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Find the vertical and horizontal components of 500 N force when it is inclined at: (a) 60 (b) 90° (c) 150° to the level ground.

Physics

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Answers

3 parts
a

250 N250 \text{ N}

433.01 N433.01 \text{ N}

b

0 N0 \text{ N}

500 N500 \text{ N}

c

−433.01 N-433.01 \text{ N}

250 N250 \text{ N}

Step-by-step solution

6 steps

To find the horizontal and vertical components of a force FF inclined at an angle θ\theta to the horizontal, we use the following formulas:
Horizontal component (FxF_x) = Fcos⁡θF \cos \theta
Vertical component (FyF_y) = Fsin⁡θF \sin \theta
Given force F=500F = 500 N.

Part a· 2 steps

When the force is inclined at 60∘60^\circ

  1. Calculate the horizontal component.

    Fx=Fcos⁡θ=500 N×cos⁡(60∘)F_x = F \cos \theta = 500 \text{ N} \times \cos(60^\circ) Fx=500 N×0.5=250 NF_x = 500 \text{ N} \times 0.5 = 250 \text{ N}

  2. Calculate the vertical component.

    Fy=Fsin⁡θ=500 N×sin⁡(60∘)F_y = F \sin \theta = 500 \text{ N} \times \sin(60^\circ) Fy=500 N×32≈500 N×0.8660F_y = 500 \text{ N} \times \frac{\sqrt{3}}{2} \approx 500 \text{ N} \times 0.8660 Fy≈433.01 NF_y \approx 433.01 \text{ N}

    The components are: Horizontal = 250 N\boxed{250 \text{ N}}, Vertical = 433.01 N\boxed{433.01 \text{ N}}.

Part b

0 N0 \text{ N}

When the force is inclined at 90∘90^\circ

  1. Calculate the horizontal component.

    Fx=Fcos⁡θ=500 N×cos⁡(90∘)F_x = F \cos \theta = 500 \text{ N} \times \cos(90^\circ) Fx=500 N×0=0 NF_x = 500 \text{ N} \times 0 = 0 \text{ N}

  2. Calculate the vertical component.

    Fy=Fsin⁡θ=500 N×sin⁡(90∘)F_y = F \sin \theta = 500 \text{ N} \times \sin(90^\circ) Fy=500 N×1=500 NF_y = 500 \text{ N} \times 1 = 500 \text{ N}

    The components are: Horizontal = 0 N\boxed{0 \text{ N}}, Vertical = 500 N\boxed{500 \text{ N}}.

Part c

−433.01 N-433.01 \text{ N}

When the force is inclined at 150∘150^\circ

  1. Calculate the horizontal component.

    Fx=Fcos⁡θ=500 N×cos⁡(150∘)F_x = F \cos \theta = 500 \text{ N} \times \cos(150^\circ) Fx=500 N×(−32)≈500 N×(−0.8660)F_x = 500 \text{ N} \times \left(-\frac{\sqrt{3}}{2}\right) \approx 500 \text{ N} \times (-0.8660) Fx≈−433.01 NF_x \approx -433.01 \text{ N}

  2. Calculate the vertical component.

    Fy=Fsin⁡θ=500 N×sin⁡(150∘)F_y = F \sin \theta = 500 \text{ N} \times \sin(150^\circ) Fy=500 N×0.5=250 NF_y = 500 \text{ N} \times 0.5 = 250 \text{ N}

    The components are: Horizontal = −433.01 N\boxed{-433.01 \text{ N}}, Vertical = 250 N\boxed{250 \text{ N}}.

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