The figure below shows a metal rule L and weight 7N pivoted at the other shown below. Calculate the angle made by the weight Plank.

We'll look at each force and decide if it tries to turn the plank clockwise (like a clock hand moving forward) or anti-clockwise (the opposite direction).

• Plank's Weight (7 N): This force pulls straight down at the middle of the plank. It will try to make the plank rotate clockwise around the pivot O.
• 4 N Force: This force is pushing downwards and to the left, at $30^\circ$ to the plank. It will also try to make the plank rotate clockwise around the pivot O.
• 8 N Force: This force is pushing downwards and to the right, at $45^\circ$ to the plank. It will try to make the plank rotate anti-clockwise around the pivot O.

We'll calculate the moment for each force around the pivot O. Remember, Moment = Force $\times$ Perpendicular Distance.

• Moment from the 4 N force ($M_1$):

  • Force: $4 \text{ N}$.
  • Distance from pivot: $L$ (we are assuming the 'l' in the diagram means the full length $L$ from the pivot, as this is common in such problems and makes it solvable).
  • Angle: The force is $30^\circ$ to the plank. The part of this force that causes rotation is $4 \sin(30^\circ)$.
  • Calculation: $$M_1 = 4 \text{ N} \times L \times \sin(30^\circ)$$ $$M_1 = 4L \times 0.5$$ $$M_1 = 2L \text{ N}\cdot\text{m}$$
  • Direction: Clockwise.

• Moment from the Plank's Weight (7 N) ($M_W$):

  • Force: $7 \text{ N}$.
  • Distance from pivot: The weight acts at the center, $\frac{L}{2}$ from the pivot.
  • Angle: The weight pulls straight down. If the plank is at an angle $\theta$ to the horizontal, the perpendicular distance from the pivot to the line where the weight acts is $\frac{L}{2} \cos(\theta)$.
  • Calculation: $$M_W = 7 \text{ N} \times \frac{L}{2} \times \cos(\theta)$$ $$M_W = 3.5 L \cos(\theta) \text{ N}\cdot\text{m}$$
  • Direction: Clockwise.

• Moment from the 8 N force ($M_2$):

  • Force: $8 \text{ N}$.
  • Distance from pivot: It's $\frac{3}{4}L$ from the pivot.
  • Angle: The force is $45^\circ$ to the plank. The part of this force that causes rotation is $8 \sin(45^\circ)$.
  • Calculation: $$M_2 = 8 \text{ N} \times \frac{3L}{4} \times \sin(45^\circ)$$ $$M_2 = 6L \times \frac{\sqrt{2}}{2}$$ $$M_2 = 3\sqrt{2} L \text{ N}\cdot\text{m}$$
  • Direction: Anti-clockwise.

For the plank to be balanced, the total clockwise moments must equal the total anti-clockwise moments:

Substitute the expressions we found for each moment:

• Notice that the length $L$ appears in every part of the equation. This is great because it means we can divide the entire equation by $L$ (since $L$ isn't zero), and it will cancel out! This means the exact length of the plank doesn't matter for finding the angle.

  $$\frac{3.5 L \cos(\theta)}{L} + \frac{2L}{L} = \frac{3\sqrt{2} L}{L}$$ $$3.5 \cos(\theta) + 2 = 3\sqrt{2}$$

• Now, we want to get $\cos(\theta)$ by itself. First, subtract 2 from both sides:

  $$3.5 \cos(\theta) = 3\sqrt{2} - 2$$

• Let's calculate the value of $3\sqrt{2}$:

  $$\sqrt{2} \approx 1.4142$$ $$3\sqrt{2} \approx 3 \times 1.4142 = 4.2426$$

• Substitute this value back into the equation:

  $$3.5 \cos(\theta) = 4.2426 - 2$$ $$3.5 \cos(\theta) = 2.2426$$

• Finally, divide by 3.5 to find $\cos(\theta)$:

  $$\cos(\theta) = \frac{2.2426}{3.5}$$ $$\cos(\theta) \approx 0.6407$$

• To find the angle $\theta$ itself, we use the inverse cosine function (often written as $\arccos$ or $\cos^{-1}$) on your calculator:

  $$\theta = \arccos(0.6407)$$ $$\theta \approx 50.15^\circ$$

So, when the plank is balanced, it makes an angle of about $50.15^\circ$ with the horizontal ground.