Conservação de movimento

Chapter

1

Vector
Problem 1.1

Analysis

(a) From the diagram, IB + CI COSO3 IBI COSO1 ICI COSO2'Multiply by IAI. = + IAIIB + CI COSO3 IAIIBI COSO1 + IAIICI COSO2. = So: A.(B + C) = A.B + A.C. (Dot product is distributive.) IAIIB + CI sin 03 n = IAIIBI sin 01 n + IAIICI sin O2n. If n is the unit vector pointing out of the page, it follows that Ax(B + e) = (AxB) + (Axe). (Cross product is distributive.)

ICI sin 82

Similarly: IB + CI sin 03 = IBI sin 01 + ICI sin O2, Mulitply by IAI n.
IBlsin81

A

(b) For the general case, see G. E. Hay's Vector and Tensor Analysis, Chapter 1, Section 7 (dot product) and Section 8 (cross product). Problem 1.2 The triple cross-product is not in general associative. For example, suppose A = ~ and C is perpendicular to A, as in the diagram. Then (B XC) points out-of-the-page, and A X(B XC) points down, and has magnitude ABC. But (AxB) = 0, so (Ax B) xC = 0 :f. Ax(BxC).

k-hB
BxC iAx(Bxe) z Problem 1.3
A

= + 1x + 1Y - H; A = /3;

B

= 1x + 1Y+

Hi B = /3. =>cosO= ~. x y

A.B = +1 + 1-1 = 1 = ABcosO = /3/3coso
10 = COS-1(t) ~ 70.5288° Problem 1.4
I

The cross-product of any two vectors in the plane will give a vector perpendicular to the plane. For example, we might pick the base (A) and the left side (B): A = -1 x + 2 y + 0 z; B = -1 x + 0 Y + 3 z.

1

2

CHAPTER

1. VECTOR ANALYSIS

x
AxB

y

Z

2 0 1= 6x + 3y + 2z. -1 0 3 This has the' right direction, but the wrong magnitude. To make a unit vector out of it, simply divide by its length: IAxBI=v36+9+4=7. '7X+ '7y + '7z . ft - IAX BI = 16 AXB 3 2
A

= I -1

I

Problem 1.5 Ax Ay Az (ByCz - BzCy) (BzCx - BxCz) (BxCy - ByCx) = x[Ay(BxCy - ByCx) - Az(BzCx - BxCz)] + yO + zO (I'll just check the x-component; the others go the same way.) = x(AyBxCy - AyByCx - AzBzCx + AzBxCz) + yO + zOo B(A.C) - C(A.B) = [Bx(AxCx + AyCy + AzCz) - Cx(AxBx + AyBy + AzBz)] x + 0 y + 0 z = x(AyBxCy + AzBxCz - AyByCx - AzBzCx) + yO + zOo They