Physics and Vectors! Rocket Taking Off.?

Question

My physics class is on-line, and i have no idea how the text explains this. Please help someone to at least explain it. A rocket fires two engines simultaneously. One produces a thrust of 745 N directly forward while the other gives a thrust of 510 N at an angle 35.6 degrees above the forward direction. Find the magnitude of the resultant force which these engines exert on the rocket.

Answer 1

Using CCW rotation from x-axis

Assuming: "an angle 35.6 degrees above the forward direction"

means: 35.6 degrees west of north.

let north=90 degrees

let "35.6 degrees above north"=90+35.6=125.6 degrees

relative to horizontal.

x components of thrust:

745*Cos( 90)+ 510* Cos(125.6) =-296.883i

sum the y components of thrust:

745* Sin(90)+ 510* Sin(125.6) =1159.681j

Resultant vector thrust = 1197.08 N "answer"

Direction: Just in case!

Angle with the positive x-axis: =104.36 degrees

need more help, let me know.

Edit: please define:"degrees above the forward direction"

need: direction +,- north, NE,angle,etc.

Answer 2

hi, basically what you want to do is first draw a picture (always helps see what u need to do) then make the vectors 'nose to tail' so the end of one vector is on the start of the other vector, then u work out the angle they make now, which turns out to be 180 - 36.5 = 144.4. now u can use cosine rule to find the magnitude so that is magnitude squared = 745 squared + 510 squared - 2 times 745 times 510 times cos 144.4 = 1433000.268 (big number) then u square root that number to get 1197.08 N <--- is your magnitude.

now for the angle , don't know if u need it or not, but you would use sine rule. sin of the angle divide 510 = sin 144.4 over 1197.08 which means with a bit of swapping around u get the angle = to 14.36 degrees,

hope that helped :D