To solve the questions in this problem, we need to calculate the axial (axially acting) and radial (radially acting) components of the given force $F = 1500 \, \text{N}$ at an angle of $60^\circ$.

Part 1: Calculate the Axial Force $F_a$

The axial component of the force $F$ along the axis of the bar can be calculated using the cosine of the angle: $F_a = F \cdot \cos(60^\circ)$

Given that:

• $F = 1500 \, \text{N}$
• $\cos(60^\circ) = 0.5$

$F_a = 1500 \, \text{N} \times 0.5 = 750 \, \text{N}$

So, the axial force $F_a$ is 750 N.

Part 2: Calculate the Radial (or Perpendicular) Force $F_r$

The radial component of the force, which acts perpendicular to the axial direction, can be calculated using the sine of the angle: $F_r = F \cdot \sin(60^\circ)$

Given that:

• $F = 1500 \, \text{N}$
• $\sin(60^\circ) = \frac{\sqrt{3}}{2} \approx 0.866$

$F_r = 1500 \, \text{N} \times 0.866 = 1299 \, \text{N}$

So, the radial force $F_r$ is approximately 1299 N.

Let me know if you need further details or have any questions.

Related Questions

1. How would the axial and radial forces change if the angle were $45^\circ$ instead of $60^\circ$?
2. What would happen to the forces if the total force $F$ were doubled to 3000 N?
3. How is the sine and cosine rule applied in calculating the forces in inclined force problems?
4. Can this force system be in equilibrium if additional forces are added? If yes, how?
5. How can the understanding of force components be useful in real-world engineering applications?

Tip

When dealing with forces at angles, always remember to split them into components using trigonometric functions (sine and cosine) for easier analysis.