Vectors - Projections and Components

[Altairs]

I am not sure if I am posting this in the right forum or not.

I had impression that there is just no difference between projection of a vector and its components until I took the Statics course. We are following The Engineering Mechanics : Statics book by Meriam and Krage. I got stuck up in the very beginning. the problem is that before this book, in elementary courses the trick of using costheta for projection and component worked very well. now it seems to be giving way.

In Sample problem 2/4 there are two forces at an agle of 50 to each other acting on a hinge. The question asks to find the projection of resultant on the line of force of the force making the lower side of parallelogram. Here, I took Rcostheta and the anser came wrong. the solution given is the Force(downward) + Force(upward)costheta(other). why didn't taking simply cos theta work when it would give the component in the direction of that force? Does it mean that components and projections aren't same always? How will I know what to do in a particular situation?

In contrast to above ,in another problem there is a 800lb force at 60 from an axis 'a' and the axis 'b' is at 135 from axis 'a'. In this case the projection is simply 800cos 60 while for component you have to make that parallelogram and that stuff. Why can't I simply find the component also by taking 800cos60 and 800cos75?

Why are the above two cases at 180 of each other?

I am very confused. And Statics is my Primary Course.

 

[pkleinod]

I had impression that there is just no difference between projection of a vector and its components...

There is a big difference. The components of a vector are defined with respect to some basis.
Suppose [tex]\sigma_1[/tex], [tex]\sigma_2[/tex],[tex]\sigma_3[/tex] are a set of orthonormal basis vectors in 3D. Then any vector [tex]x[/tex] in 3D may be written in terms of these:

[tex]x = x_1\sigma_1 + x_2\sigma_2 + x_3\sigma_3[/tex].

The scalars [tex]x_i[/tex] are the rectangular components of the vector
[tex]x[/tex] with respect to the basis {[tex]\sigma_1,\sigma_2,\sigma_3[/tex]}. They are obtained from the dot product of the vector with the corresponding unit basis vector:

[tex]x_i = x\cdot \sigma_i = |x|\cos{(x,\sigma_i)}}[/tex]

The projection of vector [tex]x[/tex] onto some other vector [tex]b[/tex] is a vector:

[tex]x_b = (x\cdot b) b /|b|^2 = |x| \cos{(x,b)} (b/|b|)[/tex].

Notice that the magnitude of [tex]x[/tex] and of [tex]b[/tex] appear in
the definition of the projection. You could of course say that the
component of [tex]x[/tex] with respect to the unit vector [tex]b/|b|[/tex]
is [tex]|x|\cos{(x,b)}[/tex].

 

[Shooting Star]

In contrast to above ,in another problem there is a 800lb force at 60 from an axis 'a' and the axis 'b' is at 135 from axis 'a'. In this case the projection is simply 800cos 60 while for component you have to make that parallelogram and that stuff. Why can't I simply find the component also by taking 800cos60 and 800cos75?

Why are the above two cases at 180 of each other?

I am very confused. And Statics is my Primary Course.

If you take the two forces 800cos 60 and 800cos 75, which you feel are components, would the resultant add up to 800, and would the angle of the resultant be at an angle of 60 deg from axis 'a'? You can try it out.

The projection of a vector F on any direction is always F(cos theta), where theta is the angle it makes with that direction.

If a force is a resultant of any two forces, then those two forces are the components of the former.

Just picture a simple case first. There are two forces P and Q, and R is the resultant. Then these forces P and Q are the components of R. P and Q need not be perpendicular to each other.

Suppose the angle between P and R is A and between P and Q is B. If you take the projection of R along the direction of P and Q, they are RcosA and Rcos(B-A). But these two don't, after vector addition, give back the resultant R, whereas P and Q do.

Since we generally use mutually perpendicular axes like the Cartesian axes, the projection on each axis and the components along the axes turn out to be the same.

 

[Altairs]

The projection of a vector F on any direction is always F(cos theta), where theta is the angle it makes with that direction.

But, this thing doesn't work on the first question with the resultant R. I'll post the whole question.

In the picture. There is a bracket on which two forces are acting. One is F1 (100N) at an angle of 30deg above the horizontal. Second, F2 (80N) is acting at an angle 20deg below horizontal.

The question reads :-

"Forces F1 and F2 avt on the bracket as shown. Determine the projection Fb of their resultant R onto the b-axis"

a and b-axis are the axis in which the forces F1 and F2 are acting.

this is an example, Page 31. Meriam, Krage

 

[Shooting Star]

In the picture. There is a bracket on which two forces are acting. One is F1 (100N) at an angle of 30deg above the horizontal. Second, F2 (80N) is acting at an angle 20deg below horizontal.

The question reads :-

"Forces F1 and F2 avt on the bracket as shown. Determine the projection Fb of their resultant R onto the b-axis"

a and b-axis are the axis in which the forces F1 and F2 are acting.

The answer should be (F2+F1*cos 50). How much did you get, and what is given as the correct answer?

 

[Altairs]

Ya it is the answer. My question...Why doesn't Rcos20 give the required ans when, as you said that simply the cos theta gives the projection? This is exactly where my confusion lies.

 

[Shooting Star]

Where did you get the totally wrong idea that the angle between R and b-axis was 20 deg? You have been the victim of a misleading diagram. If the resultant R were in the horizontal direction in the plane of F1 and F2, then R(cos 20) indeed would have given you the correct answer.

If R is the resultant of P and Q, then the projection of R in any direction is the sum of the projections of P and Q in the same direction. I have simply applied this rule. Are things clearer now?

 

[Altairs]

ohhhhhhhh...I get it...Sorrrry...If I take the real angle and take cos then I'll get the answer...thanks...