To find the magnitude and direction of the third force that produces equilibrium at the point, we need to use the condition for equilibrium: the sum of all forces acting on the point must be zero.

This means the vector sum of the three forces must be zero.

Step 1: Resolve the two given forces into components.

We have two forces:

1. A force of 33 N acting at 90° (which is along the vertical axis).
2. A force of 44 N acting at 60° (which is at an angle to the horizontal axis).

Force 1 (33 N at 90°):

This force is purely vertical, so its components are:

• $F_{1x} = 0$
• $F_{1y} = 33 \, \text{N}$

Force 2 (44 N at 60°):

We resolve this force into its horizontal and vertical components.

• $F_{2x} = 44 \cdot \cos(60^\circ) = 44 \cdot \frac{1}{2} = 22 \, \text{N}$
• $F_{2y} = 44 \cdot \sin(60^\circ) = 44 \cdot \frac{\sqrt{3}}{2} \approx 38.07 \, \text{N}$

Step 2: Find the components of the third force (F3).

For equilibrium, the sum of forces in both the horizontal and vertical directions must be zero.

Horizontal direction (x-axis):

$F_{3x} = -(F_{1x} + F_{2x}) = -(0 + 22) = -22 \, \text{N}$

Vertical direction (y-axis):

$F_{3y} = -(F_{1y} + F_{2y}) = -(33 + 38.07) = -71.07 \, \text{N}$

Step 3: Determine the magnitude and direction of the third force.

The magnitude $F_3$ is given by the Pythagorean theorem: $F_3 = \sqrt{F_{3x}^2 + F_{3y}^2} = \sqrt{(-22)^2 + (-71.07)^2} \approx \sqrt{484 + 5055.7} \approx \sqrt{5540.7} \approx 74.5 \, \text{N}$

To find the direction (angle $\theta$) of the third force, we use: $\theta = \tan^{-1}\left(\frac{F_{3y}}{F_{3x}}\right) = \tan^{-1}\left(\frac{-71.07}{-22}\right) \approx \tan^{-1}(3.23) \approx 72^\circ$

Since both components are negative, the third force acts in the third quadrant.

Final Answer:

• The magnitude of the third force is approximately 74.5 N.
• The direction is approximately 72° below the negative x-axis (or 72° south of west).

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Do you need further clarification on any part of this solution?

Here are 5 related questions:

1. How would you calculate the components of a force acting at different angles?
2. What is the condition for equilibrium in terms of force components?
3. Can we use a similar method to solve for multiple forces in non-rectangular directions?
4. How does the direction of the third force change if the angles of the original forces are different?
5. What would happen if the forces were not acting at 90° and 60° angles?

Tip: Always break forces down into components (horizontal and vertical) before summing them, as this simplifies the problem significantly!