To solve the problem, we need to consider the equilibrium of forces at point C. The forces acting at point C are the tensions in cables $AC$ and $BC$, and the applied force $P$ at angle $\alpha$.

Step 1: Analyze the forces

The tensions $T_{AC}$ and $T_{BC}$ have maximum values of 1200 N and 600 N, respectively. The angles for these tensions relative to the horizontal are given as:

• $\theta_{AC} = 35^\circ$
• $\theta_{BC} = 50^\circ$

Step 2: Set up the equilibrium equations

For static equilibrium, the sum of forces in both the horizontal ($x$) and vertical ($y$) directions must be zero:

Horizontal Direction ($x$-axis):

$T_{AC} \cos(35^\circ) + T_{BC} \cos(50^\circ) = P \cos(\alpha)$

Vertical Direction ($y$-axis):

$T_{AC} \sin(35^\circ) + T_{BC} \sin(50^\circ) = P \sin(\alpha)$

Step 3: Substitute maximum tensions

Given:

• $T_{AC} = 1200 \, \text{N}$
• $T_{BC} = 600 \, \text{N}$

Substituting these into the equations:

Horizontal Direction:

$1200 \cos(35^\circ) + 600 \cos(50^\circ) = P \cos(\alpha)$

Vertical Direction:

$1200 \sin(35^\circ) + 600 \sin(50^\circ) = P \sin(\alpha)$

Step 4: Solve for $P$ and $\alpha$

To find $P$ and $\alpha$, you divide the vertical equilibrium equation by the horizontal equilibrium equation:

$\tan(\alpha) = \frac{1200 \sin(35^\circ) + 600 \sin(50^\circ)}{1200 \cos(35^\circ) + 600 \cos(50^\circ)}$

Solving for $\alpha$:

$\alpha = \tan^{-1}\left(\frac{1200 \sin(35^\circ) + 600 \sin(50^\circ)}{1200 \cos(35^\circ) + 600 \cos(50^\circ)}\right)$

Finally, substitute $\alpha$ back into either the horizontal or vertical equation to find $P$.

Let's calculate these values.

Calculation

1. Calculate $\tan(\alpha)$:

   $\tan(\alpha) = \frac{1200 \times 0.5736 + 600 \times 0.7660}{1200 \times 0.8192 + 600 \times 0.6428}$

2. Solve for $\alpha$:

   $\alpha = \tan^{-1}(0.651)$

3. Find $P$:

   Use the horizontal equation to find $P$:

   $P = \frac{1200 \times 0.8192 + 600 \times 0.6428}{\cos(\alpha)}$

Let's calculate these values numerically.The calculations yield the following results:

• Angle $\alpha$: $39.987^\circ$
• Magnitude of the largest force $P$: $1786.318 \, \text{N}$

These are the values for the largest force $P$ that can be applied at point $C$, and the corresponding angle $\alpha$.

Do you have any questions or would you like more details on any part of this solution?

Here are some related questions you might explore:

1. How would the results change if the tension in cable $BC$ was increased to 800 N?
2. What would be the effect on $P$ if $\alpha$ was set to $45^\circ$ from the beginning?
3. How does the system behave if the angle of $T_{AC}$ was reduced by $5^\circ$?
4. What if the tension in cable $AC$ were reduced by $200 \, \text{N}$? What would happen to $P$?
5. How can this analysis be adapted if cable $BC$ was at a different angle, say $40^\circ$?

Tip: When dealing with equilibrium problems, always ensure all force components are resolved correctly into horizontal and vertical components before applying equilibrium equations.