Let's break down and solve the two problems shown in the image step by step.

Problem 1:

• A force of 100 N makes an angle $\theta$ with the x-axis, and its y-component is given as 30 N. You are asked to find the x-component of the force ($F_x$) and the angle $\theta$.

We know:

• The magnitude of the force: $F = 100 \, \text{N}$
• The y-component of the force: $F_y = 30 \, \text{N}$

Using trigonometric relationships, the components of the force can be written as:

• $F_x = F \cdot \cos(\theta)$
• $F_y = F \cdot \sin(\theta)$

We are given $F_y = 30 \, \text{N}$, so: $F_y = F \cdot \sin(\theta) = 100 \cdot \sin(\theta)$ $30 = 100 \cdot \sin(\theta)$ Solving for $\sin(\theta)$: $\sin(\theta) = \frac{30}{100} = 0.3$ Now, find $\theta$: $\theta = \sin^{-1}(0.3) \approx 17.46^\circ$

Next, we can find the x-component using $F_x = F \cdot \cos(\theta)$: $F_x = 100 \cdot \cos(17.46^\circ)$ $F_x \approx 100 \cdot 0.954 \approx 95.4 \, \text{N}$

So, the x-component of the force is approximately $F_x \approx 95.4 \, \text{N}$ and the angle $\theta \approx 17.46^\circ$.

Problem 2:

• Four forces act on an object and intersect at point O. The forces are:
  • 110 N at $30^\circ$
  • 100 N at $45^\circ$
  • 160 N at $20^\circ$
  • 80 N at $0^\circ$

You are asked to find the resultant force graphically.

This problem involves vector addition, where we resolve each force into its x and y components.

1. Force 1 (110 N at $30^\circ$):

   • $F_{1x} = 110 \cdot \cos(30^\circ)$
   • $F_{1y} = 110 \cdot \sin(30^\circ)$

2. Force 2 (100 N at $45^\circ$):

   • $F_{2x} = 100 \cdot \cos(45^\circ)$
   • $F_{2y} = 100 \cdot \sin(45^\circ)$

3. Force 3 (160 N at $20^\circ$):

   • $F_{3x} = 160 \cdot \cos(20^\circ)$
   • $F_{3y} = 160 \cdot \sin(20^\circ)$

4. Force 4 (80 N at $0^\circ$):

   • Since this force is horizontal, its components are:
   • $F_{4x} = 80 \, \text{N}$
   • $F_{4y} = 0 \, \text{N}$

After calculating the components, sum the x and y components separately to find the resultant force components: $F_{x_{total}} = F_{1x} + F_{2x} + F_{3x} + F_{4x}$ $F_{y_{total}} = F_{1y} + F_{2y} + F_{3y} + F_{4y}$

Finally, the resultant force is: $F_{res} = \sqrt{F_{x_{total}}^2 + F_{y_{total}}^2}$ and its direction: $\theta_{res} = \tan^{-1}\left(\frac{F_{y_{total}}}{F_{x_{total}}}\right)$

Would you like me to proceed with the detailed calculations for the second problem?

5 Related Questions:

1. How do we determine the components of a force in different coordinate systems?
2. Why is trigonometry essential for resolving forces?
3. How do you graphically represent the resultant of multiple forces?
4. What is the significance of the angle between force vectors in determining their resultant?
5. Can we use vector algebra to solve problems involving forces more efficiently than graphical methods?

Tip:

Always double-check angle references in physics problems to ensure you're using the correct trigonometric functions for resolution into components.