Two tugs are towing a ship at Constant speed against a total resistance of 50KN. The angles between the tow ropes and the direction of moment are 30degree and 60degree respectively. What is the tension in each of the tow ropes?

50 = T1 cos 30 + T2 cos 60

T1 sin 30 = T2 cos 60

Positive

Assignment

To find the tension in each of the tow ropes, we can analyze the forces acting on the ship and use the principles of vector addition.

First, let's draw a diagram to visualize the situation:

```
F1
-------------------
\ /
\ Ship /
\ /
\ /
\ /
\ /
\ /
F2
```

Here, F1 represents the tension in the first tow rope, F2 represents the tension in the second tow rope, and the angles 30 degrees and 60 degrees are indicated.

Now, let's resolve the forces acting on the ship:

The vertical forces must balance out since the ship is not moving up or down. Therefore, the vertical components of F1 and F2 must cancel each other out:

F1 sin(30) + F2 sin(60) = 0

Simplifying this equation, we get:

(1/2)F1 + (√3/2)F2 = 0

Similarly, the horizontal forces must balance out since the ship is moving at a constant speed. Therefore, the horizontal components of F1 and F2 must add up to the total resistance:

F1 cos(30) + F2 cos(60) = 50kN

Simplifying this equation, we get:

(√3/2)F1 + (1/2)F2 = 50kN

Now we have a system of equations:

(1/2)F1 + (√3/2)F2 = 0
(√3/2)F1 + (1/2)F2 = 50kN

To solve this system of equations, we can multiply the first equation by (√3/2) and the second equation by (1/2) to eliminate the coefficients:

[(√3/2)(1/2)F1 + (√3/2)(√3/2)F2] + [(1/2)(√3/2)F1 + (1/2)(1/2)F2] = 0 + 25kN

Simplifying this equation, we get:

(3/4)F1 + (3/4)F2 = 25kN

We can now solve this equation for either F1 or F2. Let's solve it for F1:

F1 = (25kN - (3/4)F2) / (3/4)

Finally, substitute the value of F1 in the second equation [(√3/2)F1 + (1/2)F2 = 50kN] to obtain the value of F2. Once F2 is found, you can substitute it back in the first equation to find F1.