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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.

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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=35F = 35 N
Angle to the horizontal axis, θ=30°\theta = 30°

The horizontal component (FxF_x) is given by Fcos⁡θF \cos \theta.
The vertical component (FyF_y) is given by Fsin⁡θF \sin \theta.

  1. Calculate the horizontal component.

    Fx=Fcos⁡θF_x = F \cos \theta Fx=35 N×cos⁡(30°)F_x = 35 \text{ N} \times \cos(30°) Fx=35 N×32F_x = 35 \text{ N} \times \frac{\sqrt{3}}{2} Fx≈35 N×0.8660F_x \approx 35 \text{ N} \times 0.8660 Fx≈30.31 NF_x \approx 30.31 \text{ N}
  2. Calculate the vertical component.

    Fy=Fsin⁡θF_y = F \sin \theta Fy=35 N×sin⁡(30°)F_y = 35 \text{ N} \times \sin(30°) Fy=35 N×12F_y = 35 \text{ N} \times \frac{1}{2} Fy=17.5 NF_y = 17.5 \text{ N}

    (1) Horizontal component: 30.31 N\boxed{\text{30.31 N}}
    (2) Vertical component: 17.5 N\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=35F = 35 N remains constant. The horizontal component is Fx=Fcos⁡θF_x = F \cos \theta.

    1.3.1 is increased to 60°?

  3. Calculate the new horizontal component.

    Fx=35 N×cos⁡(60°)F_x = 35 \text{ N} \times \cos(60°) Fx=35 N×0.5F_x = 35 \text{ N} \times 0.5 Fx=17.5 NF_x = 17.5 \text{ N}
  4. 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°?

  5. Calculate the new horizontal component.

    Fx=35 N×cos⁡(90°)F_x = 35 \text{ N} \times \cos(90°) Fx=35 N×0F_x = 35 \text{ N} \times 0 Fx=0 NF_x = 0 \text{ N}
  6. 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 Fy=Fsin⁡θF_y = F \sin \theta. We want Fy=0F_y = 0 N.

    Fsin⁡θ=0 NF \sin \theta = 0 \text{ N}

    Since the force FF read off the spring balance is 35 N (which is not zero), then sin⁡θ\sin \theta must be zero.

    sin⁡θ=0\sin \theta = 0

    This occurs when the angle θ\theta is 0°0° (or 180°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 0°\boxed{\text{0°}}.

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