Explain the concept of dimensional homogeneity in physical equations and determine the base units of K in the equation E = 1/2 rho A K v3 * N.

$$X_C = \frac{1}{2\pi f C}$$ $$X_C = \frac{1}{2\pi (50 \, \text{Hz}) (125 \times 10^{-6} \, \text{F})}$$ $$X_C = \frac{1}{0.0125\pi} \, \Omega$$ $$X_C \approx \frac{1}{0.03927} \, \Omega$$ $$X_C \approx 25.46 \, \Omega$$

$$Z = \sqrt{R^2 + (X_L - X_C)^2}$$ $$Z = \sqrt{(80 \, \Omega)^2 + (40\pi \, \Omega - \frac{1}{0.0125\pi} \, \Omega)^2}$$ $$Z = \sqrt{(80)^2 + (125.66 - 25.46)^2}$$ $$Z = \sqrt{6400 + (100.20)^2}$$ $$Z = \sqrt{6400 + 10040.04}$$ $$Z = \sqrt{16440.04}$$ $$Z \approx 128.22 \, \Omega$$

The impedance $Z$ is approximately $\boxed{\text{128.22 } \Omega}$.

$$I_{\text{rms}} = \frac{V_{\text{rms}}}{Z}$$ $$I_{\text{rms}} = \frac{220 \, \text{V}}{128.22 \, \Omega}$$ $$I_{\text{rms}} \approx 1.716 \, \text{A}$$

The RMS current is approximately $\boxed{\text{1.72 A}}$.

c) Step 1: Calculate the resonant frequency $f_0$.

$$f_0 = \frac{1}{2\pi\sqrt{LC}}$$ $$f_0 = \frac{1}{2\pi\sqrt{(0.4 \, \text{H})(125 \times 10^{-6} \, \text{F})}}$$ $$f_0 = \frac{1}{2\pi\sqrt{50 \times 10^{-6}}}$$ $$f_0 = \frac{1}{2\pi\sqrt{0.00005}}$$ $$f_0 = \frac{1}{2\pi (0.007071)}$$ $$f_0 = \frac{1}{0.044428}$$ $$f_0 \approx 22.51 \, \text{Hz}$$

The resonant frequency $f_0$ is approximately $\boxed{\text{22.5 Hz}}$.

At resonance, $X_L = X_C$, so the impedance $Z$ becomes equal to the resistance $R$.

$$Z_{\text{resonance}} = R = 80 \, \Omega$$ $$I_{\text{rms, resonance}} = \frac{V_{\text{rms}}}{R}$$ $$I_{\text{rms, resonance}} = \frac{220 \, \text{V}}{80 \, \Omega}$$ $$I_{\text{rms, resonance}} = 2.75 \, \text{A}$$

The current at resonance is $\boxed{\text{2.75 A}}$.

Question 3

a) The moment of a force (also known as torque) is the turning effect of a force about a pivot point. It is calculated as the product of the magnitude of the force and the perpendicular distance from the pivot to the line of action of the force. Its unit is Newton-meter (Nm).

b) One condition necessary for a body to be in equilibrium on a plane is that the net force acting on the body must be zero. This means the vector sum of all forces acting on the body in any direction (e.g., horizontal and vertical) must be zero.

c) Figure 1 shows a uniform plank of weight $6 \, \text{N}$ and length $a$, inclined at an angle $\theta$.
The system is in equilibrium.
Let's analyze the forces and distances from the pivot point. The pivot is located at a distance $\frac{a}{4}$ from the left end.

• The weight of the plank, $6 \, \text{N}$, acts at its center of gravity, which is at $\frac{a}{2}$ from the left end.
• The $16 \, \text{N}$ force acts downwards at the left end.
• The $4 \, \text{N}$ force acts downwards at the right end.

For equilibrium, the sum of clockwise moments about the pivot must equal the sum of anticlockwise moments about the pivot.

Distances from the pivot:

• For the $16 \, \text{N}$ force: The distance from the pivot is $\frac{a}{4}$. This force creates an anticlockwise moment.
• For the $6 \, \text{N}$ weight: The center of gravity is at $\frac{a}{2}$ from the left end. The pivot is at $\frac{a}{4}$. So, the distance from the pivot is $\frac{a}{2} - \frac{a}{4} = \frac{a}{4}$. This force creates a clockwise moment.
• For the $4 \, \text{N}$ force: The distance from the pivot is $a - \frac{a}{4} = \frac{3a}{4}$. This force creates a clockwise moment.

The forces are acting vertically, but the plank is inclined at an angle $\theta$. When calculating moments, we need the perpendicular distance from the pivot to the line of action of the force. If the forces are vertical and the plank is inclined, the effective lever arm for each force will be the horizontal component of the distance from the pivot to the point where the force is applied. However, since all forces are parallel (vertical) and the pivot is a point, we can consider the distances along the plank, and the $\sin\theta$ factor for each moment will cancel out. Alternatively, we can consider the perpendicular distance to the line of action of the force.

Let's assume the forces are perpendicular to the plank for simplicity, or that the angle $\theta$ is accounted for by the support. Given the diagram, the forces are shown as vertical. The distances $a/4$, $a/2$, $a$ are along the plank. The perpendicular distance from the pivot to the line of action of a vertical force at a point $x$ along the plank is $x \cos\theta$. However, since all forces are vertical, the $\cos\theta$ term will cancel out when equating moments. So we can use the distances along the plank directly.

Anticlockwise moment:

$$M_{\text{anti-clockwise}} = 16 \, \text{N} \times \frac{a}{4}$$

Clockwise moments:

$$M_{\text{clockwise}} = (6 \, \text{N} \times \frac{a}{4}) + (4 \, \text{N} \times \frac{3a}{4})$$

For equilibrium:

$$M_{\text{anti-clockwise}} = M_{\text{clockwise}}$$ $$16 \times \frac{a}{4} = (6 \times \frac{a}{4}) + (4 \times \frac{3a}{4})$$

Divide by $a$ (since $a \neq 0$):

$$\frac{16}{4} = \frac{6}{4} + \frac{12}{4}$$ $$4 = \frac{6+12}{4}$$ $$4 = \frac{18}{4}$$ $$4 = 4.5$$

This equation $4 = 4.5$ is false, which indicates that the system as described with the given forces and distances cannot be in equilibrium if the pivot is the only support providing an upward force. The question asks to "Calculate $\theta$ given that the system is in equilibrium." This implies that the angle $\theta$ might be relevant to the forces or distances, or there's another force not explicitly shown.

Let's re-examine the diagram. The $60^\circ$ angle is shown at the left end, but it's not clear what it refers to. The angle $\theta$ is the inclination of the plank. The forces $16 \, \text{N}$ and $4 \, \text{N}$ are shown as vertical. The weight $6 \, \text{N}$ is also vertical. The pivot provides an upward normal force.

If the question implies that the $16 \, \text{N}$ and $4 \, \text{N}$ forces are components or that the $60^\circ$ angle is relevant to the $16 \, \text{N}$ force, it's not clearly stated. Assuming the forces are as depicted (vertical) and the distances are along the plank, the calculation above shows an imbalance.

However, the question asks to calculate $\theta$. This suggests $\theta$ must play a role. The only way $\theta$ would play a role in the moment calculation (if forces are vertical) is if the distances were perpendicular to the forces, which means the horizontal distances from the pivot.
Let the pivot be at $x=0$.
The $16 \, \text{N}$ force is at $x = -\frac{a}{4} \cos\theta$. (This is the horizontal distance from the pivot to the line of action of the force).
The $6 \, \text{N}$ force is at $x = \frac{a}{4} \cos\theta$.
The $4 \, \text{N}$ force is at $x = \frac{3a}{4} \cos\theta$.

Anticlockwise moment: $16 \times (\frac{a}{4} \cos\theta)$
Clockwise moments: $(6 \times \frac{a}{4} \cos\theta) + (4 \times \frac{3a}{4} \cos\theta)$

Equating moments:

$$16 \times \frac{a}{4} \cos\theta = 6 \times \frac{a}{4} \cos\theta + 4 \times \frac{3a}{4} \cos\theta$$

Since $a \neq 0$ and $\cos\theta \neq 0$ (for a plank to be inclined), we can divide by $a \cos\theta$:

$$\frac{16}{4} = \frac{6}{4} + \frac{12}{4}$$ $$4 = 1.5 + 3$$ $$4 = 4.5$$

This still leads to the same contradiction. This means that the system, as drawn with the given forces and distances, cannot be in equilibrium. There might be an error in the problem statement or the diagram.

Let's consider if the $60^\circ$ angle is relevant. If the $16 \, \text{N}$ force is acting at $60^\circ$ to the plank, then its vertical component would be $16 \sin(60^\circ)$ or $16 \cos(60^\circ)$ depending on the angle definition. However, the arrow clearly shows a vertical downward force.

Given the contradiction, it's possible that the $6 \, \text{N}$ is the weight of the plank, and the $16 \, \text{N}$ and $4 \, \text{N}$ are external forces. The question asks to calculate $\theta$. If the system is in equilibrium, the sum of moments must be zero. Since the calculation consistently leads to $4 = 4.5$, it implies that the given forces and distances are not balanced.

If we assume there is a mistake in the problem and the question intends for the system to be in equilibrium, then $\theta$ cannot be determined from the moment equation alone if it cancels out. $\theta$ would only be relevant if the forces were not purely vertical or if the distances were measured differently.

Let's assume the $60^\circ$ is the angle of the $16 \, \text{N}$ force with the horizontal, and it's acting at the left end. This is not how it's drawn.

Given the problem as stated, and the diagram, the moment equation leads to a contradiction. If the system is in equilibrium, then the sum of moments must be zero.
Let's re-evaluate the distances from the pivot.
Pivot is at $a/4$ from the left end.
Force 1: $16 \, \text{N}$ at 0 (left end). Distance from pivot: $a/4$. Moment: $16 \times a/4$ (anticlockwise).
Force 2: $6 \, \text{N}$ (weight of plank) at $a/2$. Distance from pivot: $a/2 - a/4 = a/4$. Moment: $6 \times a/4$ (clockwise).
Force 3: $4 \, \text{N}$ at $a$ (right end). Distance from pivot: $a - a/4 = 3a/4$. Moment: $4 \times 3a/4$ (clockwise).

Sum of anticlockwise moments = $16 \times \frac{a}{4} = 4a$
Sum of clockwise moments = $6 \times \frac{a}{4} + 4 \times \frac{3a}{4} = \frac{6a}{4} + \frac{12a}{4} = \frac{18a}{4} = 4.5a$

For equilibrium, $4a = 4.5a$. This implies $0.5a = 0$, which means $a=0$. But $a$ is the length of the plank, so $a \neq 0$.
This confirms the contradiction. The system cannot be