Let the vertices of the equilateral triangle be O, A, and B. Let the side length be $L = 0.5 \text{ m}$. We place the origin at vertex O and the x-axis along the side OA. The coordinates of the vertices are:

• O: (0, 0)
• A: ($L$, 0) = (0.5, 0)
• B: The x-coordinate of B is $L \cos 60^\circ = L/2 = 0.5/2 = 0.25 \text{ m}$. The y-coordinate of B is $L \sin 60^\circ = L (\sqrt{3}/2) = 0.5 (\sqrt{3}/2) = 0.25\sqrt{3} \text{ m}$. So, B: (0.25, $0.25\sqrt{3}$).

Let the masses be $m_1 = 100 \text{ g}$, $m_2 = 150 \text{ g}$, $m_3 = 200 \text{ g}$, located at O, A, and B respectively.

Total mass $M = m_1 + m_2 + m_3 = 100 + 150 + 200 = 450 \text{ g}$.

The coordinates of the centre of mass (X, Y) are:

$X = \frac{m_1 x_1 + m_2 x_2 + m_3 x_3}{M} = \frac{100(0) + 150(0.5) + 200(0.25)}{450} = \frac{0 + 75 + 50}{450} = \frac{125}{450} \text{ m}$.

$X = \frac{125}{450} = \frac{25 \times 5}{25 \times 18} = \frac{5}{18} \text{ m}$.

$Y = \frac{m_1 y_1 + m_2 y_2 + m_3 y_3}{M} = \frac{100(0) + 150(0) + 200(0.25\sqrt{3})}{450} = \frac{0 + 0 + 50\sqrt{3}}{450} = \frac{50\sqrt{3}}{450} \text{ m}$.

$Y = \frac{50\sqrt{3}}{450} = \frac{50\sqrt{3}}{50 \times 9} = \frac{\sqrt{3}}{9} \text{ m}$.

The centre of mass is located at coordinates ($\frac{5}{18} \text{ m}$, $\frac{\sqrt{3}}{9} \text{ m}$).

The centre of mass does not coincide with the geometric centre (centroid) of the triangle because the masses at the vertices are not equal. The centroid would be at the average of the coordinates: $(\frac{0+0.5+0.25}{3}, \frac{0+0+0.25\sqrt{3}}{3}) = (\frac{0.75}{3}, \frac{0.25\sqrt{3}}{3}) = (0.25, \frac{0.25\sqrt{3}}{3})$. Our calculated CM is at X = 5/18 $\approx$ 0.278 m and Y = $\sqrt{3}/9 \approx$ 0.192 m, whereas the centroid is at X = 0.25 m and Y = $0.25\sqrt{3}/3 \approx$ 0.144 m. The CM is shifted towards the heavier masses (at A and B).

Consider a triangular lamina (a thin, flat triangle) LMN. Assume it has uniform mass density. We can use symmetry to find the centre of mass. Subdivide the triangle into many narrow strips parallel to the base MN.

For each strip, since it is thin and uniform, its centre of mass is at its midpoint. If we connect the midpoints of all such parallel strips, we get the median LP (where P is the midpoint of MN). Since the centre of mass of each strip lies on the median LP, the centre of mass of the entire triangle must lie on the median LP.

Similarly, we can subdivide the triangle into strips parallel to the other two sides, LN and LM. The centre of mass must also lie on the medians MQ and NR, respectively.

Since the centre of mass must lie on all three medians simultaneously, it must be located at the point where the three medians intersect. This point is the centroid of the triangle.

The centre of mass of a uniform triangular lamina is at its centroid.

We can divide the L-shaped lamina into simpler parts whose centers of mass are easy to find. The lamina has overall dimensions of 2m x 2m with a 1m x 1m square cut out from one corner. Let's place the origin at the corner where the shape is formed. The L-shape can be thought of as being formed by three 1m x 1m squares of uniform density.

Since the lamina is uniform and has a total mass of 3 kg, each 1m x 1m square has a mass of 1 kg.

The centre of mass of each uniform square is at its geometric center.

• Square 1 (bottom left): Coordinates of CM, C₁ = (1/2 m, 1/2 m). Mass $m_1 = 1 \text{ kg}$.
• Square 2 (bottom right): Coordinates of CM, C₂ = (1 + 1/2 m, 1/2 m) = (3/2 m, 1/2 m). Mass $m_2 = 1 \text{ kg}$.
• Square 3 (top left): Coordinates of CM, C₃ = (1/2 m, 1 + 1/2 m) = (1/2 m, 3/2 m). Mass $m_3 = 1 \text{ kg}$.

Now we find the centre of mass of these three point masses ($m_1, m_2, m_3$) located at ($C_1, C_2, C_3$). Total mass $M = m_1 + m_2 + m_3 = 1 + 1 + 1 = 3 \text{ kg}$.

$X = \frac{m_1 x_1 + m_2 x_2 + m_3 x_3}{M} = \frac{1(1/2) + 1(3/2) + 1(1/2)}{3} = \frac{1/2 + 3/2 + 1/2}{3} = \frac{5/2}{3} = \frac{5}{6} \text{ m}$.

$Y = \frac{m_1 y_1 + m_2 y_2 + m_3 y_3}{M} = \frac{1(1/2) + 1(1/2) + 1(3/2)}{3} = \frac{1/2 + 1/2 + 3/2}{3} = \frac{5/2}{3} = \frac{5}{6} \text{ m}$.

The centre of mass of the L-shaped lamina is located at coordinates ($\frac{5}{6} \text{ m}$, $\frac{5}{6} \text{ m}$).

We could have guessed that the centre of mass lies on the line y=x because the L-shape is symmetrical about the line y=x. Any point (x,y) on the shape corresponds to a point (y,x) on the shape if the line segment connecting them is perpendicular to y=x and bisected by y=x. While not strict point reflection symmetry about the origin, the shape's symmetry about the y=x line implies X=Y for the CM.

If the three squares had different masses, say $m_1$, $m_2$, $m_3$, you would use the same formula for X and Y, but with the different mass values. For example, $X = \frac{m_1(1/2) + m_2(3/2) + m_3(1/2)}{m_1 + m_2 + m_3}$.

Given vectors: $\mathbf{a} = 3\hat{\mathbf{i}} + 4\hat{\mathbf{j}} - 5\hat{\mathbf{k}}$ and $\mathbf{b} = -2\hat{\mathbf{i}} + \hat{\mathbf{j}} + 3\hat{\mathbf{k}}$.

Scalar Product (Dot Product):

$\mathbf{a} \cdot \mathbf{b} = (3\hat{\mathbf{i}} + 4\hat{\mathbf{j}} - 5\hat{\mathbf{k}}) \cdot (-2\hat{\mathbf{i}} + \hat{\mathbf{j}} + 3\hat{\mathbf{k}})$

Using $\hat{\mathbf{i}} \cdot \hat{\mathbf{i}} = 1$, $\hat{\mathbf{j}} \cdot \hat{\mathbf{j}} = 1$, $\hat{\mathbf{k}} \cdot \hat{\mathbf{k}} = 1$, and all other dot products of unit vectors are 0:

$\mathbf{a} \cdot \mathbf{b} = (3)(-2) + (4)(1) + (-5)(3) = -6 + 4 - 15 = -17$.

The scalar product is -17.

Vector Product (Cross Product):

$\mathbf{a} \times \mathbf{b} = (3\hat{\mathbf{i}} + 4\hat{\mathbf{j}} - 5\hat{\mathbf{k}}) \times (-2\hat{\mathbf{i}} + \hat{\mathbf{j}} + 3\hat{\mathbf{k}})$

Using the determinant form:

$\mathbf{a} \times \mathbf{b} = \begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 3 & 4 & -5 \\ -2 & 1 & 3 \end{vmatrix}$

$\mathbf{a} \times \mathbf{b} = \hat{\mathbf{i}} [(4)(3) - (-5)(1)] - \hat{\mathbf{j}} [(3)(3) - (-5)(-2)] + \hat{\mathbf{k}} [(3)(1) - (4)(-2)]$

$\mathbf{a} \times \mathbf{b} = \hat{\mathbf{i}} [12 - (-5)] - \hat{\mathbf{j}} [9 - 10] + \hat{\mathbf{k}} [3 - (-8)]$

$\mathbf{a} \times \mathbf{b} = \hat{\mathbf{i}} [12 + 5] - \hat{\mathbf{j}} [9 - 10] + \hat{\mathbf{k}} [3 + 8]$

$\mathbf{a} \times \mathbf{b} = 17\hat{\mathbf{i}} - (-1)\hat{\mathbf{j}} + 11\hat{\mathbf{k}}$

$\mathbf{a} \times \mathbf{b} = 17\hat{\mathbf{i}} + \hat{\mathbf{j}} + 11\hat{\mathbf{k}}$.

The vector product is $17\hat{\mathbf{i}} + \hat{\mathbf{j}} + 11\hat{\mathbf{k}}$.

Equation (6.36) is $\omega = \omega_0 + \alpha t$. This is the kinematic equation for angular velocity as a function of time, assuming constant angular acceleration $\alpha$. We start from the definition of angular acceleration:

$\alpha = \frac{d\omega}{dt}$.

Given that the angular acceleration $\alpha$ is uniform (constant).

Rearrange the equation: $d\omega = \alpha dt$.

Integrate both sides. Assume that at time $t=0$, the angular velocity is $\omega_0$, and at time $t$, the angular velocity is $\omega$:

$\int_{\omega_0}^{\omega} d\omega = \int_{0}^{t} \alpha dt$.

Since $\alpha$ is constant, it can be taken out of the integral on the right side:

$\int_{\omega_0}^{\omega} d\omega = \alpha \int_{0}^{t} dt$.

Evaluate the integrals:

$[\omega]_{\omega_0}^{\omega} = \alpha [t]_{0}^{t}$.

$\omega - \omega_0 = \alpha (t - 0)$.

$\omega - \omega_0 = \alpha t$.

Rearrange to solve for $\omega$:

$\omega = \omega_0 + \alpha t$.

This is the kinematic equation relating angular velocity, initial angular velocity, angular acceleration, and time for motion with constant angular acceleration. The other kinematic equations for rotation can be similarly derived by integrating angular velocity with respect to time.

Initial angular speed $\omega_0 = 1200 \text{ rpm}$ (revolutions per minute). Final angular speed $\omega = 3120 \text{ rpm}$. Time taken $t = 16 \text{ s}$. Acceleration is uniform.

First, convert angular speeds from rpm to rad/s (SI units). 1 revolution = $2\pi$ radians, 1 minute = 60 seconds.

$\omega_0 = 1200 \frac{\text{rev}}{\text{min}} \times \frac{2\pi \text{ rad}}{1 \text{ rev}} \times \frac{1 \text{ min}}{60 \text{ s}} = \frac{1200 \times 2\pi}{60} \text{ rad/s} = 20 \times 2\pi \text{ rad/s} = 40\pi \text{ rad/s}$.

$\omega = 3120 \frac{\text{rev}}{\text{min}} \times \frac{2\pi \text{ rad}}{1 \text{ rev}} \times \frac{1 \text{ min}}{60 \text{ s}} = \frac{3120 \times 2\pi}{60} \text{ rad/s} = 52 \times 2\pi \text{ rad/s} = 104\pi \text{ rad/s}$.

(i) What is the angular acceleration ($\alpha$), assuming it to be uniform? We use the kinematic equation $\omega = \omega_0 + \alpha t$.

Rearrange to solve for $\alpha$: $\alpha = \frac{\omega - \omega_0}{t}$.

$\alpha = \frac{104\pi \text{ rad/s} - 40\pi \text{ rad/s}}{16 \text{ s}} = \frac{64\pi \text{ rad/s}}{16 \text{ s}} = 4\pi \text{ rad/s}^2$.

The angular acceleration is $4\pi \text{ rad/s}^2$.

(ii) How many revolutions does the engine make during this time? We need to find the total angular displacement ($\theta$). We can use the kinematic equation $\theta = \theta_0 + \omega_0 t + \frac{1}{2}\alpha t^2$. Assume initial angular position $\theta_0 = 0$.

$\theta = 0 + (40\pi \text{ rad/s})(16 \text{ s}) + \frac{1}{2}(4\pi \text{ rad/s}^2)(16 \text{ s})^2$.

$\theta = 640\pi \text{ rad} + \frac{1}{2}(4\pi)(256) \text{ rad}$.

$\theta = 640\pi + 2\pi(256) = 640\pi + 512\pi = 1152\pi \text{ rad}$.

The total angular displacement is $1152\pi$ radians. To find the number of revolutions, divide by $2\pi$ radians per revolution.

Number of revolutions = $\frac{1152\pi \text{ rad}}{2\pi \text{ rad/revolution}} = \frac{1152}{2} = 576$ revolutions.

The engine makes 576 revolutions during this time.

Alternatively, use the average angular velocity method: $\theta = \theta_0 + \bar{\omega} t$. For uniform acceleration, $\bar{\omega} = \frac{\omega_0 + \omega}{2}$.

$\bar{\omega} = \frac{40\pi + 104\pi}{2} = \frac{144\pi}{2} = 72\pi \text{ rad/s}$.

$\theta = 0 + (72\pi \text{ rad/s})(16 \text{ s}) = 1152\pi \text{ rad}$.

Number of revolutions = $1152\pi / (2\pi) = 576$. This confirms the result.

Given position vector $\mathbf{r} = \hat{\mathbf{i}} - \hat{\mathbf{j}} + \hat{\mathbf{k}}$ and force vector $\mathbf{F} = 4\hat{\mathbf{i}} - 5\hat{\mathbf{j}} + 3\hat{\mathbf{k}}$.

The torque $\boldsymbol{\tau}$ about the origin is given by the vector product $\boldsymbol{\tau} = \mathbf{r} \times \mathbf{F}$.

Using the determinant form:

$\boldsymbol{\tau} = \begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & -1 & 1 \\ 4 & -5 & 3 \end{vmatrix}$

$\boldsymbol{\tau} = \hat{\mathbf{i}} [(-1)(3) - (1)(-5)] - \hat{\mathbf{j}} [(1)(3) - (1)(4)] + \hat{\mathbf{k}} [(1)(-5) - (-1)(4)]$

$\boldsymbol{\tau} = \hat{\mathbf{i}} [-3 - (-5)] - \hat{\mathbf{j}} [3 - 4] + \hat{\mathbf{k}} [-5 - (-4)]$

$\boldsymbol{\tau} = \hat{\mathbf{i}} [-3 + 5] - \hat{\mathbf{j}} [-1] + \hat{\mathbf{k}} [-5 + 4]$

$\boldsymbol{\tau} = 2\hat{\mathbf{i}} + \hat{\mathbf{j}} - \hat{\mathbf{k}}$.

The torque about the origin is $2\hat{\mathbf{i}} + \hat{\mathbf{j}} - \hat{\mathbf{k}}$.

Consider a single particle of mass $m$ moving with a constant velocity $\mathbf{v}$. We want to show its angular momentum about any arbitrary point O is constant. Let the position vector of the particle relative to O at time $t$ be $\mathbf{r}$. The linear momentum of the particle is $\mathbf{p} = m\mathbf{v}$. The angular momentum about O is $\mathbf{l} = \mathbf{r} \times \mathbf{p} = \mathbf{r} \times m\mathbf{v}$. Since $\mathbf{v}$ is constant, $m\mathbf{v}$ is also constant.

The equation of motion for the particle is simply $\frac{d\mathbf{v}}{dt} = \mathbf{0}$ (since velocity is constant). By Newton's Second Law, $\mathbf{F} = m\mathbf{a} = m(\mathbf{0}) = \mathbf{0}$. So, the net force on the particle is zero.

The time rate of change of angular momentum is equal to the torque acting on the particle: $\frac{d\mathbf{l}}{dt} = \boldsymbol{\tau}$.

The torque about the origin O is $\boldsymbol{\tau} = \mathbf{r} \times \mathbf{F}_{net}$. Since $\mathbf{F}_{net} = \mathbf{0}$, the torque $\boldsymbol{\tau} = \mathbf{r} \times \mathbf{0} = \mathbf{0}$.

Since the total torque acting on the particle is zero, the time rate of change of its angular momentum is zero: $\frac{d\mathbf{l}}{dt} = \mathbf{0}$.

This means that the angular momentum $\mathbf{l}$ is a constant vector throughout the motion.

Alternatively, consider the magnitude $l = rp \sin \theta$. As the particle moves with constant velocity $\mathbf{v}$, its path is a straight line. The quantity $r \sin \theta$ represents the perpendicular distance from the point O to the line of motion of the particle. This perpendicular distance remains constant as the particle moves along the straight line. The linear momentum $p = mv$ is also constant in magnitude and direction. Thus, the magnitude $l = (r \sin \theta) p$ is constant. The direction of $\mathbf{l} = \mathbf{r} \times \mathbf{p}$ is perpendicular to the plane containing O and the line of motion, and this direction is also constant. Therefore, the angular momentum vector remains constant.

A couple consists of two forces $\mathbf{F}$ and $-\mathbf{F}$ acting at points A and B respectively on a rigid body. Let $\mathbf{r}_1$ and $\mathbf{r}_2$ be the position vectors of points A and B relative to some arbitrary origin O. The moment of the couple about the origin O is the sum of the torques of the two forces about O.

Torque of force $\mathbf{F}$ at B about O is $\boldsymbol{\tau}_B = \mathbf{r}_2 \times \mathbf{F}$.

Torque of force $-\mathbf{F}$ at A about O is $\boldsymbol{\tau}_A = \mathbf{r}_1 \times (-\mathbf{F}) = -\mathbf{r}_1 \times \mathbf{F}$.

The total moment of the couple about O is $\boldsymbol{\tau}_{total} = \boldsymbol{\tau}_A + \boldsymbol{\tau}_B = -\mathbf{r}_1 \times \mathbf{F} + \mathbf{r}_2 \times \mathbf{F}$.

Using the distributive property of the cross product: $\boldsymbol{\tau}_{total} = (\mathbf{r}_2 - \mathbf{r}_1) \times \mathbf{F}$.

The vector $(\mathbf{r}_2 - \mathbf{r}_1)$ is the displacement vector from point A to point B, $\vec{AB}$. Let $\mathbf{d} = \mathbf{r}_2 - \mathbf{r}_1$.

So, $\boldsymbol{\tau}_{total} = \mathbf{d} \times \mathbf{F}$.

The moment of the couple is the vector product of the displacement vector between the points of application of the forces and the force vector $\mathbf{F}$. This expression $\mathbf{d} \times \mathbf{F}$ depends only on the relative position of the points of application of the forces ($\mathbf{d}$) and the force vector $\mathbf{F}$, not on the choice of the origin O. Therefore, the moment of a couple does not depend on the point about which the moments are taken.

Let the bar be AB, with length $L = 70 \text{ cm} = 0.70 \text{ m}$. Mass of the bar $M_{bar} = 4.00 \text{ kg}$. The bar is uniform and homogeneous, so its center of gravity (CG) is at its midpoint G, which is at $L/2 = 70/2 = 35 \text{ cm}$ from either end (A or B). The weight of the bar acts at G: $W_{bar} = M_{bar} g = (4.00 \text{ kg})g$. Let's use $g=9.8 \text{ m/s}^2$ unless specified as 10 for numerical problems. The header of this chapter says use $g=10$. So $W_{bar} = 4.00 \times 10 = 40.0 \text{ N}$.

Two knife-edges K₁ and K₂ are placed 10 cm from each end. K₁ is 10 cm from A, and K₂ is 10 cm from B. The bar is supported on these knife-edges, so they exert upward normal reaction forces R₁ and R₂ on the bar at K₁ and K₂ respectively.

A load of mass $M_{load} = 6.00 \text{ kg}$ is suspended at P, which is 30 cm from one end (let's say from end A). The weight of the load acts downwards at P: $W_{load} = M_{load} g = (6.00 \text{ kg})(10 \text{ m s}^{-2}) = 60.0 \text{ N}$.

The bar is in mechanical equilibrium, meaning it is in both translational and rotational equilibrium.

For translational equilibrium, the net force in the vertical direction is zero. Upward forces = Downward forces. R₁ and R₂ are upwards. $W_{bar}$ and $W_{load}$ are downwards.

$R_1 + R_2 - W_{bar} - W_{load} = 0$.

$R_1 + R_2 = W_{bar} + W_{load} = 40.0 \text{ N} + 60.0 \text{ N} = 100.0 \text{ N}$. (Equation 1)

For rotational equilibrium, the net torque about any point is zero. Let's choose the point K₁ as the pivot to calculate moments. Distances from K₁:

• Distance of R₁ from K₁ = 0 (R₁ acts at K₁). Moment = 0.
• Distance of $W_{bar}$ (acting at G) from K₁: G is 35 cm from A. K₁ is 10 cm from A. So K₁G = 35 - 10 = 25 cm = 0.25 m. $W_{bar}$ acts downwards. Moment about K₁ due to $W_{bar}$ is clockwise (tending to rotate the bar clockwise about K₁), so it's negative: $-W_{bar} \times 0.25$.
• Distance of $W_{load}$ (acting at P) from K₁: P is 30 cm from A. K₁ is 10 cm from A. So K₁P = 30 - 10 = 20 cm = 0.20 m. $W_{load}$ acts downwards. Moment about K₁ due to $W_{load}$ is clockwise, so negative: $-W_{load} \times 0.20$.
• Distance of R₂ (acting at K₂) from K₁: K₂ is 10 cm from B. B is 70 cm from A. K₂ is at $70-10 = 60$ cm from A. K₁ is 10 cm from A. So K₁K₂ = 60 - 10 = 50 cm = 0.50 m. R₂ acts upwards. Moment about K₁ due to R₂ is anticlockwise, so positive: $+R_2 \times 0.50$.

For rotational equilibrium about K₁:

$0 - W_{bar}(0.25) - W_{load}(0.20) + R_2(0.50) = 0$.

$-40.0(0.25) - 60.0(0.20) + 0.50 R_2 = 0$.

$-10.0 - 12.0 + 0.50 R_2 = 0$.

$-22.0 + 0.50 R_2 = 0$.

$0.50 R_2 = 22.0$.

$R_2 = \frac{22.0}{0.50} = 44.0 \text{ N}$.

Substitute the value of R₂ back into Equation 1:

$R_1 + 44.0 \text{ N} = 100.0 \text{ N}$.

$R_1 = 100.0 - 44.0 = 56.0 \text{ N}$.

The reactions at the knife-edges are R₁ = 56.0 N and R₂ = 44.0 N.

(Let's check if the moment about G is zero. Distances from G: K₁ is 25cm left (-0.25m). R₁ upwards. Moment is anticlockwise, $+R_1(0.25)$. P is 5cm right (+0.05m). $W_{load}$ downwards. Moment is anticlockwise, $+W_{load}(0.05)$. K₂ is 25cm right (+0.25m). R₂ upwards. Moment is clockwise, $-R_2(0.25)$. $W_{bar}$ is at G, moment is 0. Net moment about G: $R_1(0.25) + W_{load}(0.05) - R_2(0.25) = 56.0(0.25) + 60.0(0.05) - 44.0(0.25) = 14.0 + 3.0 - 11.0 = 17.0$. This should be zero. Error in distances or sign convention in the example solution explanation/figure distances. Let's re-check distances from K₁ to G, P, K₂. A=0, K1=0.1m, G=0.35m, P=0.3m, B=0.7m, K2=0.6m. K1G = 0.35-0.1 = 0.25m. K1P = 0.3-0.1 = 0.2m. K1K2 = 0.6-0.1 = 0.5m. These distances are correct. Let's check moments about G. Distances from G: K₁ is 25cm left (-0.25m). R₁ upwards. Torque is $(+R_1 \times 0.25)$ (anticlockwise). P is 5cm left (-0.05m). $W_{load}$ downwards. Torque is $(-W_{load} \times 0.05)$ (clockwise). K₂ is 25cm right (+0.25m). R₂ upwards. Torque is $(+R_2 \times 0.25)$ (anticlockwise). Net moment about G: $R_1(0.25) - W_{load}(0.05) + R_2(0.25) = 56.0(0.25) - 60.0(0.05) + 44.0(0.25) = 14.0 - 3.0 + 11.0 = 22.0$. This is not zero. The distances from G should be vector distances along the rod. Moments are $r \times F$. If the rod is along the x-axis, forces are in the y-direction. $r_i = (x_i - x_G)\hat{i}$. $F_i = F_i \hat{j}$. Torque is $(x_i - x_G) \hat{i} \times F_i \hat{j} = (x_i - x_G) F_i (\hat{i} \times \hat{j}) = (x_i - x_G) F_i \hat{k}$. Moments are along z-axis. Take the z-component (magnitude). Origin at A=0. G=0.35m, K1=0.1m, P=0.3m, K2=0.6m. Forces: R1 at 0.1m, -Wload at 0.3m, -Wbar at 0.35m, R2 at 0.6m. Moments about A: $R_1(0.1) - W_{load}(0.3) - W_{bar}(0.35) + R_2(0.6) = 0$. $R_1(0.1) - 60(0.3) - 40(0.35) + R_2(0.6) = 0.1R_1 - 18 - 14 + 0.6R_2 = 0.1R_1 + 0.6R_2 = 32$. From R1+R2=100, R1 = 100-R2. $0.1(100-R2) + 0.6R2 = 32. 10 - 0.1R2 + 0.6R2 = 32. 0.5R2 = 22. R2 = 44N$. $R1 = 100-44=56N$. This matches the results R1=54.88N, R2=43.12N if g=9.8. With g=10, R1=56N, R2=44N. So calculation taking moments about A is correct. The example solution's calculation taking moments about G seems to have sign errors or distance errors. Let's recheck moments about G (origin at G=0). K1 at -0.25m, P at -0.05m, K2 at +0.25m. Forces: R1 at -0.25m, -Wload at -0.05m, R2 at +0.25m. Moment = $x_i F_i$. $R_1$ upwards is positive moment. $-R_1(0.25)$ (R1 is at -0.25). No, moment is $r \times F$. If F is in y direction and r in x direction, $r=x\hat{i}$, $F=F\hat{j}$, moment is $xF(\hat{i} \times \hat{j}) = xF\hat{k}$. So moment magnitude is $x_i F_i$. $x_i$ is coordinate. $F_i$ is force component. $F_i$ is +ve for upwards, -ve for downwards. Coordinates are -0.25 (K1), -0.05 (P), +0.25 (K2). Forces are +R1, -Wload, +R2. Net moment about G: $(-0.25)R_1 + (-0.05)(-W_{load}) + (0.25)R_2 = -0.25R_1 + 0.05W_{load} + 0.25R_2 = 0$. $-0.25R_1 + 0.05(60) + 0.25R_2 = 0. -0.25R_1 + 3 + 0.25R_2 = 0. 0.25(R_2 - R_1) = -3. R_2 - R_1 = -12$. R1 - R2 = 12. This matches the example solution equation (iv) which gives R1-R2 = 11.76N (using g=9.8, 1.2*9.8 = 11.76). So the calculation using moments about G is also correct based on the equations derived in the example. My R1=56, R2=44 satisfies R1-R2=12. R1+R2=100. Yes. The numbers R1=54.88, R2=43.12 resulted from R1+R2=98, R1-R2=11.76 (using g=9.8). So it seems the example is consistent with g=9.8 used for R1+R2=98, but then used g=10 for the final forces. Let's use g=10 consistently).

Using g=10 throughout, R1+R2=100N. R1-R2=12N. Adding: 2R1=112, R1=56N. R2=100-56=44N.

The reactions are R₁ = 56.0 N and R₂ = 44.0 N.