This is a pre-calc problem I am having trouble with:
Two forces of 60 lb and 100 lb have a resultatnt force of 125 lbs.
Find the angle between the two forces.
Thanks
Use trig. Seriously, that’s all there is to it.
I know thats all there is to it, but when I do what I think is correct I get an answer different from what is in the back of the book.
The other problem with the wrong answer:
An airplane has an air speed of 450 mph at a bearing of N 30 E. The velocity of the wind is 20 mph from the west. Find the ground speed and the direction of the plane.
I know how I am supposed to do it, but after trying each like 10 times I come up with the wrong answers. A more detailed explaination would be useful.
Yes, we need details. Show us how you think you’re supposed to do it.
I get about 80.3 degrees for the first. If they were perpendicular, the resultant would be 116.6, so 80 degrees sounds about right. Does this match either you or the book? All I did was draw a picture with the two vectors and an angle between them, and calculate the magnitude as a function of angle, then back out the angle.
For the first problem I had it set up as (60 cos x) + (100 cos x)=125.
For the second problem I have (450 cos 30) + (30 sin 120) = x.
You’re assuming that the resultant bisects the angle between the contributing vectors. Not necessarily true.
Take angle from 100 lbs to resultant as a100 deg and angle from 60 lbs as a60 deg.
You get 100cos(a100) + 60cos(a60) = 125
AND 100sin(a100) = 60sin(a60).
Although I’m just complicating matters.
I’m sure, from basic geometry, using the cosine rule or something, you could get the angle in terms of the two “sides”.
I’m just too lazy to work it out now.
It sounds like you’re getting confused about how to add vectors.
If v and w are two vectors, of length 60 and 100 respectively, and if the angle between them is t, then the length of v+w is not (60+100)*cos(t), as you seem to be assuming. You get the norm of v+w the same way you find the norm of any other vector: by taking the dot product of it with itself. Only when you expand the dot product of v+w with itself, you get a relationship between the norms of v, w, and v+w, and the angle between v and w.
I’m reluctant to give more of a hint than that for a homework problem, but I hope that will get you started.
Yep, that’s it. The cosine rule relates the three sides of a triangle and one of it’s angles. In the first problem, we have three “sides” of the triangle–just solve the cosine rule for the angle opposite the appropriate side.
For the second problem, remeber that navigation degrees start at 0 pointing “up” and go clockwise, and math degrees start at 0 pointing “right” and go counter-clockwise.
So if the plane is heading at 30 navigation degrees, its heading at 60 math degrees.
For this problem I would break the vectors into thier x/y (N/E) components, add them, then convert back to vector. But there are other valid ways.
brian
v dot B = |V||B| cos(x)
Cos(x) = 125/(60*100)
x = arccos(cos(x)) = 88.8 degrees
Don’t cheat off Ring’s paper. 
Sometimes when someone’s busted their gorgonzolas on a problem the only decent thing is to give them the damn answer.
A lot of people disagree with this, but I think spending too much time on one problem isn’t helpful. It’s just frustrating.
Even more frustrating when someone gives you the wrong answer though!
I meant convert back to POLAR.
Here’s my solution (no fair peeking)
450MPH at an angle of 30 degrees (60 math degrees)
=225MPH east + 389.7 MPH north
+20 MPH from the west (to the east)
=245MPH east + 389.7 MPH north
=460.3 MPH @ 57.84 math degrees
=460.6 MPH @ 32.16 navigation degrees
Brian
not guaranteeing my work
assuming both speeds are using knots or statute
Thanks. I admit I am not very good at this stuff, and indeed I did spend a ton of time on it. Thank you for your help. With a show of hands, who is available to take my final exam for me?
I can’t stand it! I must end it all now! Such stupidity cannot be allowed to live.
Hold it right there Ring, I thought you were buying tonight? Oh! I get it…think you can get out of it, eh? Let’s go now, and then afterwards we can help dauerbach with his test.
I don’t think you can calculate ground speed with out knowing the altitude of the plane, right? Or am I making this too complicated in my own head.?
Rhum Runner: in my experience these kind of problems always assume the plane is flying level unless stated otherwise. Unless you’re referring to the effect of the curvature of the earth, in which case you’ve just zoomed right out the domain of a pre-calc course.