To solve this problem, we need to find the resultant of the four forces acting in different directions and then calculate the equilibrant force, which will be equal in magnitude but opposite in direction to the resultant.

Step 1: Break the forces into components

Let's assume the $x$-axis is aligned with the 70 N and 50 N forces, and the angles are measured from these axes.

• The 70 N force acts along the positive $x$-axis.

• The 50 N force acts along the negative $x$-axis.

• The 20 N force is at an angle of 40° with the 50 N force (negative $x$-axis). So, its components are: $F_{20x} = 20 \cos(40°), \quad F_{20y} = 20 \sin(40°)$ Since this force is at 40° from the negative $x$-axis, the $x$-component will be negative.

• The 60 N force is at an angle of 40° with the 70 N force (positive $x$-axis). So, its components are: $F_{60x} = 60 \cos(40°), \quad F_{60y} = 60 \sin(40°)$ Since this force is at 40° from the positive $x$-axis, both components are positive.

Step 2: Sum the components in the $x$- and $y$-directions

1. In the $x$-direction: $F_x = 70 - 50 - 20 \cos(40°) + 60 \cos(40°)$

2. In the $y$-direction: $F_y = 60 \sin(40°) - 20 \sin(40°)$

Step 3: Calculate the magnitudes of components

Let’s calculate these values.

Step 4: Calculate the resultant force magnitude

Once the components $F_x$ and $F_y$ are known, the resultant force can be found using the Pythagorean theorem: $R = \sqrt{F_x^2 + F_y^2}$

Step 5: Find the direction of the resultant force

The angle $\theta$ of the resultant force relative to the $x$-axis is: $\theta = \tan^{-1} \left(\frac{F_y}{F_x}\right)$

Step 6: Calculate the equilibrant force

The equilibrant force will have the same magnitude as the resultant force but will point in the opposite direction. Its angle will be: $\theta_{\text{equilibrant}} = \theta + 180°$

I will now calculate these values for you.### Result:

1. The components of the resultant force are:

   • $F_x = 50.64 \, \text{N}$
   • $F_y = 25.71 \, \text{N}$

2. The magnitude of the resultant force is: $R = 56.79 \, \text{N}$

3. The direction of the resultant force is: $\theta_{\text{res}} = 26.92^\circ \, \text{above the positive x-axis}$

4. The equilibrant force has the same magnitude but acts in the opposite direction: $\theta_{\text{equilibrant}} = 206.92^\circ$

Would you like further details on the process, or any additional clarifications?

Here are 5 related questions:

1. How do you resolve forces at different angles into components?
2. Why do opposite forces cancel each other in some cases?
3. How does the direction of the equilibrant force relate to the resultant force?
4. What role does trigonometry play in calculating force components?
5. How would the result change if the angles were different?

Tip: Always draw a diagram when working with force vectors to visually represent directions and angles.