Explain the operation of a vapor absorption refrigeration system and list two methods of starting a synchronous motor.

$$F_x = F \cos \theta = 20 \cos 30^\circ \, \mathrm{N}$$ $$\cos 30^\circ = \frac{\sqrt{3}}{2}$$ $$F_x = 20 \times \frac{\sqrt{3}}{2} = 10\sqrt{3} \, \mathrm{N}$$ $$F_y = F \sin \theta = 20 \sin 30^\circ \, \mathrm{N}$$ $$\sin 30^\circ = \frac{1}{2}$$

$F_y = 20 \times \frac{1}{2} = 10 \, \mathrm{N}$ (upward)

Vertical equilibrium: $N + F_y = mg$

$$N = mg - F_y$$ $$mg = 4 \times 10 = 40 \, \mathrm{N}$$ $$N = 40 - 10 = 30 \, \mathrm{N}$$

$$F_\mathrm{net} = F_x - f_k = 10\sqrt{3} - 6 \, \mathrm{N}$$ $$\sqrt{3} \approx 1.732$$ $$10\sqrt{3} \approx 17.32 \, \mathrm{N}$$ $$F_\mathrm{net} \approx 17.32 - 6 = 11.32 \, \mathrm{N}$$

$$a = \frac{F_\mathrm{net}}{m} = \frac{11.32}{4} = 2.83 \, \mathrm{m/s^2}$$

Or exactly:

$$a = \frac{10\sqrt{3} - 6}{4} \, \mathrm{m/s^2}$$

Final answer: $a \approx 2.83 \, \mathrm{m/s^2}$

2.83 m/s^{2}