Question

1. find the unknown force f and state the magnitude of the reaction at the pivot in each of the situations shown in below

i.
ii.

Explanation:

Analyze situation i rotational equilibrium

Using the Torque and Rotational Equilibrium knowledge point, we sum torques about the pivot.
Taking counterclockwise torques as positive:

$$ \sum \tau = (7\text{ N} \times 3\text{ m}) - (3\text{ N} \times 2\text{ m}) - (F \times 5\text{ m}) = 0 $$

$$ 21 - 6 - 5F = 0 \implies 15 = 5F \implies F = 3\text{ N} $$

Analyze situation i translational equilibrium

Using the Static Equilibrium knowledge point, we balance vertical forces.
Let \(R_1\) be the upward reaction force at the pivot:

$$ \sum F_y = R_1 - 7\text{ N} - 3\text{ N} - F = 0 $$

Substitute \(F = 3\text{ N}\):

$$ R_1 = 7 + 3 + 3 = 13\text{ N} $$

Analyze situation ii rotational equilibrium

Using the Torque and Rotational Equilibrium knowledge point, we sum torques about the pivot.
Taking counterclockwise torques as positive:

$$ \sum \tau = (6\text{ N} \times 2\text{ m}) - (4\text{ N} \times 1\text{ m}) - (F \times 4\text{ m}) + (2\text{ N} \times 5\text{ m}) = 0 $$

$$ 12 - 4 - 4F + 10 = 0 \implies 18 - 4F = 0 \implies F = 4.5\text{ N} $$

Analyze situation ii translational equilibrium

Using the Static Equilibrium knowledge point, we balance vertical forces.
Let \(R_2\) be the upward reaction force at the pivot:

$$ \sum F_y = R_2 - 6\text{ N} - 4\text{ N} - F + 2\text{ N} = 0 $$

Substitute \(F = 4.5\text{ N}\):

$$ R_2 - 6 - 4 - 4.5 + 2 = 0 \implies R_2 = 12.5\text{ N} $$

Answer:

Question i

• Unknown force \(F\): <blank>\(3\text{ N}\)</blank>
• Reaction force at the pivot: <blank>\(13\text{ N}\)</blank>

Question ii

• Unknown force \(F\): <blank>\(4.5\text{ N}\)</blank>
• Reaction force at the pivot: <blank>\(12.5\text{ N}\)</blank>