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dauerbach
02-04-2003, 10:35 AM
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

asterion
02-04-2003, 10:42 AM
Use trig. Seriously, that's all there is to it.

dauerbach
02-04-2003, 11:07 AM
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.

RM Mentock
02-04-2003, 11:19 AM
Originally posted by dauerbach
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.

ZenBeam
02-04-2003, 11:21 AM
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.

dauerbach
02-04-2003, 11:58 AM
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.

II Gyan II
02-04-2003, 12:15 PM
Originally posted by dauerbach
[B]For the first problem I had it set up as (60 cos x) + (100 cos x)=125.

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.

Orbifold
02-04-2003, 12:26 PM

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.

RM Mentock
02-04-2003, 12:28 PM
Originally posted by Gyan9
I'm sure, from basic geometry, using the cosine rule or something, you could get the angle in terms of the two "sides".
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.

N9IWP
02-04-2003, 12:33 PM
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.

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

Ring
02-04-2003, 12:42 PM
v dot B = |V||B| cos(x)

Cos(x) = 125/(60*100)

x = arccos(cos(x)) = 88.8 degrees

RM Mentock
02-04-2003, 12:48 PM
Don't cheat off Ring's paper. :)

Ring
02-04-2003, 12:56 PM
Originally posted by RM Mentock
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.

RM Mentock
02-04-2003, 01:03 PM
Even more frustrating when someone gives you the wrong answer though! :)

N9IWP
02-04-2003, 01:44 PM
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

dauerbach
02-04-2003, 01:51 PM
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?

Ring
02-04-2003, 01:58 PM
Originally posted by RM Mentock
Even more frustrating when someone gives you the wrong answer though! :)

I can't stand it! I must end it all now! Such stupidity cannot be allowed to live.

RM Mentock
02-04-2003, 02:21 PM
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.

Rhum Runner
02-04-2003, 03:48 PM
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.?

Orbifold
02-04-2003, 03:52 PM
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. :)

asterion
02-04-2003, 03:56 PM
Originally posted by Ring
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.

Yeah, I wasn't attempting to be a smart-ass, but I don't like to give answers without some idea of what they've done so far.