Here are the solutions to the problems: 1.2 Determine (1) the horizontal component and (2) the vertical component of a force that is exerted by the spring balance on the rubber band. Use a scale where 1 cm = 1 N. From the previous problem, we have: Magnitude of the force, $F = 35$ N Angle to the horizontal axis, $\theta = 30°$ The horizontal component ($F_x$) is given by $F \cos \theta$. The vertical component ($F_y$) is given by $F \sin \theta$. Step 1: Calculate the horizontal component. $$F_x = F \cos \theta$$ $$F_x = 35 \text{ N} \times \cos(30°)$$ $$F_x = 35 \text{ N} \times \frac{\sqrt{3}}{2}$$ $$F_x \approx 35 \text{ N} \times 0.8660$$ $$F_x \approx 30.31 \text{ N}$$ Step 2: Calculate the vertical component. $$F_y = F \sin \theta$$ $$F_y = 35 \text{ N} \times \sin(30°)$$ $$F_y = 35 \text{ N} \times \frac{1}{2}$$ $$F_y = 17.5 \text{ N}$$ (1) Horizontal component: $\boxed{\text{30.31 N}}$ (2) Vertical component: $\boxed{\text{17.5 N}}$ 1.3 By keeping the reading on the spring balance the same, how will the magnitude of the horizontal component of the force change if the angle measured in the figure: The force $F = 35$ N remains constant. The horizontal component is $F_x = F \cos \theta$. 1.3.1 is increased to 60°? Step 1: Calculate the new horizontal component. $$F_x = 35 \text{ N} \times \cos(60°)$$ $$F_x = 35 \text{ N} \times 0.5$$ $$F_x = 17.5 \text{ N}$$ Step 2: Compare with the original horizontal component ($30.31$ N). The horizontal component will decrease from $30.31$ N to $17.5$ N. 1.3.2 is increased to 90°? Step 1: Calculate the new horizontal component. $$F_x = 35 \text{ N} \times \cos(90°)$$ $$F_x = 35 \text{ N} \times 0$$ $$F_x = 0 \text{ N}$$ Step 2: Compare with the original horizontal component ($30.31$ N). The horizontal component will decrease to 0 N. 1.4 When will the vertical component of the force that is read off the spring balance have a magnitude of 0 N? The vertical component is $F_y = F \sin \theta$. We want $F_y = 0$ N. $$F \sin \theta = 0 \text{ N}$$ Since the force $F$ read off the spring balance is $35$ N (which is not zero), then $\sin \theta$ must be zero. $$\sin \theta = 0$$ This occurs when the angle $\theta$ is $0°$ (or $180°$, etc.). In the context of an angle with the horizontal, this means the force is applied purely horizontally. The vertical component will have a magnitude of 0 N when the angle to the horizontal axis is $\boxed{\text{0°}}$.