Math Problem Statement

Two tugboats pull a barge by means of 2 horizontal ropes at the same elevation, as shown. The tensions in these ropes are F1 = 280 kN and F2 = 336 kN respectively which makes an angle of 60o between them. If the direction of the motion of the barge should be to the east direction,

  1. Compute the direction of the pull F1.
  2. Compute the direction of the pull F2.
  3. Compute the resultant pull on the barge

Solution

To solve this problem, we will break down the forces into their components and then determine the resultant force vector. Here's the step-by-step approach:

Step 1: Represent the Forces as Vectors

Given:

  • Force F1=280kNF_1 = 280 \, \text{kN}
  • Force F2=336kNF_2 = 336 \, \text{kN}
  • The angle between F1F_1 and F2F_2 is 6060^\circ.

Let:

  • θ1\theta_1 be the angle of F1F_1 with respect to the east (x-axis).
  • θ2\theta_2 be the angle of F2F_2 with respect to the east (x-axis).

We are required to determine θ1\theta_1, θ2\theta_2, and the resultant force RR.

Step 2: Resolve the Forces into Components

Assume:

  • F1F_1 is at an angle θ1\theta_1 to the x-axis.
  • F2F_2 is at an angle θ2\theta_2 to the x-axis.

Components of F1F_1:

F1x=F1cos(θ1)F_{1x} = F_1 \cos(\theta_1) F1y=F1sin(θ1)F_{1y} = F_1 \sin(\theta_1)

Components of F2F_2:

F2x=F2cos(θ2)F_{2x} = F_2 \cos(\theta_2) F2y=F2sin(θ2)F_{2y} = F_2 \sin(\theta_2)

Step 3: Apply the Conditions

  1. The barge moves directly east, so the total y-component of the resultant force should be zero: F1y+F2y=0.F_{1y} + F_{2y} = 0.

  2. Solve for θ1\theta_1 and θ2\theta_2 using the condition above.

Step 4: Compute the Resultant Force

The resultant force RR can be found by summing the x-components of both forces: Rx=F1x+F2x.R_x = F_{1x} + F_{2x}.

Now, let's calculate these values:Here are the results of the calculations:

  1. Direction of the pull F1F_1: θ168.95\theta_1 \approx 68.95^\circ (counterclockwise from the east direction).
  2. Direction of the pull F2F_2: θ2128.95\theta_2 \approx 128.95^\circ (counterclockwise from the east direction).
  3. Resultant pull on the barge: R534.21kNR \approx 534.21 \, \text{kN}.

These angles and the resultant force ensure that the barge moves directly eastward.

Would you like more details on the calculations or have any questions? Here are some related questions that might help:

  1. How would the result change if the angle between the forces was different?
  2. What would happen if one of the forces was increased or decreased?
  3. How can the equilibrium conditions be applied to find unknown forces?
  4. What role does vector resolution play in real-world physics problems?
  5. How can you determine the angle between two forces using their components?

Tip: Always ensure the equilibrium of forces in both x and y directions when dealing with such vector problems to simplify calculations.

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Math Problem Analysis

Mathematical Concepts

Vector Resolution
Trigonometry
Statics

Formulas

F_{1x} = F_1 cos(θ1)
F_{1y} = F_1 sin(θ1)
F_{2x} = F_2 cos(θ2)
F_{2y} = F_2 sin(θ2)
Resultant Force R = sqrt(Rx^2 + Ry^2)

Theorems

Equilibrium Condition for Forces
Trigonometric Angle Resolution

Suitable Grade Level

Grades 10-12