Calculate Fnet & Direction: 2 Cars & Big Airplane

[homevolend]

Homework Statement

2 cars at an airport are pulling a big airplane with two steel cables. Car 1 pulls with force 51000N at 30° angle. Car 2 pulls with force 68000N at 60° angle, both pull in forward direction. find fnet and direction

[PLAIN]http://img16.imageshack.us/img16/8035/planek.png

The Attempt at a Solution

Well we need to finding F1x F1y and F2x F2y.

I found Fx already by doing:


F1x=51000cos30
F1x=44167.3N

F2x=68000cos60
F2x=34000N

So Fx= F1x+f2x
Fx=78167.3 N


But I don't know how to find Fy and the direction, but I know I need fy first before I find the direction. I know for Fy you use the sin instead of cos but I don't know how to find Fy.. need some help please.


thanks!

 

[fizzynoob]

Its just simple trig and breaking down the force into the components. You cannot add vectors directly, but you can add there components together.

Sum of forces in Y direction = F1sin(30) - F2sin(60) //the minus is due to opposite direction

 

[homevolend]

Ok I did this but does y turn out negative that's what I got.. thanks for the help. I know how to find direction and fnet now, just not sure on the negative part.

thanks

 

[fizzynoob]

yes, you get a negative answer but that negative is there just to show direction. So in this case the net force in Y is directed downward.

So Fy = -3.3*10^4 (j) j = y direction

Fx = -8.81 * 10^4 (i) i = x direction (I put the negative since the forces are pulling to the left, negative x direction)Adding the components will give you a vector describing the net force

Fnet = Fx + Fy = -8.81*10^4 (i)-3.3*10^4 (j)

Looking at the picture, doesn't that vector make sense? You have 68000 Newtons pulling downward at 60 degrees, there for the force has to be downward. X direction is easy, obviously the forces is to the left

To find the magnitude of Fnet you simple take the square root of the sum of the squared components. Its Pythagorean theorem basically