% Ch 2. Assignment #5 due 10/3
clear, clc
date
ans = 26-Sep-2007
% Given: Velocity(v) = [0, 10, 0]. Positin(r) = [2, 10t+3, 0], t = time. % Momentum(L) = m(r cross v). m = 5 v = [0, 10, 0] t = [0:0.5:5] f = 10*t+3 P = [2*ones(length(t),1), f', zeros(length(t),1)] % a. % b. I used f(length(t)) because we want to find the location when % t = 5s, and 5 is the last number in the vector f. r_at_5s = [2, f(length(t)), 0] L = m*cross(r_at_5s, v) % c.
m =
5
v =
0 10 0
t =
Columns 1 through 9
0 0.5000 1.0000 1.5000 2.0000 2.5000 3.0000 3.5000 4.0000
Columns 10 through 11
4.5000 5.0000
f =
3 8 13 18 23 28 33 38 43 48 53
P =
2 3 0
2 8 0
2 13 0
2 18 0
2 23 0
2 28 0
2 33 0
2 38 0
2 43 0
2 48 0
2 53 0
r_at_5s =
2 53 0
L =
0 0 100
% Given: M = (r cross F)*n, where: M is the magnitude of the moment, % F is a force vector, r is the position vector, and n is the unit % vector in the dirrection of the line. F = [10, -5, 4] r = [-3, 7, 2] n = [6, 8, -7] M = dot(cross(r,F),n)
F =
10 -5 4
r =
-3 7 2
n =
6 8 -7
M =
869
% Area of parallelogram is the magnitude of the cross product of two % of its sides. A = [7, 0, 0] B = [1, 3, 0] sqrt(sum(cross(A,B).*cross(A,B))) % Area of parallelogram. % Simply by looking at the cross product (which is [0,0,21]), it is % obvious that its magnitude will be 21.
A =
7 0 0
B =
1 3 0
ans =
21
f = [14, 6, 3, 9] g = [5, 7, 4] [Quotient, Remainder] = deconv(f,g)
f =
14 6 3 9
g =
5 7 4
Quotient =
2.8000 -2.7200
Remainder =
-0.0000 0.0000 10.8400 19.8800
f = [8, -9, 0, -7]
g = [10, 5, -3, -7]
% Evaluate each polynomial when x = 5.
f_at_5 = polyval(f,5)
g_at_5 = polyval(g,5)
f_at_5/g_at_5
f =
8 -9 0 -7
g =
10 5 -3 -7
f_at_5 =
768
g_at_5 =
1353
ans =
0.5676