Ch 14 problems¶
In [1]:
from numpy import *
dtr = pi/180
g = 9.81 # m/s2
gfps = 32.2 # ft/s2
Example 14.2
In [2]:
m = 3500/gfps
Q = 10*dtr
v = 20 # ft/s
mu = 0.5
In [3]:
s = v**2/(2*gfps)/(mu*cos(Q)-sin(Q))
print('skid distance s = {:.2f} ft'.format(s))
skid distance s = 19.49 ft
In [4]:
m = 1600
Q = 5*dtr
v = 10
s = v**2/(2*g)/(mu*cos(Q)-sin(Q))
print('skid distance s = {:.2f} m'.format(s))
skid distance s = 12.40 m
In [5]:
v,a = 12,6
e = 0.8
m = 50
Pout = m*v/2*(0.5*a+g)
P = m*v/(2*e)*(0.5*a+g)
print('power required is Pin = {:.2f} W'.format(P))
Pout
power required is Pin = 4803.75 W
Out[5]:
3843.0
In [6]:
v,a = 10,5
e = 0.8
m = 40
P = m*v/(2*e)*(0.5*a+g)
print('power required is Pin = {:.2f} W'.format(P))
power required is Pin = 3077.50 W
In [7]:
DsA = 3 # ft
mu = 0.2
WA = 60 # lb
WB = 10 # lb
Q = arctan(3/4)
f = mu*WA*cos(Q)
mA,mB = WA/gfps, WB/gfps
vA = sqrt( (-f+WA*sin(Q)-2*WB)*DsA / (0.5*mA+2*mB) )
print('speed of A = {:.3f} ft/s'.format(vA))
speed of A = 3.516 ft/s
In [8]:
F,W = 30,10 # lb
B = 2 # ft
A = sqrt(6**2+4.5**2)
h = A-B
Dy = 4.5 # ft
m = W/gfps
vB = sqrt( 2/m*( F*h - W*Dy ))
print('speed at B = {:.3f} ft/s'.format(vB))
speed at B = 27.799 ft/s
In [9]:
mu = 0.1
WA,WB = 60, 40 # lb
a,b = 60*dtr, 30*dtr # rad
DsB = -2 # ft
mA,mB = WA/gfps, WB/gfps
NA = WA*cos(a)
NB = WB*cos(b)
vA = sqrt( DsB*(mu*NA/2 - WA*sin(a)/2 + mu*NB +WB*sin(b) )/( mA/2 +2*mB ) )
print('speed of A = {:.3f} ft/s'.format(vA))
speed of A = 0.771 ft/s
In [10]:
k = 100 # lb/ft
Ds = 2 # ft
Dy = 3 # ft
W = 10 # lb
Q = arctan(3/4)
m = W/gfps
vB = sqrt( 2/m*(0.5*k*Ds**2 - W*Dy) )
A,B,C = -0.5*gfps, vB*sin(Q), Dy
tf_roots = ( (-B+sqrt(B**2-4*A*C))/(2*A),
(-B-sqrt(B**2-4*A*C))/(2*A) )
tf = tf_roots[1] # take positive root
d = vB * cos(Q) * tf
print('distance d = {:.3f} ft'.format(d))
distance d = 36.242 ft
In [11]:
W = 150 # lb
vA = 5 # ft/s
H = 53 # ft
K = pi/100 # 1/ft
xB = 35 # ft
yB = H*cos(K*xB)
vB = sqrt( vA**2 + 2*gfps*H*( 1 - cos(K*xB)) )
print('at point B:')
print('\t speed = {:.3f} ft/s'.format(vB))
rho = ( 1 + (H*K*sin(K*xB))**2)**(3/2) / abs( H*K**2*cos(K*xB) )
m = W/gfps
Q = arctan(H*K*sin(K*xB))
N = W*( cos(Q) - vB**2/rho/gfps )
print('\t normal force = {:.3f} lb'.format(N))
at = gfps*sin(Q)
print('\t rate of speed increase = {:.3f} ft/s2'.format(at))
at point B: speed = 43.458 ft/s normal force = 47.357 lb rate of speed increase = 26.701 ft/s2
In [12]:
F = 300 # N
m = 75 # kg
Ds = sqrt(8**2+6**2) - sqrt(4+36)
vB = sqrt(2/m*F*Ds)
print('speed at point B = {:.3f} m/s'.format(vB))
speed at point B = 5.423 m/s
F14-12¶
In [13]:
vP = 12 # m/s
aP = 6 # m.s2
m = 100 # kg
e = 0.6
Pin = 1/e*m/2*(g+aP/2)*vP
print('Pin = {:.2f} W'.format(Pin))
Pin = 12810.00 W
In [14]:
P = 100*550 # ft.lb/s
W = 2500 # lb
v = 30 # ft/s
Q = arcsin( P/W/v )
print('angle = {:.3f} deg '.format(Q/dtr))
angle = 47.167 deg
In [15]:
vf = 25 # m/s
tf = 30 # s
m = 2e3 # kg
Q = arctan(1/10)
e = 0.8
a = vf/tf
Pout_max = m*(a+g*sin(Q))*a*tf
P_max = Pout_max / e
print('max power = {:.3f} kW '.format(P_max/1e3))
print('ave power = {:.3f} kW '.format(P_max/1e3/2))
max power = 113.092 kW ave power = 56.546 kW
In [16]:
W = 1000 # lb
F = 500 # lb
H = 15 # ft
e = 0.65
m = W/gfps
v1 = sqrt(2*H/m*(3*F-W))
print('speed = {:.3f} fts'.format(v1))
P = 3*F*v1/e
print('power supplied = {:.3f} ft.lb/s = {:.3f} hp'.format(P,P/550))
speed = 21.977 fts power supplied = 50716.756 ft.lb/s = 92.212 hp
Note that this solution and Hibbeler's make the assumption that the contact force of the road on the wheels does work, done at a rate $P_{\rm{out}}$. But actually no work is done by the static friction force (like the skater pushing off the wall). In this case one can cut off the wheels and say there's work $ \tau \omega $ (for torque $\tau$) done there, and it's equal to $\tau \omega = (Fr)(\frac{v}{r}) = Fv $.
In [17]:
m = 2.3e3 # kg
v1 = 28 # m/s
a = 5 # m/s2
e = 0.68
A = 0.3 # N.s2/m2
P = (A*v1**3 + m*a*v1)/e
print('power to engine = {:.3f} kW'.format(P/1e3))
power to engine = 483.214 kW
In [18]:
W = 49 # lb
A,B = 40,1 # lb, lb/ft2
Q = 30*dtr # rad
mu = 0.2
k = 20 # lb/ft
D = 1.5 # ft
m = W/gfps
v1 = sqrt( 2/m*( -.5*k*D**2 - mu*W*D + (A*D+B*D**3/3)*(cos(Q)-mu*sin(Q)) ))
print('speed = {:.3f} ft/s '.format(v1))
P = (A+B*D**2)*cos(Q)*v1
print('power = {:.3f} ft.lb = {:.3f} hp '.format(P,P/550))
42.25, (A+B*D**2)*v1
speed = 3.556 ft/s power = 130.126 ft.lb = 0.237 hp
Out[18]:
(42.25, 150.25664334472313)
In [19]:
phi = 30*dtr # rad
Q = 60*dtr # rad
rho = 2 # m
m = 40 # kg
F = m*g/(4*cos(phi))*(3-2*cos(Q))
print('force on each post F = {:.3f} N'.format(F))
force on each post F = 226.552 N
In [20]:
W = 30 # lb
kB,kC = 2400,1200 # lb/ft
DsB,DsC = 1/3, 0.5 # ft
h = 1/(2*W)*( kB*DsB**2 + kC*DsC**2 )
print('height h = {:.3f} ft = {:.3f} in'.format(h,h*12))
height h = 9.444 ft = 113.333 in
14-69¶
In [21]:
m = 5 # kg
vA = 8 # m/s
l = 0.2 # m
l0 = 0.1 #m
k = 50 # N/m
dB = 2*l-l0
dA = sqrt(2)*l-l0
VA = .5*k*(dA**2) + m*g*l
VB = .5*k*(dB**2)
TA = .5*m*vA**2
vB = sqrt( vA**2 -k/m*(dB**2-dA**2) + 2*g*l )
dV = VB-VA
print('vB = {:.2f} m/s'.format(vB))
VA,VB,TA
vB = 8.21 m/s
Out[21]:
(10.645786437626908, 2.2500000000000004, 160.0)
In [22]:
Q = arccos(2/3)
print('angle = {:.3f} deg '.format(Q/dtr))
angle = 48.190 deg
In [23]:
M = 600/200
D = 30 # ft
Q = 20*dtr # rad
vc = sqrt( gfps*D*(-1+2*M*sin(Q))/(1/4+M))
print('angle = {:.3f} ft/s '.format(vc))
angle = 17.684 ft/s
Review Problem 14.7
In [24]:
m = 0.25
l0 = 100e-3
h,r = 0.2,0.4
k = 150 # N/m
v = sqrt( 2*g*(r+h) + k/m*( (r+h-l0)**2 - (r-l0)**2 ) )
print('speed = {:.2f} m/s'.format(v))
speed = 10.38 m/s
In [25]:
m = 0.5
l0 = 100e-3
h,r = 0.3,0.5
k = 100 # N/m
v = sqrt( 2*g*(r+h) + k/m*( (r+h-l0)**2 - (r-l0)**2 ) )
print('speed = {:.4f} m/s'.format(v))
speed = 9.0386 m/s