Equilibrium of Forces Acting on Point Mass: Cross Product with a

[jamesbob]

Three forces with magnitudes F_a, F_b, F_c act on a point mass, pulling in unit directions a, b, c, respectively. Thr forces are in 'equilibrium' which means that

F_aa + F_bb + F_cc = 0​

By taking the cross product with a, show that

F_b(a \times b) = F_c(c \times a)​

and find two similar equations (involving F_a and F_c, and F_a and F_b, respectively).

My Work

I'm quite stuck here, i don't think i understand fully. The only mere working i can muster is:

a \times (a \times b) = (a.b)a - (a.a)c

a \times (c \times a) = (a.a)c - (a.c)a

Can anyone help me out?

 

[assyrian_77]

That is the wrong approach. It is much simpler. What can you say about this crossproduct: \hat{u}\times\hat{u}=? Also, remember the anticommutation of a crossproduct.

 

[jamesbob]

Is \hat{u}\times\hat{u}=0?

im not sure what the anticommutation of a crossproduct is unless its the dot product. Is it something to do with that in each instance the a's will "cross out" and become zero, leaving something more useful? Sorry, I'm really quite lost.

 

[assyrian_77]

Ok sorry, should have expressed it differently. By the anticommutation I meant this: \hat{u}\times\hat{v}=-\hat{v}\times\hat{u}. Surely this is something you've had in class, right?

 

[0rthodontist]

Yes, u x u = 0.

When the question says, "simplify by taking the cross product with a," it means just take the cross product with a of each side of your original equation.

 

[HallsofIvy]

F_aa + F_bb + F_cc = 0
so
a \times (F_aa+ F_bb+ F_cc)= F_a a \times a+F_b a \times b+ F_c a \times c= 0
= F_b a \times b+ F_c a \times c= 0
Oh, and remember that a x c= - c x a.

 

[jamesbob]

Damn, i replied to this last week sometime but it seems to not have worked, gutted! Took ages to do, il try remmeber it...

I see now that you've to take the cross product with the
F_aa + F_bb + F_cc That helps a lot and was all i really needed to be shown. So continuing from HallsofIvy (even tho its prety much done [tanks very mucg btw ] ) you get

F_b(a \times b) - F_c(c \times a) = 0 so F_b(a \times b) = F_c(c \times a)

For the 2 similar equations i got:

b \times (F_aa + F_bb + F_cc) = F_a(b \tiems a) + F_b(b \times b) + F_c(b \times c) = 0 \Rightarrow F_a(b \times a) = F_c(c \times b)

and finally,

F_a(c \times a) = F_b(b \times c)

All seem right?

Next the question asks to use one or more of these equations to show that the directions of the three forces must all lie in the same plane.

Also, explain why the magnitude of each force is proportional to the sine of the angle between the directions of the other two forces.

For the 2nd part I am guessing its using the fact that a \times b = [|a||b|\sin\theta]n

But for the first part i can't see how to use those equations. Can anyone help me (again ) ?

 

[matt grime]

Assume for now that all of the F_a etc are nonzero, then axb, axc and bxc are all multiples of each other. What does that tell you about a,b, and c? If anyone of the F_. are zero then you can do the same kind of analysis, but it just needs you to keep track of things a little more.

 

[jamesbob]

Ok, but i don't see how they are multiples of each other. Assuming they are, i still can;t make the connection between a b and c. Does it mean they are all equal?

 

[matt grime]

They are certainly not equal, and if you can't relate what you've done to this then I don't know what else to suggest. You have shown that F_b(a \times b) = F_c(c \times a)and

F_a(c \times a) = F_b(b \times c)

I don't know what else to point out that *you* have written.

I just cut and pasted these from a post *you* wrote...

 

[jamesbob]

Hmm, i know your trying to make sure i do the work which i deffinitely agree is a good thing, i learn a lot better plus after all its my work. Could i ask, for a normal set of equations, would i show them all to be in the same plane by proving a common point is present? For here i see that an a x c is equal to both a b x c and an a x b for different F_. is this like a common point?

 

[matt grime]

common points? i don't even know what those are nor what they have to do with anything in this question.

axb is a multiple of axc, and axb is a multiple of bxc, and bxc is a multiple of axc. just work with them...

if uxv=uxw then ux(v-w)=0

already the answer has almost been explained before this.

 

[jamesbob]

Ok, got it. Taking the scaler triple product with F_b(a \times b) = F_c(c \times a) gives

F_b(a \times b).c = F_c(c \times a).c \rightarrow F_b[c,a,b] = F_c[(c \times c).a] ( noting c \times c = 0) \Rightarrow F_b[a,b,c] = 0 Therefore a,b,c = 0 and so a,b,c are coplanar