A 314 kg crate is pulled by two horizontal forces. The first force is 255 N at a 43.5 degree angle and the second is 360 N at a 212 degree angle. What is the direction of the acceleration of the crate?



Answer :

Answer:

187.22°

Explanation:

According to Newton's second law of motion, the net force on an object is equal to the mass times the acceleration. By resolving the forces into their x and y components, the components of the acceleration vector can be found. We can then use trigonometry to find the direction.

The x and y components of a force are Fₓ = F cos θ and Fᵧ = F sin θ.

The x component of the first force is:

F₁ₓ = F₁ cos θ₁

F₁ₓ = 255 cos 43.5

F₁ₓ = 184.97 N

The y component of the first force is:

F₁ᵧ = F₁ sin θ₁

F₁ᵧ = 255 sin 43.5

F₁ᵧ = 175.53 N

The x component of the second force is:

F₂ₓ = F₂ cos θ₂

F₂ₓ = 360 cos 212

F₂ₓ = -305.30 N

The y component of the second force is:

F₂ᵧ = F₂ sin θ₂

F₂ᵧ = 360 sin 212

F₂ᵧ = -190.77 N

The x component of the acceleration is:

∑Fₓ = maₓ

F₁ₓ + F₂ₓ = maₓ

184.97 + -305.30 = 314 aₓ

aₓ = -0.3832 m/s²

The y component of the acceleration is:

∑Fᵧ = maᵧ

F₁ᵧ + F₂ᵧ = maᵧ

175.53 + -190.77 = 314 aᵧ

aᵧ = -0.0485 m/s²

Using trigonometry, the angle of the acceleration vector relative to the +x axis is:

tan θ = aᵧ / aₓ

tan θ = -0.0485 / -0.3832

tan θ = 0.1267

θ = 187.22°

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