Undergraduate Mechanics (Problem with force expressed as basis vectors)

[s3a]

Homework Statement

The problem along with its solution is attached as Problem 1-2.jpg.

Homework Equations

Norm of a vector.

The Attempt at a Solution

Starting from the final answer of the solution, sqrt((-0.625)^2 + (0.333)^2) == 0.708176532 != 1. Did the book do something wrong? I ask because, a unit vector should have a magnitude of 1. Also, sin(ϕ) = sin(60°) should equal 0.866025404 rather than 0.883 and cos(ϕ) = cos(60°) should equal 0.5 instead of 0.470. Am I missing something here?

Any input would be greatly appreciated!
Thanks in advance!

 

[gneill]

It looks like they're not requiring the basis vectors to be unit vectors (although they've specified e₁ to be of unit length). Also, the angle $\phi$ is actually not precisely 60°, which you can work out from the calculated side lengths of the small triangle.

 

[s3a]

Thanks for pointing those out; I've now confirmed the numbers are correct in the solutions. Just to make sure I understood you correctly, it is wrong that they put a caret (^) on top of the e_2 basis vector, right? Other than that everything is correct, right?

 

[gneill]

Thanks for pointing those out; I've now confirmed the numbers are correct in the solutions. Just to make sure I understood you correctly, it is wrong that they put a caret (^) on top of the e_2 basis vector, right? Other than that everything is correct, right?

In general the caret on a variable can simply be taken to mean that it is a vector. Some publications use it to imply that the vector is a unit vector of a basis set in order to distinguish them from other vector variables. It is not a mandatory restriction, and in fact there's no mathematical necessity for basis vectors to be measured in units that make them have length "1", even if it is a practical convenience. So while it may not be conventional or esthetically pleasing to have the caret on the e₂ basis vector when it's not of unit length, it's not technically incorrect.

 

[s3a]

Thanks again :).