Adding vectors and angles using trig

[talaroue]

Homework Statement

300 lb force is to be resolved into components along a-a(prime) and b-b(prime).
a.) Determine the angle by trigonometry knowing that the compnent along line a-a(prime) is to be 240 lb.

b.) What is the corresponding value of the component b-b(prime)

in the picture i changed a-a(prime) to x-x(prime) and b-b(prime) to y-y(prime)

Homework Equations

sin(theata 1)/A=sin(theata 2)

The Attempt at a Solution

I don't know how to do this seeing as it is tilted?

 

[Doc Al]

I'm having trouble understanding the problem and the diagram. I see a vector, a horizontal line label x at one end and x' at the other, and a tilted line labeled y' at one end. Can you restate the problem?

 

[talaroue]

i took a different picture

 

[talaroue]

This is straight from the book...

"The 300-lb force is to be resolved into components along the lines a-a' and b-b'

a.) Determine the angle alpha by trigonometry knowing that the component along line a-a' is to be 240-lb

b.0What is the corresponding value of the component along b-b'

 

[Doc Al]

i took a different picture

Where is it?

 

[talaroue]

 

[talaroue]

sorry it took me a while to upload it to photobucket, then onto here sorry

 

[talaroue]

does this help?

 

[Doc Al]

does this help?

Yes--it's clear now.

Since you're given the component of the force along a-a', what must alpha be? That's step one. Then you'll have all the angles needed.

 

[talaroue]

Ok, i thought the angle of a-a' would have been 180 since its a straight line. But then its wrong. Then i thought maybe its 120 but then after working through it again it is wrong.

 

[Doc Al]

Ok, i thought the angle of a-a' would have been 180 since its a straight line. But then its wrong. Then i thought maybe its 120 but then after working through it again it is wrong.

Answer this: How would you find the x-component of a vector making an angle of alpha with the x-axis? It's the same problem.

 

[talaroue]

cos(alpha)=a-a'/300

is that what you are asking?

 

[talaroue]

which would make alpha=36 deg which is wrong.

 

[Doc Al]

cos(alpha)=a-a'/300

is that what you are asking?

I would write it as:
F_a-a' = F cos(alpha)
240 = 300 cos(alpha)

 

[talaroue]

right so alpha=cos^-1(.8)=36 deg, the anwser is 76.1 deg...which doesn't make sense

 

[Doc Al]

which would make alpha=36 deg which is wrong.

cos^-1(240/300) = 36.9 deg

Why do you say that's wrong?

 

[talaroue]

because in the back of the book it says its 76.1

 

[Doc Al]

because in the back of the book it says its 76.1

What's the answer given for b? Is it consistent?

 

[talaroue]

the answer for b is 336 lb

 

[Doc Al]

because in the back of the book it says its 76.1

OK. I understand what they want. The axes are not orthogonal, thus to find the a-a' component you must draw a line parallel to b-b' that intersects the tip of the 300-lb vector. You'll get a triangle, two sides of which are given (240 and 300). You'll be able to use some trig to find alpha. (The answer is correct, now that I understand it. :uhh:)

(Taking Fcos(alpha) is only good for orthogonal coordinates, not skewed. Sorry about that!)

And to then find the b-b' component, you'll draw a line parallel to a-a' and get another triangle.

 

[talaroue]

so then how do you find the correct angle that they did?

 

[Doc Al]

so then how do you find the correct angle that they did?

Using some trig. (Law of sines, for one.)

 

[talaroue]

which i orginally had, but the problem was I couldn't figure out what angle i use for axis a-a'

 

[Doc Al]

which i orginally had, but the problem was I couldn't figure out what angle i use for axis a-a'

The line that you'll draw parallel to b-b' will also make a 60 degree angle with a-a'. So one of the angles in that triangle is 60. Then use the law of sines to find one other angle. Then find angle alpha.

 

[talaroue]

sin(60)/300=sin(theata)/240

i see now so then i get 43.9

180-60-43.8=76.1!
and then from there i can find b-b'

thank you so much your a life saver