Question

Two ropes are attached to either side of a 100.0 kg wagon as shown below. The rope on the right is pulled at an angle 40.0° relative to the horizontal with a tension of 120.0N. The rope on the left is pulled directly to the west with a tension of 80.0N.
1. Find the x and y components of the tension forces applied by both ropes. Show all work.
2. Find the net force in the x (horizontal) direction assuming no friction. Show all work.
3. Find the net acceleration of the wagon in the horizontal direction. Show all work.

Thanks in advance for any help, it's appreciated

Answer 1

The acceleration of the wagon is found by applying Newton's Second Law of motion.

1. The responses for question 1 are;

• x-component of the tension in the rope on the right is approximately 91.93 N

• y-component of the tension in the rope on the right is approximately 71.135 N

• x-component of the tension in the rope on the left is -80.0 N

• y-component of the tension in the rope on the left is 0

2. The net force in the x-direction is approximately 11.93 N

3. The net acceleration of the wagon in the horizontal direction is approximately 0.1193 m/s².

Reasons:

The given parameters are;

Mass of the wagon, m = 100.0 kg

Angle of inclination to the horizontal of the rope to the right, θ = 40.0°

Tension in the rope on the right = 120.0 N

Direction in which the rope on the left is pulled = To the west

Tension in the rope on the left = 80.0 N

1. The x and y component of the tension in the rope on the right are;

x-component = 120.0 N × cos(40.0°) ≈ 91.93 N

y-component = 120.0 N × sin(40.0°) ≈ 77.135 N

The x and y component of the tension in the rope on the left are;

x-component = 80.0 N × cos(180°) = -80.0 N

y-component = 80.0 N × sin(180°) = 0.0 N

2. The net force in the horizontal direction, Fₓ, is found as follows;

Fₓ = The x-component of the rope on the left + The x-component of the rope on the right

Which gives;

Fₓ = 91.93 N - 80.0 N = 11.93 N

3. The net acceleration of the block is given as follows;

According to Newton's Second Law of motion, we have;

Force in the horizontal direction, Fₓ = Mass of wagon, m × Acceleration of the wagon in the horizontal direction, aₓ

Fₓ = m × aₓ

Therefore;

$\displaystyle a_x = \frac{F_x}{m} \approx \frac{11.93 \, N}{100.0 \, kg} = \mathbf{0.1193 \ m/s^2}$

• The acceleration of the wagon in the horizontal direction, aₓ ≈ 0.1193 m/s².

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