the sum of the forces FA and FB exerted by the horizental cables is parrallel to line L and FA=4500N determine?

[Erik C]

Two snowcats tow a housing unit to a new location. The sum of teh two forces FA and FB exerted on the unit by the horizontal cables is parallel to line L and FA=4500N. Determin Fb and the magnitude of FA +FB

the angle between FA and line L is 50 degrees. The angle between Fb and line L is 30 degrees

[sojsail]

Understanding what's going on helps to understand how to work it out. Line L must point to where you want the unit to go. Cat A must be a bit weaker because it has to pull more to the side to keep the thing going straight.

You can resolve forces FA and FB into forces parallel and perpendicular to L. Call these components FAP, FAL, FBP, FBL (P for perpendicular to L and L for in the L direction).
FAP = FA*sin50
FAL = FA*cos50
FBP = FB*FBsin30
FBL = FB*cos30

Since it's moving along L, the 2 perpendicular components must cancel each other out. So
FA*sin50 = FB*FBsin30
From that, you can get the answer the 1st question and then the 2nd question.

[electron1]

Two snowcats tow a housing unit to a new location. The sum of teh two forces FA and FB exerted on the unit by the horizontal cables is parallel to line L and FA=4500N. Determin Fb and the magnitude of FA +FB

the angle between FA and line L is 50 degrees. The angle between Fb and line L is 30 degrees

Yea, a vector problem
The component of FA parallel to line l = FA * cos 50°
The component of FB parallel to line l = FB * cos 30°

The sum of these 2 forces = 4500 N
FA * cos 50° + FB * cos 30° = 4500

Since the housing unit moves parallel to line L, and does not move perpendicular to line L; the perpendicular components of FA = FB

The component of FA perpendicular to line l = FA * sin 50°
The component of FB perpendicular to line l = FB * sin 30°

2 equations
Eq. #1. FA * sin 50° = (FB * sin 30°)
Eq. #2. FA * cos 50° + FB * cos 30° = 4500

Solve Eq.#1 for FA:
FA = (FB * sin 30°) ÷ sin 50°
FA = FB * (sin 30° ÷ sin 50°)

Substitute into Eq. #2
[FB * (sin 30° ÷ sin 50°) * cos 50°] + FB * cos 30° = 4500

FB * (0.4195 + 0.866) = 4500

0.1.2855 FB = 4500
FB = 3500 N
FA = 3500 * (sin 30° ÷ sin 50°)
FA = 2284.5 N

Check:
FA * sin 50° + (FB * sin 30°) = 0
2284.5 * sin 50° = (3500 * sin 30°)
1750 = 1750

2284.5 * cos 50° + 3500 * cos 30° = 4500
1468.4 + 3031= 4499.5( rounding error)