100 points

a man pulls a crate with a rope. the crate slides along the floor in the horizontal direction (x-direction). the man exerts a force of 60 n on the rope, and the rope is at an angle 0. describe how the force components change as the angle increases from 0° to 90° and use your graph to explain your answer. give a detailed explanation of the forces at 0 = 0°,45°, and 90°. show a sample calculation at one angle for both components.

The changes in the force components(f_x and f_y) as the angle θ increases from 0° to 90° is that;

The horizontal force component f_x decreases from 60 N to 0 N while the vertical force component f_y increases from 0 N to 60 N.

       We are told the man exerts a force of 60N at an angle of θ. If we resolve the force into the horizontal(x - component) and vertical(y - component), we have;

Horizontal Component; f_x = 60cosθ

Vertical Component; f_y = 60sinθ

We want to find f_x and f_y when θ = 0°,45°, and 90°.

Thus;

When θ = 0;

cos 0° = 1 and sin 0° = 0

∴ f_x = 60 × 1

f_x = 60 N

f_y = 60 × 0

f_y = 0 N

When θ = 45°;

cos 45° = 0.7071 and sin 45° = 0.7071

Thus;

f_x = 60 × 0.7071

f_x = 42.43 N

f_y = 60 × 0.7071

f_y = 42.43 N

When θ = 90°;

cos 90° = 0 and sin 90° = 1

∴ f_x = 60 × 0

f_x = 0 N

f_y = 60 × 1

f_y = 60 N

       From the above calculations, we can see that as θ increases from 0° to 90°, the horizontal force component f_x decreases from 60 N to 0 N while the vertical force component f_y increases from 0 N to 60 N.

       Lastly, the graph of sin θ and cos θ has been attached showing how their values increase or decrease as angle increases from 0 to 90°.

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Explanation:

Spilling the force of 60N on the rope at an angle θ in x-y direction:

fx = Fcosθ = 60cosθ

fy = Fsinθ = 60sinθ

cosθ and sinθ both vary from 0 to 1 between 0° to 90° but they go in opposite direction:

cos0° = 1 and cos90° = 0

sin0° = 0 and sin90° = 1

sin45° = cos45° = 0.7071

When θ increases from 0° to 90°,

θ=0°, fx = 60cos0° = 60N, fy = 60sin0° = 0N

θ=90°, fx = 60cos90° = 0N, fy = 60sin90° = 60N

θ=45°, fx = 60cos45° = fy = 60sin45° = 42.43N

it's a b and