A force applied to an object is given, in Newtons, by the vectorF~ = 45 N ^{ 10

Question

A force applied to an object is given, in Newtons, by the vectorF~ = 45 N ^{ 10 N |^+ 20 N k^

(a) Let us call the component of the force in the x-direction Fx; and the component in the y-direction Fy; and the component in the z-direction Fz: By visually inspecting the expression for the force above, Ond the values for Fx; Fy;and Fz: (Let us call this the "inspection" method for Onding the components of a vector.)

(b) Find the same answers you found in part (a) above using the "projection" method. That is, show that the component of F~ in the x direction is given by the dot product of F~ with the unit vector ^{ in the x direction.

F x = F~ ^{

and similarly, show that Fy = F~ |^ and Fz = F~ k^: (If you need a refresher on dot-products, see Chapter 1.10. This is also called a scalar product. The rules are that for two perpendicular unit vectors the dot product is zero, i.e. for example, ^{ |^ = 0, but the dot product of a unit vector with itself is 1, i.e. for example ^{ ^{ = 1 )

(c) Show that the magnitude of the vector can be determined from the dot product of the vector with itself, since

~2~~ F =FF

What is the magnitude of the force F~ in Newtons?
(d) Write down an expression for a dimensionless unit vector u^ that points

in this same direction as F~:
(e) Use your answers to (c) and (d) to rewrite F~ as a product of the magni-

tude of the force and a unit vector indicating which direction the force is acting, i.e.,

F~ = F~ u^:
Show that when you multiply F~ by u^ and distribute the terms, you get back

the same expression for F~ that we started with at the beginning of this problem.

Explanation / Answer

F = 45 i + 10 j + 20 k

F = Fx i + Fy j + Fz k

(A) Fx = 45 N , Fy = 10 N , Fz = 20 N  

(B) Fx = F.i = (45i + 10j + 20k).(1i + 0j + 0k)

= (45 x 1) + (10 x 0 ) + (20 x 0)

= 45 N  

Fy = F.j = (10 x 1) = 10 N  

Fz = F.k = (20 x 1) = 20 N  

(C) F.F = |F||F| cos(theta)

angle between same vector is 0 deg .

F.F = F^2 cos0 = F^2

magnitude of force, |F| = sqrt[ F.F ]

{ |F| = sqrt[ (45 x 45) + (10 x 10) + (20 x 20)] = 50.25 N }

(D) unit vector, F^ = F / |F|

F^ = (F) / sqrt[ F.F]

{ F^ = (45i + 10j + 20k) / 50.25

= 0.8955 i + 0.199j + 0.398k }

(e) F = |F| . F^ = (sqrt(F.F)) (F / sqrt(F.F))

F = F

{ F = (50.25 N ) (0.8955 i + 0.199j + 0.398k )

F = 45i + 10j + 20k N }

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