Question

A ship is towed through a narrow channel by applying forces to three ropes attached to its bow. Determine the magnitude and orientation θ of the force F→ so that the resultant force is in the direction of line a and the magnitude of F→ is as small as possible. Given: A = 5.5 kN.

Answer 1

This question is solved using an available similar problem as data provided for the forces was not given.

Repeat the same steps outlined for your problem.

Regards.

Answer:

F = 1.598 KN , Q = 90 degree (+ y-axis)

Explanation:

Sum of Forces in x-direction to the left (+)

2 cos (30) + 3cos (60) + F*cos (Q) = F_a   ..... 1

Sum of Forces in y-direction to the up (+)

2 sin (30) + F*sin (Q) - 3 sin (60)  ...... 2

Using Eq 2 and solve:

F*sin (Q) = 1.598 KN

F_min when sin (Q) is max, max possible value of sin(Q) = 1 @ Q = 90 degrees.

Hence,

F_min = 1.598 KN

Using Eq 1 @ Q = 90 degrees and F = 1.598 KN:

F_a = 2 cos (30) + 3cos (60)  = 3.2 KN