Torque and Couple Questions in English

(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 $\vec{r} = 2\hat{i} + 2\hat{j} + 1\hat{k}$ and $\vec{F} = 3\hat{i} + 7\hat{j} + 4\hat{k}$.
$\vec{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 2 & 1 \\ 3 & 7 & 4 \end{vmatrix}$
$\vec{\tau} = \hat{i}(2 \times 4 - 1 \times 7) - \hat{j}(2 \times 4 - 1 \times 3) + \hat{k}(2 \times 7 - 2 \times 3)$
$\vec{\tau} = \hat{i}(8 - 7) - \hat{j}(8 - 3) + \hat{k}(14 - 6)$
$\vec{\tau} = 1\hat{i} - 5\hat{j} + 8\hat{k}$.

(C) The radius of the wheel is $r = 10\,cm = 0.1\,m$.
$1$. Torque due to $12\,N$ force at $S$: This force is tangential,so $\tau_1 = F_1 \times r = 12 \times 0.1 = 1.2\,Nm$ (anticlockwise).
$2$. Torque due to $5\,N$ force at $P$: The perpendicular distance from $C$ to the line of action of the force is $r \sin(30^\circ) = 0.1 \times 0.5 = 0.05\,m$. Thus,$\tau_2 = 5 \times 0.05 = 0.25\,Nm$ (anticlockwise).
$3$. Torque due to $8\,N$ force at $Q$: The line of action passes through the center $C$,so $\tau_3 = 0$.
$4$. Torque due to $4\,N$ force at $R$: This force is tangential,so $\tau_4 = F_4 \times r = 4 \times 0.1 = 0.4\,Nm$ (clockwise).
Net torque $\tau_{net} = \tau_1 + \tau_2 - \tau_4 = 1.2 + 0.25 - 0.4 = 1.05\,Nm$ (anticlockwise).

(C) The torque $\vec{\tau}$ about the centre of mass is given by $\vec{\tau} = \vec{r} \times \vec{N}$,where $\vec{r}$ is the position vector of the point of application of the normal force $\vec{N}$ relative to the centre of mass.
If the normal force acts directly through the centre of mass,the torque is zero.
However,in many physical situations,such as a block on an inclined plane or a body undergoing toppling,the normal force acts at a point shifted from the centre of mass.
In such cases,the position vector $\vec{r}$ is non-zero and not parallel to the normal force $\vec{N}$,resulting in a non-zero torque.
Therefore,the torque due to the normal reaction about the centre of mass may be non-zero depending on the point of application of the force.

(B) The moment of a force about a point is given by $\tau = \pm F \cdot d$, where $d$ is the perpendicular distance from the pivot to the line of action of the force. By convention, anticlockwise moments are positive $(+)$ and clockwise moments are negative $(-)$.
$(i)$ For force $F$: The force $F$ acts downwards at a distance $2a$ from $A$. This creates a clockwise moment.
Moment $= -F \cdot (2a) = -2Fa$.
$(ii)$ For force $2F$: The force $2F$ acts upwards at a distance $a$ from $A$. This creates an anticlockwise moment.
Moment $= +(2F) \cdot a = +2Fa$.
$(iii)$ For force $3F$: The force $3F$ acts at $B$ (distance $4a$ from $A$) at an angle of $30^{\circ}$ to the rod. The perpendicular distance from $A$ is $d = 4a \sin 30^{\circ} = 4a \cdot (1/2) = 2a$. This force creates an anticlockwise moment.
Moment $= +(3F) \cdot (2a) = +6Fa$.
Thus, the moments are $-2Fa, +2Fa, +6Fa$.

(A) The torque $\vec{\tau}$ is defined as the cross product of the position vector $\vec{r}$ and the force vector $\vec{F}$:
$\vec{\tau} = \vec{r} \times \vec{F}$
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 cross product using the determinant method:
$\vec{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 7 & 3 & 1 \\ -3 & 1 & 5 \end{vmatrix}$
Expanding the determinant along the first row:
$\vec{\tau} = \hat{i}(3 \times 5 - 1 \times 1) - \hat{j}(7 \times 5 - 1 \times (-3)) + \hat{k}(7 \times 1 - 3 \times (-3))$
$\vec{\tau} = \hat{i}(15 - 1) - \hat{j}(35 + 3) + \hat{k}(7 + 9)$
$\vec{\tau} = 14\hat{i} - 38\hat{j} + 16\hat{k}$

(B) The radius of the wheel is $r = 20\, cm = 0.2\, m$.
Torque $\tau = r \cdot F \cdot \sin \theta$,where $\theta$ is the angle between the position vector and the force vector.
$1$. For the $4\, N$ force at $A$: The force is tangential,so $\theta = 90^{\circ}$. It produces an anticlockwise torque: $\tau_A = 0.2 \times 4 \times \sin 90^{\circ} = 0.8\, N-m$ (anticlockwise).
$2$. For the $8\, N$ force at $B$: The angle between the radial line and the force is $30^{\circ}$. It produces a clockwise torque: $\tau_B = 0.2 \times 8 \times \sin 30^{\circ} = 1.6 \times 0.5 = 0.8\, N-m$ (clockwise).
$3$. For the $6\, N$ force at $C$: The force is directed towards the center,so $\theta = 0^{\circ}$. $\tau_C = 0.2 \times 6 \times \sin 0^{\circ} = 0\, N-m$.
$4$. For the $9\, N$ force at $D$: The force is tangential,so $\theta = 90^{\circ}$. It produces a clockwise torque: $\tau_D = 0.2 \times 9 \times \sin 90^{\circ} = 1.8\, N-m$ (clockwise).
Net torque $\tau_{net} = \tau_A - \tau_B - \tau_D = 0.8 - 0.8 - 1.8 = -1.8\, N-m$.
The negative sign indicates a clockwise direction. Thus,the net torque is $1.8\, N-m$ (clockwise).

(C) The torque $\vec \tau$ is defined as the cross product of the position vector $\vec r$ and the force vector $\vec F$:
$\vec \tau = \vec r \times \vec F$
Given $\vec r = 3\hat i + 2\hat j + 3\hat k$ and $\vec F = 2\hat i - 3\hat j + 4\hat k$,we calculate the determinant:
$\vec \tau = \begin{vmatrix} \hat i & \hat j & \hat k \\ 3 & 2 & 3 \\ 2 & -3 & 4 \end{vmatrix}$
Expanding the determinant:
$\vec \tau = \hat i(2 \times 4 - 3 \times (-3)) - \hat j(3 \times 4 - 3 \times 2) + \hat k(3 \times (-3) - 2 \times 2)$
$\vec \tau = \hat i(8 + 9) - \hat j(12 - 6) + \hat k(-9 - 4)$
$\vec \tau = 17\hat i - 6\hat j - 13\hat k$

(A) The torque $\vec{\tau}$ is given by the cross product of the position vector $\vec{r}$ and the force vector $\vec{F}$ as $\vec{\tau} = \vec{r} \times \vec{F}$.
Given $\vec{r} = \hat{i} - \hat{j} + \hat{k}$ and $\vec{F} = 7 \hat{i} + 3 \hat{j} - 5 \hat{k}$.
Using the determinant method:
$\vec{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 1 \\ 7 & 3 & -5 \end{vmatrix}$
$\vec{\tau} = \hat{i}((-1)(-5) - (1)(3)) - \hat{j}((1)(-5) - (1)(7)) + \hat{k}((1)(3) - (-1)(7))$
$\vec{\tau} = \hat{i}(5 - 3) - \hat{j}(-5 - 7) + \hat{k}(3 + 7)$
$\vec{\tau} = 2 \hat{i} + 12 \hat{j} + 10 \hat{k}$.

(N/A) Consider a couple acting on a rigid body as shown in the figure. The forces $F$ and $-F$ act respectively at points $B$ and $A$. These points have position vectors $r_1$ and $r_2$ with respect to the origin $O$.
Let us calculate the moments of the forces about the origin $O$.
The moment of the couple is the sum of the moments of the two forces forming the couple:
$\text{Moment} = r_1 \times (-F) + r_2 \times F$
$= r_2 \times F - r_1 \times F$
$= (r_2 - r_1) \times F$
From the triangle law of vector addition,we have $r_1 + AB = r_2$,which implies $AB = r_2 - r_1$.
Substituting this into the expression for the moment,we get:
$\text{Moment} = AB \times F$
Since $AB$ is the vector representing the separation between the two forces,the moment of the couple depends only on the forces and their separation,not on the origin $O$ chosen to calculate the moments. Thus,the moment of a couple is independent of the point about which the moments are taken.

(N/A) The role of torque in rotational motion is similar to the role of force in translational motion.
Consider a force $\vec{F}$ acting on a particle $P$ having a position vector $\vec{r}$ with respect to the origin $O$. The angle between $\vec{r}$ and $\vec{F}$ is $\theta$. The vector product of $\vec{r}$ and $\vec{F}$ is defined as the torque $\vec{\tau}$ acting on the particle with respect to the origin $O$.
$\therefore \vec{\tau} = \vec{r} \times \vec{F}$
The magnitude of torque is given by:
$\tau = r F \sin \theta$
where $|\vec{r}| = r$ and $|\vec{F}| = F$.
Since $\tau = r F \sin \theta$,we can rewrite this as:
$\tau = (r \sin \theta) F = r_{\perp} F$
where $r_{\perp} = r \sin \theta$ is the perpendicular distance of the line of action of the force from the origin.
Alternatively:
$\tau = r (F \sin \theta) = r F_{\perp}$
where $F_{\perp} = F \sin \theta$ is the component of the force perpendicular to the position vector.
Thus,torque is the moment of force with respect to point $O$.

(N/A) couple is a pair of equal and opposite forces acting on a body with different lines of action. $A$ couple produces rotation without translation.
Illustration $1$:
As shown in figure $(a)$,when we open the lid of a bottle by turning it,our fingers apply a couple to the lid. The lid is in translational equilibrium (net force is zero) but not in rotational equilibrium.
Illustration $2$:
As shown in figure $(b)$,the Earth's magnetic field exerts equal and opposite forces on the poles of a compass needle. The force on the North pole is towards the north,and the force on the South pole is towards the south.
When the needle is not aligned in the North-South direction,the lines of action of these two forces are different. Hence,a couple acts on the needle due to the Earth's magnetic field,causing it to undergo rotational motion without translational motion.

(B) couple is defined as a pair of two equal 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 translational motion.
However,the two forces produce a net torque about any point,which causes the body to rotate.
Therefore,a couple produces pure rotational motion.

(N/A) The role of moment of inertia and torque in the rotational motion of a rigid body is analogous to the role of mass and force in translational motion.
In the rotational motion of a rigid body,only the components of torque parallel to the fixed axis of rotation should be considered,as these components are responsible for the rotation of the body relative to that axis.
The components of torque perpendicular to the axis of rotation tend to rotate the axis itself from its position. To cancel the effect of these perpendicular components,an equal and opposite reaction torque is generated by the supports,keeping the axis steady. Therefore,the perpendicular components of torque are not considered in the calculation of rotational motion about a fixed axis.
In summary,for the calculation of torque about a fixed axis:
$(1)$ We consider only those forces that lie in planes perpendicular to the axis.
$(2)$ We consider only those components of the position vectors that are perpendicular to the axis.

(B) For an object to undergo rotational motion about a fixed axis, it must experience a torque $(\tau)$.
Torque is defined as $\vec{\tau} = \vec{r} \times \vec{F}$, where $\vec{r}$ is the position vector from the axis of rotation to the point of application of the force, and $\vec{F}$ is the applied force.
If the line of action of the force passes through the axis of rotation, the perpendicular distance (lever arm) is zero, resulting in zero torque.
Therefore, to produce rotational motion, the force must have a line of action that does not intersect the axis of rotation, ensuring a non-zero torque is applied.

(A) The torque $\vec{\tau}$ is defined as the cross product of the position vector $\vec{r}$ and the force vector $\vec{F}$,given by $\vec{\tau} = \vec{r} \times \vec{F}$.
When calculating the torque about a specific axis,we consider the component of the position vector $\vec{r}$ that is perpendicular to the axis of rotation.
Any component of the position vector $\vec{r}$ that lies along the axis of rotation is parallel to the axis.
Since the cross product of any vector with a vector parallel to the axis of rotation results in a component that is perpendicular to the axis,the parallel components of $\vec{r}$ do not contribute to the torque component along that axis.
Therefore,only the perpendicular components of the position vector are relevant for determining the torque about a given axis.

(C) In linear motion,force $(F)$ is responsible for changing the state of motion of an object,as described by Newton's second law $(F = ma)$.
In rotational motion,the physical quantity that plays an analogous role to force is torque $(\tau)$.
Torque is defined as the rotational equivalent of force and is responsible for changing the angular state of motion of an object,given by the relation $\tau = I\alpha$,where $I$ is the moment of inertia and $\alpha$ is the angular acceleration.

(C) The direction of torque $\vec{\tau} = \vec{r} \times \vec{F}$ is determined by the right-hand screw rule.
If we rotate a right-handed screw from the direction of the position vector $\vec{r}$ towards the direction of the force vector $\vec{F}$ through the smaller angle,the direction in which the screw advances gives the direction of the torque vector.
Alternatively,it is perpendicular to the plane containing both $\vec{r}$ and $\vec{F}$.

(C) For the rotation of a rigid body about the $Z$-axis,only the $Z$-component of the torque is responsible.
The torque vector is given by $\vec{\tau} = \vec{r} \times \vec{F}$.
The $Z$-component of the torque is expressed as $\tau_{z} = (x F_{y} - y F_{x})$.
Therefore,the component responsible for rotation about the $Z$-axis is $\tau_{z}$.

(N/A) The moment of a couple is defined as the product of the magnitude of one of the forces and the perpendicular distance between the lines of action of the two forces.
Formula: $\tau = F \times d$
Where:
$\tau$ is the moment of the couple,
$F$ is the magnitude of one of the forces,
$d$ is the perpendicular distance between the two forces.

(B) Torque is defined as the rotational analogue of force. It measures the effectiveness of a force in changing the rotational state of an object about an axis. Mathematically,it is the cross product of the position vector and the force vector,given by $\vec{\tau} = \vec{r} \times \vec{F}$. Therefore,it represents the rotational effect of force.

(B) The torque $\tau$ is given by the formula $\tau = \vec{r} \times \vec{F}$,where $\vec{r}$ is the position vector (length of the handle) and $\vec{F}$ is the applied force.
By increasing the length of the handle,the magnitude of the position vector $\vec{r}$ increases.
Since $\tau \propto r$,increasing the length of the handle increases the torque applied to the screw.
This makes it easier to rotate the screw with less effort.

(B) The torque $\tau$ is given by the formula $\tau = \vec{r} \times \vec{F}$,where $\vec{r}$ is the position vector from the axis of rotation (hinge) to the point where the force $\vec{F}$ is applied.
By placing the handle at the end opposite to the hinges,the distance $r$ is maximized.
Since $\tau = rF \sin \theta$,increasing $r$ results in a larger torque for the same amount of applied force.
This makes it easier to open or close the door or window with minimal effort.

(B) No,it is not necessarily zero about any arbitrary point.
The torque $\vec{\tau}$ about a point $P$ is given by $\vec{\tau}_P = \sum (\vec{r}_i - \vec{r}_P) \times \vec{F}_i$.
If we shift the point to $P'$,the new torque is $\vec{\tau}_{P'} = \sum (\vec{r}_i - \vec{r}_{P'}) \times \vec{F}_i = \sum (\vec{r}_i - \vec{r}_P + \vec{r}_P - \vec{r}_{P'}) \times \vec{F}_i$.
This simplifies to $\vec{\tau}_{P'} = \vec{\tau}_P + (\vec{r}_P - \vec{r}_{P'}) \times \sum \vec{F}_i$.
Since the net force $\sum \vec{F}_i \neq 0$,the torque $\vec{\tau}_{P'}$ will only be zero if the vector $(\vec{r}_P - \vec{r}_{P'})$ is parallel to the net force vector $\sum \vec{F}_i$.

(N/A) No,the weight of the door does not cause any torque about the vertical axis of rotation.
The torque $\vec{\tau}$ is defined as the cross product of the position vector $\vec{r}$ and the force vector $\vec{F}$,given by $\vec{\tau} = \vec{r} \times \vec{F}$.
In this case,the vertical axis of rotation is the $Y$-axis. The weight $W$ of the door acts vertically downwards,which is along the negative $Y$-axis (i.e.,$\vec{F} = -W \hat{j}$).
The position vector $\vec{r}$ of any point on the door lies in the $XY$ plane. Since the force vector $\vec{F}$ is parallel to the axis of rotation ($Y$-axis),the line of action of the weight passes through the axis of rotation.
Mathematically,the perpendicular distance from the axis of rotation to the line of action of the force is zero. Therefore,the torque $\vec{\tau} = \vec{r} \times \vec{F} = 0$ about the vertical axis.

(B) The torque $\vec{\tau}$ about a point $\vec{r}_2$ is given by $\vec{\tau} = (\vec{r}_1 - \vec{r}_2) \times \vec{F}$.
First,calculate the position vector relative to the point of rotation:
$\vec{r} = \vec{r}_1 - \vec{r}_2 = (4\hat{i} + 3\hat{j} - \hat{k}) - (\hat{i} + 2\hat{j} + \hat{k}) = 3\hat{i} + \hat{j} - 2\hat{k}$.
Now,calculate the cross product $\vec{\tau} = \vec{r} \times \vec{F}$:
$\vec{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & -2 \\ 1 & 2 & 3 \end{vmatrix}$
$= \hat{i}(1(3) - (-2)(2)) - \hat{j}(3(3) - (-2)(1)) + \hat{k}(3(2) - 1(1))$
$= \hat{i}(3 + 4) - \hat{j}(9 + 2) + \hat{k}(6 - 1) = 7\hat{i} - 11\hat{j} + 5\hat{k}$.
The magnitude of the torque is $|\vec{\tau}| = \sqrt{7^2 + (-11)^2 + 5^2} = \sqrt{49 + 121 + 25} = \sqrt{195}$.
Comparing this with $\sqrt{x}$,we get $x = 195$.

(D) 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: $\vec{F} = 3 \hat{j} \text{ N}$ and $\vec{r} = 2 \hat{k} \text{ m}$.
$\vec{\tau} = \vec{r} \times \vec{F} = (2 \hat{k}) \times (3 \hat{j})$.
Using the properties of unit vector cross products: $\hat{k} \times \hat{j} = -\hat{i}$.
Therefore,$\vec{\tau} = 6 (\hat{k} \times \hat{j}) = 6(-\hat{i}) = -6 \hat{i} \text{ Nm}$.

(D) The coordinates of point $P$ are $(5, 5 \sqrt{3})$ cm. The coordinates of point $O$ are $(0, 0)$ and point $Q$ are $(10, 0)$.
Given force $\overrightarrow{F} = 4 \hat{i} - 3 \hat{j}$.
Position vector of $P$ with respect to $O$ is $\overrightarrow{r}_1 = 5 \hat{i} + 5 \sqrt{3} \hat{j}$.
Torque about $O$ is $\vec{\tau}_O = \overrightarrow{r}_1 \times \overrightarrow{F} = (5 \hat{i} + 5 \sqrt{3} \hat{j}) \times (4 \hat{i} - 3 \hat{j}) = (-15 \hat{k} - 20 \sqrt{3} \hat{k}) = (-15 - 20 \sqrt{3}) \hat{k}$.
Position vector of $P$ with respect to $Q$ is $\overrightarrow{r}_2 = (5-10) \hat{i} + 5 \sqrt{3} \hat{j} = -5 \hat{i} + 5 \sqrt{3} \hat{j}$.
Torque about $Q$ is $\vec{\tau}_Q = \overrightarrow{r}_2 \times \overrightarrow{F} = (-5 \hat{i} + 5 \sqrt{3} \hat{j}) \times (4 \hat{i} - 3 \hat{j}) = (15 \hat{k} + 20 \sqrt{3} \hat{k}) = (15 + 20 \sqrt{3}) \hat{k}$.
Thus,the values are $(-15 - 20 \sqrt{3})$ and $(15 + 20 \sqrt{3})$.

(B) The position vector $\vec{r}$ is the vector from the point of rotation $(2, 3, 4)$ to the point of application of the force.
The intersection of the plane $x = 2$ and the $x$-axis is the point $(2, 0, 0)$.
Thus,$\vec{r} = (2 - 2)\hat{i} + (0 - 3)\hat{j} + (0 - 4)\hat{k} = -3\hat{j} - 4\hat{k}$.
The force is $\vec{F} = 4\hat{i} + 3\hat{j} + 4\hat{k}$.
The torque $\vec{\tau}$ is given by $\vec{\tau} = \vec{r} \times \vec{F}$.
$\vec{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & -3 & -4 \\ 4 & 3 & 4 \end{vmatrix}$.
$\vec{\tau} = \hat{i}(-12 - (-12)) - \hat{j}(0 - (-16)) + \hat{k}(0 - (-12))$.
$\vec{\tau} = 0\hat{i} - 16\hat{j} + 12\hat{k}$.
The magnitude of the torque is $|\vec{\tau}| = \sqrt{0^2 + (-16)^2 + 12^2} = \sqrt{256 + 144} = \sqrt{400} = 20$.

(B) The torque $\vec{\tau}$ is given by the cross product of the position vector $\vec{r}$ and the force vector $\vec{F}$:
$\vec{\tau} = \vec{r} \times \vec{F}$
Using the determinant form for the cross product:
$\vec{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 2 \\ 3 & 4 & -2 \end{vmatrix}$
Expanding the determinant:
$\vec{\tau} = \hat{i} [(1)(-2) - (2)(4)] - \hat{j} [(2)(-2) - (2)(3)] + \hat{k} [(2)(4) - (1)(3)]$
$\vec{\tau} = \hat{i} [-2 - 8] - \hat{j} [-4 - 6] + \hat{k} [8 - 3]$
$\vec{\tau} = -10 \hat{i} + 10 \hat{j} + 5 \hat{k}$

(C) The torque $\vec{\tau}$ is given by the cross product of the position vector $\vec{r}$ and the force vector $\vec{F}$:
$\vec{\tau} = \vec{r} \times \vec{F}$
Using the determinant form for the cross product:
$\vec{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 2 & 1 \\ 5 & 3 & -7 \end{vmatrix}$
Expanding the determinant:
$\vec{\tau} = \hat{i} [(2)(-7) - (1)(3)] - \hat{j} [(2)(-7) - (1)(5)] + \hat{k} [(2)(3) - (2)(5)]$
$\vec{\tau} = \hat{i} [-14 - 3] - \hat{j} [-14 - 5] + \hat{k} [6 - 10]$
$\vec{\tau} = -17 \hat{i} + 19 \hat{j} - 4 \hat{k}$

(B) The net torque $\tau_{\text{net}}$ is the sum of the torques produced by each force about the central axis.
$1$. Torque due to force $F$ at the outer rim (at $45^{\circ}$): The tangential component of this force is $F \cos(45^{\circ})$. The torque is $\tau_1 = R \cdot F \cos(45^{\circ}) = R \cdot F \cdot \frac{1}{\sqrt{2}} \approx 0.707 RF$.
$2$. Torque due to force $F$ at the inner axle: This force is applied tangentially at radius $R/2$. The torque is $\tau_2 = \frac{R}{2} \cdot F = 0.5 RF$.
$3$. Torque due to force $2F$ at the outer rim: This force is applied tangentially at radius $R$. The torque is $\tau_3 = R \cdot 2F = 2 RF$.
All these torques act in the same rotational sense (counter-clockwise). Therefore,the net torque is:
$\tau_{\text{net}} = \tau_1 + \tau_2 + \tau_3 = 0.707 RF + 0.5 RF + 2 RF = 3.207 RF$.
Rounding to the nearest value,the magnitude of the net torque is nearly $3.2 FR$.

(C) The torque (moment of force) is given by the cross product of the position vector and the force vector: $\vec{\tau} = \vec{r} \times \vec{F}$.
Given,the force vector is $\vec{F} = 20 \hat{i} \, N$.
The position vector $\vec{r}$ of the point of application $(3, 0, 0)$ with respect to the point $(0, 2, 0)$ is $\vec{r} = (3 - 0) \hat{i} + (0 - 2) \hat{j} + (0 - 0) \hat{k} = (3 \hat{i} - 2 \hat{j}) \, m$.
Now,calculating the cross product:
$\vec{\tau} = (3 \hat{i} - 2 \hat{j}) \times (20 \hat{i})$
$\vec{\tau} = 3 \hat{i} \times 20 \hat{i} - 2 \hat{j} \times 20 \hat{i}$
Since $\hat{i} \times \hat{i} = 0$ and $\hat{j} \times \hat{i} = -\hat{k}$:
$\vec{\tau} = 0 - 40 (-\hat{k}) = 40 \hat{k} \, N \cdot m$.
The magnitude of the torque is $|\vec{\tau}| = 40 \, N \cdot m$.

(D) The position vector $\vec{r}$ of the point of application relative to the point about which torque is calculated is given by $\vec{r} = \vec{r}_1 - \vec{r}_2$.
$\vec{r} = (2 \hat{i} + 4 \hat{j} + 7 \hat{k}) - (\hat{i} + 2 \hat{j} + 3 \hat{k}) = \hat{i} + 2 \hat{j} + 4 \hat{k} \, m$.
The torque $\vec{\tau}$ is defined as $\vec{\tau} = \vec{r} \times \vec{F}$.
$\vec{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 4 \\ 2 & 3 & -5 \end{vmatrix}$.
Calculating the determinant: $\vec{\tau} = \hat{i} [ (2)(-5) - (4)(3) ] - \hat{j} [ (1)(-5) - (4)(2) ] + \hat{k} [ (1)(3) - (2)(2) ]$.
$\vec{\tau} = \hat{i} [ -10 - 12 ] - \hat{j} [ -5 - 8 ] + \hat{k} [ 3 - 4 ]$.
$\vec{\tau} = -22 \hat{i} + 13 \hat{j} - \hat{k} \, N m$.

(C) The torque $\vec{\tau}$ about a point is given by $\vec{\tau} = \vec{r} \times \vec{F}$.
Here,the force $\vec{F} = -P \hat{k}$ acts at the origin $(0, 0, 0)$.
The position vector $\vec{r}$ of the origin with respect to the point $(2, -3)$ is $\vec{r} = (0 - 2)\hat{i} + (0 - (-3))\hat{j} = -2\hat{i} + 3\hat{j}$.
Now,calculate the cross product:
$\vec{\tau} = (-2\hat{i} + 3\hat{j}) \times (-P\hat{k})$
$\vec{\tau} = -P [(-2)(\hat{i} \times \hat{k}) + 3(\hat{j} \times \hat{k})]$
Since $\hat{i} \times \hat{k} = -\hat{j}$ and $\hat{j} \times \hat{k} = \hat{i}$,we get:
$\vec{\tau} = -P [(-2)(-\hat{j}) + 3(\hat{i})]$
$\vec{\tau} = -P [2\hat{j} + 3\hat{i}] = P(-3\hat{i} - 2\hat{j})$.
Comparing this with $P(a\hat{i} + b\hat{j})$,we get $a = -3$ and $b = -2$.
The ratio $\frac{a}{b} = \frac{-3}{-2} = \frac{3}{2}$.
Given $\frac{a}{b} = \frac{x}{2}$,we have $\frac{3}{2} = \frac{x}{2}$,which implies $x = 3$.

(C) The torque $\vec{\tau}$ is given by the cross product of the position vector $\vec{r}$ and the force vector $\vec{F}$.
$\vec{\tau} = \vec{r} \times \vec{F} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 2 & 1 & 2 \end{vmatrix}$
Expanding the determinant:
$\vec{\tau} = \hat{i}(1 \times 2 - 1 \times 1) - \hat{j}(1 \times 2 - 1 \times 2) + \hat{k}(1 \times 1 - 1 \times 2)$
$\vec{\tau} = \hat{i}(2 - 1) - \hat{j}(2 - 2) + \hat{k}(1 - 2)$
$\vec{\tau} = \hat{i} - 0\hat{j} - \hat{k} = \hat{i} - \hat{k}$.

(A) The torque $\vec{\tau}$ is given by the cross product of the position vector $\vec{r}$ and the force vector $\vec{F}$.
Given $\vec{r} = (1\hat{i} + 1\hat{j} + 1\hat{k}) \ m$ and $\vec{F} = (1\hat{i} - 1\hat{j} + 1\hat{k}) \ N$.
$\vec{\tau} = \vec{r} \times \vec{F} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 1 & -1 & 1 \end{vmatrix}$
Expanding the determinant:
$\vec{\tau} = \hat{i}(1 - (-1)) - \hat{j}(1 - 1) + \hat{k}(-1 - 1)$
$\vec{\tau} = \hat{i}(2) - \hat{j}(0) + \hat{k}(-2)$
$\vec{\tau} = 2\hat{i} - 2\hat{k} \ N \cdot m$
The torque in the $z$-direction is the component associated with the $\hat{k}$ unit vector,which is $-2 \ N \cdot m$.
The magnitude of this component is $|-2| = 2 \ N \cdot m$.

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

(C) Let the perpendicular distance from the center $O$ to each side of the equilateral triangle be $r$.
The torque $\tau$ due to a force $F$ is given by $\tau = F \cdot r$,where $r$ is the perpendicular distance from the axis of rotation to the line of action of the force.
Looking at the directions of the forces in the figure,the forces $\overrightarrow{F}_{1}$ and $\overrightarrow{F}_{2}$ create a torque in the same rotational sense (e.g.,clockwise) about point $O$,while $\overrightarrow{F}_{3}$ creates a torque in the opposite sense (e.g.,counter-clockwise).
For the total torque about point $O$ to be zero,the sum of the torques must be zero:
$\tau_{1} + \tau_{2} - \tau_{3} = 0$
$rF_{1} + rF_{2} - rF_{3} = 0$
Dividing by $r$ (since $r \neq 0$):
$F_{1} + F_{2} - F_{3} = 0$
$F_{3} = F_{1} + F_{2}$

(C) couple is defined as a pair of two equal and opposite forces acting at a certain distance from each other.
Since the two forces are equal in magnitude and opposite in direction,their vector sum (net force) is zero,which means there is no linear (translatory) acceleration.
However,because the forces act at different points,they create a torque about any point,which causes the body to rotate.
Therefore,a couple produces purely rotational motion.

(B) The torque $\vec{\tau}$ is given by the cross product of the position vector $\vec{r}$ and the force vector $\vec{F}$:
$\vec{\tau} = \vec{r} \times \vec{F}$
Given $\vec{r} = (4 \hat{i} - 3 \hat{j})$ and $\vec{F} = (5 \hat{i} - 10 \hat{j})$:
$\vec{\tau} = (4 \hat{i} - 3 \hat{j}) \times (5 \hat{i} - 10 \hat{j})$
Using the distributive property of the cross product:
$\vec{\tau} = 4 \hat{i} \times (5 \hat{i}) - 4 \hat{i} \times (10 \hat{j}) - 3 \hat{j} \times (5 \hat{i}) + 3 \hat{j} \times (10 \hat{j})$
Since $\hat{i} \times \hat{i} = 0$ and $\hat{j} \times \hat{j} = 0$,and knowing $\hat{i} \times \hat{j} = \hat{k}$ and $\hat{j} \times \hat{i} = -\hat{k}$:
$\vec{\tau} = 0 - 40(\hat{k}) - 15(-\hat{k}) + 0$
$\vec{\tau} = -40 \hat{k} + 15 \hat{k}$
$\vec{\tau} = -25 \hat{k} \text{ N m}$

(C) The torque $\tau$ required to open or close the door remains constant.
According to the principle of moments, $\tau = r_1 \times F_1 = r_2 \times F_2$.
Given $r_1 = 1.2 \,m$, $F_1 = 1 \,N$, and $r_2 = 0.2 \,m$.
Substituting these values: $1.2 \,m \times 1 \,N = 0.2 \,m \times F_2$.
$F_2 = \frac{1.2}{0.2} \,N = 6 \,N$.

(A) The point $P$ is given as $(1, -1)$,so its position vector is $\vec{r}_P = \hat{i} - \hat{j}$.
Since the force acts at the origin $O(0, 0)$,the position vector of the point of application relative to point $P$ is $\vec{r} = \vec{r}_O - \vec{r}_P = 0 - (\hat{i} - \hat{j}) = -\hat{i} + \hat{j}$.
The torque (moment of force) about point $P$ is given by $\vec{\tau} = \vec{r} \times \vec{F}$.
Substituting the values: $\vec{\tau} = (-\hat{i} + \hat{j}) \times (-F\hat{k})$.
Using the distributive property of cross product: $\vec{\tau} = F[(\hat{i} \times \hat{k}) - (\hat{j} \times \hat{k})]$.
Using the cross product relations $\hat{i} \times \hat{k} = -\hat{j}$ and $\hat{j} \times \hat{k} = \hat{i}$:
$\vec{\tau} = F[-\hat{j} - \hat{i}] = -F(\hat{i} + \hat{j})$.

(C) The moment of a couple is given by the product of the magnitude of one of the forces and the perpendicular distance between the lines of action of the two forces.
Let the angle between the rod and the direction of the force be $\theta = 30^{\circ}$.
The perpendicular distance between the two parallel forces $F$ acting at the ends of the rod of length $l$ is $d = l \sin \theta$.
The moment of the couple $\tau$ is given by:
$\tau = F \times d = F \times l \sin 30^{\circ}$
Given $\sin 30^{\circ} = 0.5 = \frac{1}{2}$.
Substituting the values:
$\tau = F \times l \times \frac{1}{2}$
$\tau = \frac{F l}{2}$
Rearranging to solve for $F$:
$F = \frac{2 \tau}{l}$

(A) Given: $\vec{F} = 5\hat{i} + 2\hat{j} - 5\hat{k}$ and $\vec{r} = \hat{i} - 2\hat{j} + \hat{k}$.
Torque $\vec{\tau}$ is defined as the cross product of the position vector and the force vector: $\vec{\tau} = \vec{r} \times \vec{F}$.
Using the determinant method:
$\vec{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 1 \\ 5 & 2 & -5 \end{vmatrix}$
Expanding the determinant:
$\vec{\tau} = \hat{i}((-2)(-5) - (1)(2)) - \hat{j}((1)(-5) - (1)(5)) + \hat{k}((1)(2) - (-2)(5))$
$\vec{\tau} = \hat{i}(10 - 2) - \hat{j}(-5 - 5) + \hat{k}(2 + 10)$
$\vec{\tau} = 8\hat{i} - (-10)\hat{j} + 12\hat{k}$
$\vec{\tau} = 8\hat{i} + 10\hat{j} + 12\hat{k}$.

(A) Given that,
$F_1 = A \hat{j}, r_1 = a \hat{i}$
$F_2 = B \hat{i}, r_2 = b \hat{j}$
The moment of force (torque) is given by $\tau = r \times F$.
For the first force:
$\tau_1 = r_1 \times F_1 = (a \hat{i}) \times (A \hat{j}) = a A (\hat{i} \times \hat{j}) = a A \hat{k}$
For the second force:
$\tau_2 = r_2 \times F_2 = (b \hat{j}) \times (B \hat{i}) = b B (\hat{j} \times \hat{i}) = b B (-\hat{k}) = -b B \hat{k}$
The net moment relative to $O$ is:
$\tau = \tau_1 + \tau_2 = a A \hat{k} - b B \hat{k} = (a A - b B) \hat{k}$