A light rigid bar PQ of length 5.0\text{ m} is hinged at end P and held horizontally. An u… — JAMB UTME

A light rigid bar $PQ$ of length $5.0\text{ m}$ is hinged at end $P$ and held horizontally. An upward force of $40\text{ N}$ is applied at end $Q$ at an angle of $30^\circ$ to the bar. To keep the bar in horizontal equilibrium, a vertical downward force $F$ is applied at a distance of $2.0\text{ m}$ from $P$ . What is the magnitude of the force $F$ ?

A
$20\text{ N}$
$50\text{ N}$ Answer
C
$86.6\text{ N}$
D
$100\text{ N}$

Answer

The magnitude of the force

F

required to maintain equilibrium is

50\text{ N}

The force of

40\text{ N}

applied at an angle of

30^\circ

to the bar has a perpendicular component of

40\sin 30^\circ = 20\text{ N}

. The counterclockwise moment about pivot

P

20\text{ N} \times 5.0\text{ m} = 100\text{ N}\cdot\text{m}

. For the bar to remain in equilibrium, the clockwise moment created by

F

must equal

100\text{ N}\cdot\text{m}

, giving

F \times 2.0\text{ m} = 100\text{ N}\cdot\text{m}

, which yields

F = 50\text{ N}

Step-by-Step Solution

Calculate the perpendicular component of the force applied at end

Q

F_{\perp} = 40\text{ N} \times \sin(30^\circ) = 40 \times 0.5 = 20\text{ N}

Only the force component perpendicular to the bar produces a moment about the pivot

P

Calculate the counterclockwise moment produced by the force at

Q

about pivot

P

\text{Moment}_{Q} = 20\text{ N} \times 5.0\text{ m} = 100\text{ N}\cdot\text{m}

Moment is defined as perpendicular force multiplied by distance from the pivot.

Apply the Principle of Moments about pivot

P

to find force

F

F \times 2.0\text{ m} = 100\text{ N}\cdot\text{m} \implies F = \frac{100}{2.0} = 50\text{ N}

For rotational equilibrium, total clockwise moment about

P

must equal total counterclockwise moment about

P

Key Concept

Principle of Moments and Rotational Equilibrium with Forces at an Angle

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Topics

More questions from this topic