\hat{i} - 5\hat{j} - 2\hat{k}$.

(B) The torque $\vec{\tau}$ is defined as the cross product of the position vector $\vec{r}$ and the force vector $\vec{F}$,given by $\vec{\tau} = \vec{r} \times \vec{F}$.
The magnitude of the torque is given by $\tau = rF \sin \theta$,where $\theta$ is the angle between the position vector $\vec{r}$ and the force vector $\vec{F}$.
Given:
$F = 2.0\,N$
$r = 3\,m$
Since the force $F$ is parallel to the $x-$ axis and the position vector $\vec{r}$ makes an angle of $30^\circ$ with the $x-$ axis,the angle $\theta$ between $\vec{r}$ and $\vec{F}$ is $30^\circ$.
Substituting these values into the formula:
$\tau = (3\,m)(2.0\,N) \sin 30^\circ$
$\tau = (3\,m)(2.0\,N) \left(\frac{1}{2}\right)$
$\tau = 3\,N-m$.

(B) The torque $\tau$ is defined as the product of the force $F$ and the perpendicular distance $r$ from the axis of rotation to the line of action of the force.
Given:
Force $F = 10\,N$
Radius $r = 0.5\,m$
Since the force is applied tangentially to the circumference,the perpendicular distance is equal to the radius of the disc.
Therefore,torque $\tau = F \times r = 10\,N \times 0.5\,m = 5\,N-m.$

(C) The moment of a force about a point is given by $\tau = rF \sin \phi$,where $r$ is the position vector from the pivot point $A$ to the point of application $B$,$F$ is the magnitude of the force,and $\phi$ is the angle between the position vector $\vec{r}$ and the force vector $\vec{F}$.
To maximize the moment for a fixed magnitude of force,the force must be applied perpendicular to the position vector $\vec{r} = \vec{AB}$.
Let the coordinates of $A$ be $(0, 4)$ and $B$ be $(2, 0)$. The vector $\vec{AB} = (2 - 0)\hat{i} + (0 - 4)\hat{j} = 2\hat{i} - 4\hat{j}$.
The slope of the line $AB$ is $m_{AB} = \frac{0 - 4}{2 - 0} = -2$.
For the force to be perpendicular to $AB$,the slope of the force vector $m_F$ must satisfy $m_F \cdot m_{AB} = -1$,so $m_F = -\frac{1}{-2} = \frac{1}{2}$.
The force vector makes an angle $\theta$ with the horizontal,so its slope is $\tan \theta$. Thus,$\tan \theta = \frac{1}{2}$.

(B) The total torque $\vec \tau_O$ about point $O$ is the sum of the torques due to individual forces: $\vec \tau_O = \vec r_1 \times \vec F_1 + \vec r_2 \times \vec F_2$.
For force $\vec F_1$: The position vector is $\vec r_1 = 2\hat i + 3\hat j$ and the force is $\vec F_1 = F\hat k$. Thus,$\vec r_1 \times \vec F_1 = (2\hat i + 3\hat j) \times F\hat k = 2F(\hat i \times \hat k) + 3F(\hat j \times \hat k) = -2F\hat j + 3F\hat i = 3F\hat i - 2F\hat j$.
For force $\vec F_2$: The position vector is $\vec r_2 = 6\hat j$ (as shown in the figure). The force $\vec F_2$ makes an angle of $30^\circ$ with the $y$-axis in the $XY$-plane,directed towards the negative $x$ and negative $y$ directions. Thus,$\vec F_2 = F(-\cos 30^\circ \hat i - \sin 30^\circ \hat j) = F(-\frac{\sqrt{3}}{2}\hat i - \frac{1}{2}\hat j)$.
Calculating the torque: $\vec r_2 \times \vec F_2 = (6\hat j) \times F(-\frac{\sqrt{3}}{2}\hat i - \frac{1}{2}\hat j) = -3\sqrt{3}F(\hat j \times \hat i) - 3F(\hat j \times \hat j) = 3\sqrt{3}F\hat k$.
Wait,re-evaluating the geometry from the image: The force $\vec F_2$ is at $(0, 6, 0)$ and acts in the $XY$-plane. The torque is $\vec \tau_2 = (6\hat j) \times (F \cos 30^\circ (-\hat i) + F \sin 30^\circ (-\hat j)) = 6F \cos 30^\circ \hat k = 6F(\frac{\sqrt{3}}{2})\hat k = 3\sqrt{3}F\hat k$.
Given the options,there might be a simplification or specific interpretation of the diagram. Based on the provided solution structure: $\vec \tau_O = (3\hat i - 2\hat j + 3\hat k)F$.

(A) The formula for the magnitude of torque is given by $\tau = rF \sin \theta$,where $r$ is the position vector magnitude,$F$ is the force magnitude,and $\theta$ is the angle between them.
Given: $\tau = 2.5\,Nm$,$F = 1\,N$,and $r = 5\,m$.
Substituting the values into the formula: $2.5 = 5 \times 1 \times \sin \theta$.
This simplifies to $\sin \theta = \frac{2.5}{5} = 0.5$.
Since $\sin \theta = 0.5$,the angle $\theta = \arcsin(0.5) = \frac{\pi}{6}$ radians.

(A) The position vector of the particle is $\vec{r} = (x_0 + a \cos \omega_1 t) \hat{i} + (y_0 + b \sin \omega_2 t) \hat{j}$.
At $t = 0$,$\vec{r} = (x_0 + a) \hat{i} + y_0 \hat{j}$.
The acceleration is $\vec{a} = \frac{d^2\vec{r}}{dt^2} = (-a \omega_1^2 \cos \omega_1 t) \hat{i} + (-b \omega_2^2 \sin \omega_2 t) \hat{j}$.
At $t = 0$,$\vec{a} = -a \omega_1^2 \hat{i}$.
The force acting on the particle is $\vec{F} = m \vec{a} = -m a \omega_1^2 \hat{i}$.
The torque about the origin is $\vec{\tau} = \vec{r} \times \vec{F}$.
$\vec{\tau} = [(x_0 + a) \hat{i} + y_0 \hat{j}] \times [-m a \omega_1^2 \hat{i}]$.
Using the cross product rules $\hat{i} \times \hat{i} = 0$ and $\hat{j} \times \hat{i} = -\hat{k}$,we get:
$\vec{\tau} = y_0 (-m a \omega_1^2) (\hat{j} \times \hat{i}) = y_0 (-m a \omega_1^2) (-\hat{k}) = m y_0 a \omega_1^2 \hat{k}$.

(B) couple is defined as a pair of equal and opposite forces acting at different points on a rigid body.
Since the vector sum of the forces is zero $(F_{net} = F + (-F) = 0)$,there is no net force to cause translational motion.
However,because the forces act at different points,they exert a net torque about any point,which causes the body to rotate.
Therefore,a couple produces purely rotational motion.

(B) The torque $\vec{\tau}$ about the origin is given by the cross product of the position vector $\vec{r}$ and the force vector $\vec{F}$.
Given,$\vec{r} = -\hat{i} + 2\hat{j} - 3\hat{k}$ and $\vec{F} = 2\hat{i} - \hat{j} + 3\hat{k}$.
$\vec{\tau} = \vec{r} \times \vec{F} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 2 & -3 \\ 2 & -1 & 3 \end{vmatrix}$
$= \hat{i}((2)(3) - (-3)(-1)) - \hat{j}((-1)(3) - (-3)(2)) + \hat{k}((-1)(-1) - (2)(2))$
$= \hat{i}(6 - 3) - \hat{j}(-3 + 6) + \hat{k}(1 - 4)$
$= 3\hat{i} - 3\hat{j} - 3\hat{k}$
$= 3(\hat{i} - \hat{j} - \hat{k}) \text{ N}\,\text{m}$.

(A) Given:
Position vector $\vec{r} = \hat{i} - \hat{j} + \hat{k}$
Force vector $\vec{F} = 7\hat{i} + 3\hat{j} - 5\hat{k}$
Torque $\vec{\tau}$ is defined as the cross product of position vector and force vector:
$\vec{\tau} = \vec{r} \times \vec{F}$
Calculating the cross product using the determinant method:
$\vec{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 1 \\ 7 & 3 & -5 \end{vmatrix}$
Expanding the determinant:
$\vec{\tau} = \hat{i}((-1)(-5) - (1)(3)) - \hat{j}((1)(-5) - (1)(7)) + \hat{k}((1)(3) - (-1)(7))$
$\vec{\tau} = \hat{i}(5 - 3) - \hat{j}(-5 - 7) + \hat{k}(3 + 7)$
$\vec{\tau} = 2\hat{i} - \hat{j}(-12) + 10\hat{k}$
$\vec{\tau} = 2\hat{i} + 12\hat{j} + 10\hat{k}$