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A man uses a rope to pull a crate across the floor. The rope forms an angle of 35° to the horizontal. If the man exerts a tensile force of 500 N on the rope, determine the magnitude of the force

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2 parts
2.1

409.58 N\text{409.58 N}

2.2

286.79 N\text{286.79 N}

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2. A man uses a rope to pull a crate across the floor. The rope forms an angle of 35° to the horizontal. If the man exerts a tensile force of 500 N on the rope, determine the magnitude of the force with which the man:

Given:
Magnitude of the force, F=500F = 500 N
Angle to the horizontal, θ=35°\theta = 35°

Part 2.1· 1 step

tends to cause the crate to move horizontally.

This is the horizontal component of the force.

  1. Calculate the horizontal component (FxF_x).

    Fx=Fcos⁡θF_x = F \cos \theta Fx=500 N×cos⁡(35°)F_x = 500 \text{ N} \times \cos(35°) Fx≈500 N×0.81915F_x \approx 500 \text{ N} \times 0.81915 Fx≈409.58 NF_x \approx 409.58 \text{ N}

    The magnitude of the force that tends to cause the crate to move horizontally is 409.58 N\boxed{\text{409.58 N}}.

Part 2.2

286.79 N\text{286.79 N}

tends to cause the crate to lift vertically.

This is the vertical component of the force.

  1. Calculate the vertical component (FyF_y).

    Fy=Fsin⁡θF_y = F \sin \theta Fy=500 N×sin⁡(35°)F_y = 500 \text{ N} \times \sin(35°) Fy≈500 N×0.57358F_y \approx 500 \text{ N} \times 0.57358 Fy≈286.79 NF_y \approx 286.79 \text{ N}

    The magnitude of the force that tends to cause the crate to lift vertically is 286.79 N\boxed{\text{286.79 N}}.

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