To solve Problem 19, we need to combine two forces, $800 \, \text{N}$ (vertical) and $900 \, \text{N}$ (diagonal), into a single equivalent force $R$ acting at point $A$. Additionally, we must find the distance $x$ from the left edge to point $A$.

Step-by-step solution:

1. Resolve the forces:

• The 900 N force is at an angle, so we break it into horizontal and vertical components.
• Using the geometry provided (the diagonal force makes a 2:1 right triangle), the angle $\theta$ is: $\tan \theta = \frac{2}{1} \quad \Rightarrow \quad \theta = \tan^{-1}(2) \approx 63.43^\circ$
• The components of the 900 N force are: $F_x = 900 \cos(63.43^\circ) \approx 900 \times 0.447 = 402.3 \, \text{N}$ $F_y = 900 \sin(63.43^\circ) \approx 900 \times 0.894 = 804.6 \, \text{N}$

2. Find the resultant force $R$:

• Add the horizontal components: $F_{R_x} = 402.3 \, \text{N} \quad (\text{no horizontal component from the 800 N force})$
• Add the vertical components: $F_{R_y} = 800 \, \text{N} + 804.6 \, \text{N} = 1604.6 \, \text{N}$
• The magnitude of $R$ is the vector sum: $R = \sqrt{F_{R_x}^2 + F_{R_y}^2} = \sqrt{402.3^2 + 1604.6^2} \approx 898 \, \text{N}$
• This matches the answer given in the problem.

3. Determine the location $x$:

• To find $x$, use the principle of moments (taking moments about the left edge): $\sum M_{\text{left}} = R \cdot x$
• Moments of each force:
  • The 800 N force acts at the left edge, so it creates no moment.
  • The vertical component of the 900 N force (804.6 N) creates a moment about the left edge: $M_{900\,\text{N}} = 804.6 \, \text{N} \times 2.4 \, \text{m} = 1931.04 \, \text{Nm}$
• Solving for $x$: $898 \, \text{N} \cdot x = 1931.04 \, \text{Nm} \quad \Rightarrow \quad x = \frac{1931.04}{898} \approx 1.62 \, \text{m}$
• This is the given answer for $x$.

Final Answer:

• $R = 898 \, \text{N}$
• $x = 1.62 \, \text{m}$

Would you like more details on any step, or do you have any questions?

Here are 5 related questions for further exploration:

1. How would the solution change if the 900 N force were acting at a different angle?
2. What if the rigid frame had a different length, say 3.0 meters, instead of 2.4 meters?
3. Can the principle of moments be used if there are more than two forces involved?
4. How would the result change if the 800 N force were not aligned vertically?
5. Can you solve this problem graphically and verify the algebraic result?

Tip: When resolving forces into components, always pay attention to the angles and how they relate to the force triangle.