Question

Two horses are pulling a barge with a mass 2000 kg along a canal. The cable connected to the first horse makes an angle of θ1= 40 o with respect to the direction of the canal, while the cable connected to the second horse makes an angle of θ2= -20o . Find the acceleration of the barge (magnitude and direction), starting at rest, if each horse exerts a force of magnitude 600 N. ignore forces of resistance.

Answer 1

Answer:

a = 0.5195 m/s²

θ = 9.997º ≈ 10º

Explanation:

We apply Newton's 2nd Law as follows:

∑ Fx = m*ax

∑ Fy = m*ay

Then we have

∑ Fx = F₁x + F₂x = m*ax    ⇒ 600*Cos 40º + 600*Cos (-20º) = 2000*ax

⇒   ax = 0.5117 m/s²

∑ Fy = F₁y + F₂y = m*ay    ⇒ 600*Sin 40º + 600*Sin (-20º) = 2000*ay

⇒   ay = 0.0902 m/s²

the magnitude of the acceleration of the barge is

a = √(ax² + ay²) = √((0.5117 m/s²)² + (0.0902 m/s²))= 0.5196 m/s²

and the direction is

θ = Arctan (ay / ax) = Arctan (0.0902 / 0.5117) = 9.997º ≈ 10º

Answer 2

The magnitude and direction of the acceleration are respectively; 0.5196 m/s² and 10°

What is the magnitude and direction of the acceleration?

Let us first calculate the x-components of the forces exerted by the horses;

F₁ₓ = F₁ * cosθ₁ = 600 * cos 40° = 459.63 N

F₂ₓ = F₂ * cosθ₂ = 600 * cos -20° = 563.82

The total force in the x-direction is; Fₓ = F₁ₓ + F₂ₓ

Fₓ = 459.63 + 563.82

Fₓ = 1023.45 N

Similarly, calculate the y-components of the forces exerted by the horses;

F₁y = F₁ * sinθ₁ = 600 * sin 40° = 385.67 N

F₂y = F₂ * sinθ₂ = 600 * sin -20° = -205.21 N

The total force in the y-direction is; Fy = F₁y + F₂y

Fy = 385.67 - 205.21

Fy = 180.46 N

x-component of acceleration is;

aₓ = Fₓ/m = 1023.45/2000

aₓ = 0.5117 m/s²

a_y = F_y/m = 180.46/2000

a_y = 0.0903 m/s²

Magnitude of acceleration is;

a = √(aₓ² + a_y²)

a = √(0.5117² + 0.09023²)

a = 0.5196 m/s²

Direction;

θ = tan⁻¹(aₓ/a_y) = tan⁻¹(0.5117/0.0903)

θ = 10°

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