Answers to given problems.

3.1 MB = 1.277 Nm ccw
3.3 MB = 186.6 Nm cw
3.5 α=6.12°, 33.8°
3.17 (a) λ = 1/√11(-3i -j - k)
(b) λ = 1/√13(2j + 3k)
3.19 (a) Mo = -18i + 16j - 44k
(b) Mo = -25i + 47j +38k
(c) Mo = 0
3.22 MA = (23.5Nm)i + (78.5Nm)j - (473Nm)k
3.24 MC = -(153 lbft)i + (63 lbft)j + (215 lbft)k
3.26 Mo = (2.02 Nm)i - (1.926 Nm)j - (1.315 Nm)k
3.35 P · Q = 18
P · S = 10
Q · S = 0 thus Q and S are perpendicular
3.37 θ = 43.6°
3.39 θ = 27.4°
3.71 (a) M = 12.39 Nm cw
(b) M = 12.39 Nm cw
(c) M = 12.39 Nm cw
3.75 M = 10 lbft cw
θ = 126.9°
3.80 M = 4.5 Nm
θx = 90°
θy = 177.1°
θz = 87.1°
3.82 FD = 100 N ↑
FC = 100 N ↓
3.87 (a) F = (240 N)k, M = (100 Nm)j
(b) F' = (240 N)k applied .417 m from A along side AB
(c) P = 100 N
3.89 (a) F = 48 lb ∠ 65°, M = 490 lbin cw
(b) F' = 48 lb ∠ 65° applied to the lever 17.78 in to the left of pin B
3.101 (a)
a. Ra = 600 N ↓, Ma = 1000 Nm ccw
b. Rb = 600 N ↓, Mb = 900 Nm cw
c. Rc = 600 N ↓, Mc = 900 Nm ccw
d. Rd = 400 N ↑, Md = 900 Nm ccw
e. Re = 600 N ↓, Me = 200 Nm cw
f. Rf = 600 N ↓, Mf = 800 Nm ccw
g. Rg = 1000 N ↓, Mg = 1000 Nm ccw
h. Rh = 600 N ↓, Mh = 900 Nm ccw
(b) loading c. and h. are equivalent
3.105 (a) WD = 60 lb, d = 2 ft
(b) WD = 52 lb, d = 2.31 ft
3.108 R = 72.4 lb ∠ -81.9°
M = 206 lbft
3.111 (a) R = 665 lb ∠ 79.6° applied 64.0 inches to the right of A. (b) α = 22.9°
3.120 R = -(420 N)j - (339 N)k
M = (1.125 Nm)i + (163.9 Nm)j - (109.9 Nm)k

11.98 (a) vo = 15.5 m/s
(b) d = 5.12 m
11.100 (a) 70.4 mph <= vo <= 89.4 mph
(b) α31 = 6.89°, α42 = 4.29°
11.103 d = 242 yd
11.133 an = 23.1 m/s2
11.135 (vmax)AB = 59.9 mph
11.138 t = 8.55 s
11.162 (a) v = 3πbuθ
(b) by observation, v = vmax when cosθ = 1 or θ = 2Nπ, N=0,1,2 ...
11.167 v = (b θdot)/(cos2θ)
11.170 θdot = (vo cos β)/(h) · (tan β cos θ + sin θ)2
11.171 vavg = 185.7 kph
11.172 vavg = 61.8 mph, β = 49.7°