b. The system shown in Figure Q1b below is in equilibrium. Calculate the values of the tensions in the strings A, B, C and D.

At Junction C, the 400 N weight is pulling downwards, and tensions $T_C$ and $T_D$ are pulling upwards at angles.

• Weight $W = 400 \text{ N}$ (acting vertically downwards).
• Tension $T_C$ acts at $20^\circ$ above the horizontal, to the left.
• Tension $T_D$ acts at $35^\circ$ above the horizontal, to the right.

For equilibrium, the sum of forces in the x-direction and y-direction must be zero.

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

$$T_D \cos(35^\circ) - T_C \cos(20^\circ) = 0$$ $$T_D \cos(35^\circ) = T_C \cos(20^\circ) \quad (\text{Eq. 1})$$

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

$$T_D \sin(35^\circ) + T_C \sin(20^\circ) - 400 \text{ N} = 0$$ $$T_D \sin(35^\circ) + T_C \sin(20^\circ) = 400 \text{ N} \quad (\text{Eq. 2})$$

From Eq. 1, express $T_D$ in terms of $T_C$:

$$T_D = T_C \frac{\cos(20^\circ)}{\cos(35^\circ)}$$

Substitute this expression for $T_D$ into Eq. 2:

$$\left( T_C \frac{\cos(20^\circ)}{\cos(35^\circ)} \right) \sin(35^\circ) + T_C \sin(20^\circ) = 400$$

Factor out $T_C$:

$$T_C \left( \frac{\cos(20^\circ)\sin(35^\circ)}{\cos(35^\circ)} + \sin(20^\circ) \right) = 400$$

Recall that $\frac{\sin\theta}{\cos\theta} = \tan\theta$:

$$T_C \left( \cos(20^\circ)\tan(35^\circ) + \sin(20^\circ) \right) = 400$$

Using the trigonometric identity $\cos A \tan B + \sin A = \frac{\sin(A+B)}{\cos B}$:

$$T_C \left( \frac{\sin(20^\circ+35^\circ)}{\cos(35^\circ)} \right) = 400$$ $$T_C \left( \frac{\sin(55^\circ)}{\cos(35^\circ)} \right) = 400$$

Since $\sin(55^\circ) = \cos(90^\circ - 55^\circ) = \cos(35^\circ)$:

$$T_C \left( \frac{\cos(35^\circ)}{\cos(35^\circ)} \right) = 400$$ $$T_C \times 1 = 400$$ $$T_C = 400 \text{ N}$$

Now, substitute $T_C = 400 \text{ N}$ back into the expression for $T_D$:

$$T_D = 400 \text{ N} \times \frac{\cos(20^\circ)}{\cos(35^\circ)}$$ $$T_D = 400 \text{ N} \times \frac{0.93969}{0.81915}$$ $$T_D \approx 400 \text{ N} \times 1.14715$$ $$T_D \approx 458.86 \text{ N}$$

At Junction A, tension $T_C$ (from the string connecting to C) pulls downwards and to the right, while tensions $T_A$ and $T_B$ pull horizontally and vertically, respectively.

• Tension $T_C$ acts at $20^\circ$ below the horizontal, to the right.
• Tension $T_A$ acts horizontally to the left.
• Tension $T_B$ acts vertically upwards.

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

$$T_A - T_C \cos(20^\circ) = 0$$ $$T_A = T_C \cos(20^\circ) \quad (\text{Eq. 3})$$

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

$$T_B - T_C \sin(20^\circ) = 0$$ $$T_B = T_C \sin(20^\circ) \quad (\text{Eq. 4})$$

Using $T_C = 400 \text{ N}$:

$$T_A = 400 \text{ N} \times \cos(20^\circ)$$ $$T_A = 400 \text{ N} \times 0.93969$$ $$T_A \approx 375.88 \text{ N}$$ $$T_B = 400 \text{ N} \times \sin(20^\circ)$$ $$T_B = 400 \text{ N} \times 0.34202$$ $$T_B \approx 136.81 \text{ N}$$

$$\boxed{T_A = 376 \text{ N}}$$ $$\boxed{T_B = 137 \text{ N}}$$ $$\boxed{T_C = 400 \text{ N}}$$ $$\boxed{T_D = 459 \text{ N}}$$