Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

The problem is to determine the components of $F_2$.

Problem image

Problem

Method 1

method 1

Method 2

enter image description here

My question is why do I receive different answers?

share|improve this question
    
The components with respect to horizontal and vertical axes? –  rschwieb Feb 5 '13 at 19:48
    
Along the u, v axes. –  miles t Feb 5 '13 at 19:49

2 Answers 2

up vote 2 down vote accepted

Agreed, @5PM.

Here's a general approach to vector problems that I've found useful, using the magnitude-angle notation.

There is the original vector, 150$\angle$165°, with x-component $150 *$cos$(165°)$and y-component $150 *$sin$(165°)$

This is to be broken down into two non-orthogonal "components", F$_v$$\angle$90°, and F$_u$$\angle$195°

Each of these two last can be separately broken down into x and y components in the same way as the 150 N force, above.

Then the 150 N x-component equals the $F_v$ x-component plus the $F_u$ x-component, and the 150 N y-component equals the $F_v$ y-component plus the $F_u$ y-component. Two equations, two unknowns, all angles known , so just solve. Oh, and one equation has only one unknown!

share|improve this answer
    
That's a nice way of thinking about components. Doing so allows me to calculate ("wolframalpha.com/input/?i=Solve[{uCos[195]%2BvCos[90]%3D%3D150*Co‌​s[165]%2CuSin[195]%2BvSin[90]%3D%3D150*Sin[165]}%2C{u%2Cv}") –  miles t Feb 6 '13 at 8:54

Method 1 is correct. In method 2 you calculated orthogonal projections of the force $F_2$ onto the given axes. However, the components of a force coincide with orthogonal projections only when the axes are mutually orthogonal.

share|improve this answer
    
True; that's why I think the original question is worded in a confusing manner, in that it refers to the two, non-orthogonal desired vectors as "components" of the original vector.(See the first comment on the question by rshwieb, above, and the reply by the OP referring to the u and v directions as axes) The method I proposed treated all three vectors the same, resolving each one into two orthogonal, x-y components that could then be dealt with as scalars. –  User58220 Feb 6 '13 at 16:05
    
@User58220 From mathematical point of view, there is nothing wrong in calling non-orthogonal vectors components. The vectors $u,v$ form a basis of $\mathbb R^2$; they are distinguished by this designation. Given any other vector $w$, we can uniquely decompose/expand it as $w=au+bv$. It is natural to refer to $au$ and $bv$ as components of $w$ in this basis. What else should we call them? // Note that the existence of such expansions has nothing to do with orthogonality: it makes sense in vector spaces that have no inner product, and therefore no concept of orthogonality. –  user53153 Feb 6 '13 at 16:59

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.