Answer :
Final answer:
The magnitude of the net force exerted on the donkey is 166.71 N directly ahead. The direction of the net force is 0° from straight ahead, indicating that it is pointing directly ahead without any angle to the left or right.
Explanation:
The question involves calculating the net force on a donkey being pulled by three people in different directions. We'll use vector addition to combine the forces exerted by Jack, Jill, and Jane.
First, we convert the angled forces into components. Jill's force can be broken down into horizontal and vertical components using trigonometry:
- Horizontal component (Jill): 60.9 N * cos(45°) = 43.06 N to the left
- Vertical component (Jill): 60.9 N * sin(45°) = 43.06 N upward
Jane's force, also at a 45° angle, has the following components:
- Horizontal component (Jane): 159 N * cos(45°) = 112.47 N to the right
- Vertical component (Jane): 159 N * sin(45°) = 112.47 N upward
Jack's force is directly ahead (horizontal) and there is no need to break it down:
- Horizontal component (Jack): 97.3 N directly ahead
To calculate the total horizontal force, we add the horizontal components, keeping in mind the directions:
- Total horizontal force = 97.3 N (ahead) + 112.47 N (right) - 43.06 N (left) = 166.71 N ahead
The vertical forces from Jill and Jane are equal and opposite, so they cancel each other out:
- Total vertical force = 43.06 N (upward) - 112.47 N (upward) = 0 N
Since there are no net vertical forces, the magnitude of the net force on the donkey is the total horizontal force, which is 166.71 N directly ahead.
The direction of the net force is straight ahead, which can be expressed as an angle of '0°' from straight ahead.
Answer:
most stubborn is Jill, F = 227.97 N and θ = 43º
Explanation:
As the forces give us we must use Newton's second law, in this case as the donkey does not move the acceleration is zero
Let's start by breaking down the forces
Jack
F1 = 97.3 i ^ N
Jill
F2 = 60.9 N
θ = 45º to the left
This angle measured from the positive part of the x axis is T ’= 90 + 45 = 135º
F2x = F2 cos (θ’)
F2y = F2 sin (θ’)
F2x = 60.9 cos (135)
F2y = 60.9 sin (135)
F2x = -43.06 N
F2y = 43.06 N
Jane
F3 = 159N
T = 45º
F3x = F3 cos θ
F3y = F3 sin θ
F3x = 159 cos 45
F3y = 159 sin 45
F3x = 112.43 N
F3y = 112.43 N
a) The most stubborn of the three must be the one who pulls in the opposite direction, bone Jill who is pulling to the left while the other two do it to the right
b) We perform the summary of the forces in each axis
X axis
Fx = F1 + F2x + F3x
Fx = 97.3 -43.06 +112.43
Fx = 166.67 N
Axis y
Fy = F2y + F3y
Fy = 43.06 + 112.43
Fy = 155.49 N
F = √ Fx²+Fy²
F = 166.67² + 155.49²
F = 227.97 N
We use trigonometry
tan θ = Fy / Fx
θ = tan⁻¹ (155.49 / 166.67)
θ = 43º
Regarding the x axis
