To determine the moment about point A caused by the applied force $F$, we can use the following formula for the moment:

$M = F \cdot d \cdot \sin(\theta)$

Where:

• $M$ is the moment about point A.
• $F$ is the applied force, which is given as $5 \, \text{N}$.
• $d$ is the distance from point A to the point of application of the force, which is $3 \, \text{m}$.
• $\theta$ is the angle between the force and the position vector, which is $60^\circ$.

Now we can substitute the values:

$M = 5 \, \text{N} \cdot 3 \, \text{m} \cdot \sin(60^\circ)$

We know that $\sin(60^\circ) = \frac{\sqrt{3}}{2} \approx 0.866$.

$M = 5 \cdot 3 \cdot 0.866$

Now let's calculate the moment.The moment about point A caused by the applied force is approximately $13 \, \text{Nm}$. Based on the available options, the closest answer is $12 \, \text{Nm}$.

Would you like further details on the calculations, or do you have any other questions?

Here are 5 related questions for practice:

1. How would the moment change if the angle were $90^\circ$?
2. What if the force applied was doubled but the angle remained the same?
3. How does the moment vary with respect to the distance $d$?
4. What happens to the moment if the angle $\theta$ is $0^\circ$?
5. Can the moment ever be negative, and if so, under what conditions?

Tip: Always remember that the moment depends on the perpendicular distance to the line of action of the force!