A vertical frame of two legs shown below has a weight of 50kN hanging from point A. Determine the forces in legs 'AB' and 'AC' produced by the 50kN weight.

Let D be the point directly below A on the line BC.
The height $AD = 3.25 \, \text{m}$.
The horizontal distance $BD = 1.75 \, \text{m}$.
The horizontal distance $DC = 2.5 \, \text{m}$.

The angle $\theta_1$ (angle of leg AB with the horizontal) is given by:

$$\tan(\theta_1) = \frac{AD}{BD} = \frac{3.25 \, \text{m}}{1.75 \, \text{m}}$$ $$\theta_1 = \arctan\left(\frac{3.25}{1.75}\right) \approx 61.699^\circ$$

The angle $\theta_2$ (angle of leg AC with the horizontal) is given by:

$$\tan(\theta_2) = \frac{AD}{DC} = \frac{3.25 \, \text{m}}{2.5 \, \text{m}}$$ $$\theta_2 = \arctan\left(\frac{3.25}{2.5}\right) = \arctan(1.3) \approx 52.431^\circ$$

Let $F_{AB}$ be the force in leg AB and $F_{AC}$ be the force in leg AC. The weight $W = 50 \, \text{kN}$ acts vertically downwards.

Sum of forces in the horizontal direction ($\sum F_x = 0$):

$$F_{AC} \cos(\theta_2) - F_{AB} \cos(\theta_1) = 0$$ $$F_{AC} \cos(\theta_2) = F_{AB} \cos(\theta_1) \quad (\text{Equation 1})$$

Sum of forces in the vertical direction ($\sum F_y = 0$):

$$F_{AB} \sin(\theta_1) + F_{AC} \sin(\theta_2) - W = 0$$ $$F_{AB} \sin(\theta_1) + F_{AC} \sin(\theta_2) = W \quad (\text{Equation 2})$$

Using $\theta_1 \approx 61.699^\circ$ and $\theta_2 \approx 52.431^\circ$:
$\cos(\theta_1) \approx 0.4740$
$\sin(\theta_1) \approx 0.8805$
$\cos(\theta_2) \approx 0.6100$
$\sin(\theta_2) \approx 0.7926$

From Equation 1:

$$F_{AC} (0.6100) = F_{AB} (0.4740)$$ $$F_{AC} = F_{AB} \frac{0.4740}{0.6100} \approx 0.7770 F_{AB}$$

Substitute this expression for $F_{AC}$ into Equation 2:

$$F_{AB} (0.8805) + (0.7770 F_{AB}) (0.7926) = 50 \, \text{kN}$$ $$F_{AB} (0.8805) + F_{AB} (0.6150) = 50 \, \text{kN}$$ $$F_{AB} (1.4955) = 50 \, \text{kN}$$

$$F_{AB} = \frac{50 \, \text{kN}}{1.4955} \approx 33.434 \, \text{kN}$$

Now, calculate $F_{AC}$:

$$F_{AC} = 0.7770 \times 33.434 \, \text{kN} \approx 25.979 \, \text{kN}$$

The forces in the legs are:

• Force in leg AB: $\boxed{\text{33.434 kN}}$
• Force in leg AC: $\boxed{\text{25.979 kN}}$