Three tug boats are pulling on a cruise ship with three ropes. Two pull with a force of 1800 N each at angle of [N15°E] and [S24°E]. If the net force on the ship is zero, what must the tension in the third rope be?
To find the tension in the third rope, we can first resolve the forces of the two tug boats into their x-component and y-component.
Given:
Force 1 (F1) = 1800 N at an angle of N15°E
Force 2 (F2) = 1800 N at an angle of S24°E
We can use trigonometry to find the x-component and y-component of each force.
1. Resolving Force 1:
The x-component, F1x, can be found using the cosine function:
F1x = F1 * cos(theta)
F1x = 1800 N * cos(15°)
The y-component, F1y, can be found using the sine function:
F1y = F1 * sin(theta)
F1y = 1800 N * sin(15°)
2. Resolving Force 2:
Similarly, we can find the x-component and y-component of Force 2.
The x-component, F2x, can be found using the cosine function:
F2x = F2 * cos(theta)
F2x = 1800 N * cos(180° - 24°)
The y-component, F2y, can be found using the sine function:
F2y = F2 * sin(theta)
F2y = 1800 N * sin(180° - 24°)
Now, let's calculate the values:
F1x = 1800 N * cos(15°)
F1x ≈ 1742.84 N
F1y = 1800 N * sin(15°)
F1y ≈ 464.19 N
F2x = 1800 N * cos(180° - 24°)
F2x ≈ -1507.97 N
F2y = 1800 N * sin(180° - 24°)
F2y ≈ -741.15 N
Since the net force on the ship is zero, the x-components and y-components of all forces must balance each other:
Net force in the x-direction: F1x + F2x + F3x = 0
F3x = - (F1x + F2x)
Net force in the y-direction: F1y + F2y + F3y = 0
F3y = - (F1y + F2y)
Now, we can calculate the tension in the third rope (F3):
F3 = sqrt(F3x^2 + F3y^2)
Substituting the calculated values, we get:
F3x = - (1742.84 N + (-1507.97 N)) = -234.87 N
F3y = - (464.19 N + (-741.15 N)) = 276.96 N
F3 = sqrt((-234.87 N)^2 + (276.96 N)^2)
F3 ≈ 367.50 N
Therefore, the tension in the third rope must be approximately 367.50 N.
F1 + F2 + F3 = 0
1800N[75o] + 1800N[294o] + F3 = 0
1800*cos75+i1800*sin75 + 1800*cos294+i
1800*sin294+F3 = 0, 465.87+i1738.7 + 732.1+i(-1644)+F3 = 0
1198 + i94.7+F3 = 0
F3 = -1198 - i94.7 = -1202 N.[4.52o]
Tan Ar = -94.7/-1198 = 0.07905
Ar = 4.52o = Reference angle.
A = 4.52 + 180 = 184.52o CCW = 85.5o W.
of S.(S85.5oW)