Ch 15 problems¶

In [1]:
from numpy import *
dtr = pi/180
g = 9.81     # m/s2
gfps = 32.2  # ft/s2

No description has been provided for this image

hand written solution

In [2]:
W,T = 5e3,5e3  # lb
v = 10     # ft/s
P = 8/5*T
Dt =  v/gfps / (P/W-1)
print('min time = {:.3f} s'.format(Dt))
min time = 0.518 s

hand written solution

In [3]:
A,B = 100, 5  # N, N/s2
m = 20        # kg
t = 3         # s
s = A/m*t**2 + B/6/m*t**4 - 0.5*g*t**2
print('distance = {:.3f} m'.format(s))
distance = 4.230 m

hand written solution

In [4]:
A = 400  # N/sqrt(s)
m = 200  # kg
mus = 0.5
muk = 0.4
t2 = 4   # s
t1 = ( mus*m*g/2/A )**2
print('time to move is {:.3f} s'.format(t1))
def v(t):
    return (4/3*A*t**(3/2)/m - muk*g*t)
v = v(t2) - v(t1)
print('speed = {:.3f} ft/s'.format(v))
time to move is 1.504 s
speed = 6.621 ft/s

hand written solution

In [5]:
t1,t2 = 0.1e-3,0.4e-3  # s
F1 = 1500e3            # N
m = 2      # kg
vi = 90    # m/s
I = -0.5*t2*F1
vf = vi + I/m
print('speed = {:.3f} m/s'.format(vf))
speed = -60.000 m/s

hand written solution

In [6]:
F1,F2 = 600,750  # N
t1,t2 = 5,8      # s
m = 400   # kg
v1 = F1*t1/(2*m)
s1 = F1*t1**2/(6*m)
print('at time t1:')
print('\t speed = {:.3f} m/s'.format(v1))
print('\t distance = {:.3f} m'.format(s1))
A = (F2-F1)/(t2-t1)
v2 = v1 + 1/m*( F1*t2 + 0.5*A*t2**2 - A*t1*t2 - F1*t1 + 0.5*A*t1**2 )
Dt = t2-t1
s2 = v1*Dt + 1/m*( (-F1*t1 + 0.5*A*t1**2)*Dt + 0.5*(F1-A*t1)*(t2**2-t1**2) + 1/6*(A*(t2**3-t1**3)) )
print('at time t2:')
print('\t speed = {:.3f} m/s'.format(v2))
print('\t distance = {:.3f} m'.format(s2))
at time t1:
	 speed = 3.750 m/s
	 distance = 6.250 m
at time t2:
	 speed = 8.812 m/s
	 distance = 18.562 m

hand written solution

In [7]:
A = 4e3   # N/sqrt(s)
B = 20e3  # N
m = 4e3   # kg
t1,t2 = 25,35  # s
v1 = 2*A/(3*m)*t1**(3/2)
s1 = 4*A/(15*m)*t1**(5/2)
print('at time t1:')
print('\t speed = {:.3f} m/s'.format(v1))
print('\t distance = {:.3f} m'.format(s1))
Dt = t2-t1
v2 = v1 + 1/m*B/Dt*( 0.5*t2**2 + 0.5*t1**2 - t1*t2 )
s2 = v1*Dt - 1/m*B/Dt*( 1/6*(t2**3-t1**3) + 0.5*t1**2*Dt - 0.5*t1*(t2**2-t1**2) )
print('at time t2:')
print('\t speed = {:.3f} m/s (max speed)'.format(v2))
print('\t distance = {:.3f} m'.format(s2))
at time t1:
	 speed = 83.333 m/s
	 distance = 833.333 m
at time t2:
	 speed = 108.333 m/s (max speed)
	 distance = 750.000 m

hand written solution

In [8]:
m1,m2 = 15e3,12e3  # kg
v1,v2 = 1.5,-0.75  # m/s
vf = (m1*v1+m2*v2)/(m1+m2)
DT = 0.5*(m1+m2)*vf**2 - 0.5*m1*v1**2 - 0.5*m2*v2**2
print('change in kinetic = {:.3f} kJ'.format(DT/1e3))
change in kinetic = -16.875 kJ

hand written solution

In [9]:
M,m = 4,2e-3  # kg
Q = 6*dtr     # rad
l = 1.25      # m
v1 = sqrt( 2*g*l*(1-cos(Q)))
vb = (M+m)/m*v1
print('bullet speed = {:.3f} m/s'.format(vb))
bullet speed = 733.444 m/s

This is a bad problem, I think. I doubt the friction is nonimpulsive during the 1ms impact time.

hand written solution

In [10]:
M,m = 2,20e-3  # kg
vb = 400       # m/s
mu = 0.2
v1 = m*vb/(M+m)
s = v1**2/(2*mu*g)
print('distance = {:.3f} m'.format(s))
distance = 3.997 m

hand written solution

In [11]:
k = 60  # N/m
x = 0.3 # m
l = 2   # m
m = 5   # kg
Q = arccos( 1 - k*x**2/(4*m*g*l) )
print('angle = {:.3f} deg'.format(Q/dtr))
angle = 9.516 deg

hand written solution

In [12]:
mA,mB = 26e3,14e3              # kg
vA,vB = 20e3/3600, -10e3/3600  # m/s
k = 3e6                        # N/m
vG = (mA*vA + mB*vB)/(mA+mB)
x = sqrt( 1/k*( mA*(vA-vG)**2 + mB*(vB-vG)**2 ) )
print('compression = {:.3f} m'.format(x))
compression = 0.459 m

hand written solution

In [13]:
mA,mB = 40,60  # kg
x = 2          # m
k = 180        # N/m
vA = x* sqrt(k/mA/(1+mA/mB))
print('speed of A = {:.2f} m/s'.format(vA))
vB = vA*mA/mB
print('speed of B = {:.2f} m/s'.format(vB))
speed of A = 3.29 m/s
speed of B = 2.19 m/s

hand written solution

In [14]:
mA,mB = 2e3,10e3  # kg
L = 40            # m
m = mA/mB
x = L*m/(1+m)
print('barge moves {:.3f} m'.format(x))
barge moves 6.667 m

hand written solution

In [15]:
e = 0.5
d = 100   # ft
vA = 75   # ft/s
phi = 60*dtr # rad
Q0 = 45*dtr  # rad
vfx = vA*cos(Q0)
tf = d/vfx
vfy = vA*sin(Q0) - gfps*tf
v1 = sqrt( vfx**2 + vfy**2 )
Q1 = arctan(-vfy/vfx)
alpha = pi/2 - (phi+Q1)
v2x = e*v1*cos(alpha)
v2y = v1*sin(alpha)
v2 = sqrt(v2x**2+v2y**2)
print('speed after bounce = {:.3f} ft/s'.format(v2))
vfx,vfy
Q1/dtr
speed after bounce = 31.840 ft/s
Out[15]:
8.244152025894227

hand written solution

In [16]:
vA1 = 7.5     # ft/s
Q = 30*dtr  # rad
e = 0.8
vA1x = -vA1*cos(Q)
vA2y =  vA1*sin(Q)
vA2x = 0.5*vA1x*(1-e)
vB2x = e*vA1x + vA2x
vA2 = sqrt(vA2x**2+vA2y**2)
print('vA2 = {:.3f} ft/s'.format(vA2))
print('vB2 = {:.3f} ft/s'.format(vB2x))
vA2x
vA2 = 3.806 ft/s
vB2 = -5.846 ft/s
Out[16]:
-0.6495190528383289

hand written solution

In [17]:
mA,mB = 4,2      # kg
e = 0.8     
Q = arctan(4/3)  # rad
vA,vB = 15,8     # m/s
#
m = mB/mA
vA1x = -vA*cos(Q)
vA1y =  vA*sin(Q)
vA2y = vA1y
vB1 = vB
vA2x = ( m*vB1*(1+e) + vA1x*(1-e*m) )/(1+m)
vB2 = e*(vA1x-vB1) + vA2x
print('speed of A = {:.2f} m/s'.format( sqrt(vA2x**2+vA2y**2) ))
print('speed of B = {:.2f} m/s'.format( abs(vB2) ))
speed of A = 12.06 m/s
speed of B = 12.40 m/s

hand written solution

In [18]:
H = 2.9    # ft
e = 0.8
v1 = sqrt(2*gfps*H)
p = arctan(3/4)
p2 = arctan(3/4/e)
Q = pi/2 - p2 - p
v2 = v1/5*sqrt(9+16*e**2)
tp = v2/gfps*sin(Q)
X = v2*cos(Q)*tp
Y = v2*sin(Q)*tp - 0.5*gfps*tp**2
d = 5*X/4
h = Y + 3*d/5
print('d = {:.2f} ft'.format(d))
print('h = {:.2f} ft'.format(h))
v1
Y
d = 0.95 ft
h = 0.64 ft
Out[18]:
0.06700159999999995

hand written solution

In [19]:
mA,mB = 15,10  # kg
e = 0.8
vA,vB = 10,8   # m/s
p = arctan(4/3) # rad
m = mB/mA
vB2y = -vB*sin(p)
vA2y =  vA*sin(p)
vB2x = cos(p)*(m*vB-vA-e*vB-e*vA)/(1+m)
vA2x = cos(p)*(m*vB-vA) - m*vB2x
vA2 = sqrt(vA2x**2+vA2y**2)
vB2 = sqrt(vB2x**2+vB2y**2)
print('speed of A = {:.3f} m/s'.format(vA2))
print('speed of B = {:.3f} m/s'.format(vB2))
speed of A = 8.195 m/s
speed of B = 9.385 m/s

In [20]:
rOP = array((-1, 1.5, 2))
rOA = array(( 2,-1.5, 2))
rOB = array(( 3,   3, 0))
m = 3.0     # kg
rPA = rOA - rOP
rAB = rOB - rOA
uAB = rAB / sqrt(sum(rAB**2))
v = 6*uAB   # m/s
H = cross( rPA, m*v )
print('H = ({:.2f} i {:+.2f} j {:+.2f} k) kg.m2/s'.format(*H))
H = (21.49 i +21.49 j +59.11 k) kg.m2/s

hand written solution

In [21]:
v0 = 2     # m/s
A,B = 1,2  # Nm/s2, Nm
r = 0.5    # m
m = 10     # kg
t = 3      # s
v = v0 + 1/(2*r*m)*(A/3*t**3 + B*t)
print('speed is {:.3f} m/s'.format(v))
speed is 3.500 m/s

hand written solution

In [22]:
v = 60   # ft/s
h = 8    # ft
R = 8    # ft
p = arctan(h/(2*pi*R))
t = v/(gfps*sin(p))
print('time = {:.3f} s'.format(t))
time = 11.855 s

hand written solution

In [23]:
r1,r2 = 3,2  # ft
v1 = 6       # ft/s
W = 4        # lb
vr = 2       # ft/s
m = W/gfps
v2Q = v1*r1/r2
v2 = sqrt(v2Q**2+vr**2)
print('ball speed = {:.3f} ft/s'.format(v2))
U = .5*m*(v2**2-v1**2)
print('work done = {:.3f} ft.lb'.format(U))
ball speed = 9.220 ft/s
work done = 3.043 ft.lb

hand written solution

hand written solution

In [24]:
W = 2      # lb
d,h = 4,3  # ft
k = 12     # lb/ft
l0 = 1.5   # ft
l = sqrt(d**2+h**2)
m = W/gfps
vc = sqrt(  k/m*( (l-l0)**2 - (h-l0)**2) )
print('speed at C = {:.3f} ft/s'.format(vc))
HA = -m*vc*h
print('ang momentum about A = {:.3f} slug.ft2/s'.format(HA))
speed at C = 43.955 ft/s
ang momentum about A = -8.190 slug.ft2/s

hand written solution

In [25]:
rho = 1020  # kg/m3
Q = 0.25    # m3/s
v = 50      # m/s
p,Q1,Q2 = 60*dtr,45*dtr,30*dtr
T = rho*Q*v/cos(p) * (cos(Q1)+cos(Q2))
print('tension is {:.3f} kN'.format(T/1e3))
tension is 40.115 kN

hand written solution

In [26]:
Q = 0.68    # m3/s
rho = 1e3  # kg/m3
A = 0.05   # m2
h,d = 2,1.5  # m
r = 0.12     # m
v = Q/A
Dy = rho*Q*v
Cx = rho*Q*(d-r)*v/h
Dx = rho*Q*v - Cx
print('reaction at C is {:.3f} kN'.format(Cx/1e3))
print('reaction at D is {:.3f} i {:+.3f} j kN'.format(Dx/1e3,Dy/1e3))
v
reaction at C is 6.381 kN
reaction at D is 2.867 i +9.248 j kN
Out[26]:
13.6

hand written solution

In [27]:
v0 = 14     # m/s
rho = 1.22  # kg/m3
d = 0.75    # m
m = 200     # kg
Fx = -pi*(d/2)**2*rho*v0**2
ax = -Fx/m
print('initial acceleration is {:.3f} m/s2'.format(ax))
initial acceleration is 0.528 m/s2

hand written solution

In [28]:
AA =  4/12/12  # ft2
AB = 12/12/12  # ft2
Q = 2          # ft3/s
pB = 2*12**2   # lb/ft2
rho = 62.4/gfps # slug/ft3
Fx = -pB*AB-rho*Q**2/AB
Fy = -rho*Q**2/AA
print('force at B is {:.3f} i {:+.3f} j lb (magnitude {:.3f} lb)'.format(Fx,Fy,sqrt(Fx**2+Fy**2)))
force at B is -117.019 i -279.056 j lb (magnitude 302.598 lb)

hand written solution

In [29]:
rA,rB,rC = 50e-3/2, 30e-3/2, 30e-3/2
rho,v1,p = 1e3,18,150.5e3
v2 = v1*rA**2/(rB**2+rC**2)
sinq,cosq = 3/5,4/5
Rx = rho*v2**2*pi*(rB**2+rC**2*cosq)
Ry = rho*pi*(v1**2*rA**2+v2**2*rC**2*sinq) + p*pi*rA**2
print(f'Rx = {Rx:.0f} N, Ry = {Ry/1e3:.2f} kN,')
Rx = 795 N, Ry = 1.20 kN,

hand written solution

In [30]:
dC = 0.75/12     # ft
dA = dB = 0.5/12 # ft
vA,vB = 12,25    # ft/s
pC = 40 *12**2   # lb/ft2
gamma = 62.4     # lb/ft3
rho = gamma/gfps
AC = pi*(dC/2)**2
AA = pi*(dA/2)**2
QC = AA*(vA+vB)
vC = QC/AC
QA = vA*AA
QB = vB*AA
Fex = -pC*AC - 3/5*rho*QA*vA - rho*QC*vC
Fey = 4/5*rho*QA*vA + rho*QB*vB
print('force on elbow {:.3f} i {:+.3f} j lb'.format(Fex,Fey))
force on elbow -19.508 i +1.956 j lb

hand written solution

In [31]:
F = 250   # N
R = 50    # kg/s
M,m0 = 4e3,2e3  # kg
u = 5     # m/s
a = (F+u*R)/M
v = -(F/R+u)*log(M/(M+m0))
print('velocity at time t0 is {:.3f} m/s'.format(v))
print('acceleration at time t0 is {:.3f} m/s2'.format(a))
velocity at time t0 is 4.055 m/s
acceleration at time t0 is 0.125 m/s2

hand written solution

In [32]:
lam = 5/gfps  # slub/ft
u = 4         # ft/s
print('R(t) = ({:.3f} + {:.3f} t) lb'.format(lam*u**2,lam*u*gfps))
R(t) = (2.484 + 20.000 t) lb

hand written solution

In [33]:
v = 950e3/3600  # m/s
uE = 450        # m/s
rho = 1.22      # kg/m3
Qs = 50         # m3/s
R = 0.4         # kg/s
FD = uE*(rho*Qs+R) - v*rho*Qs
print('drag force = {:.3f} kN'.format(FD/1e3))
drag force = 11.533 kN

hand written solution

In [34]:
u = 10   # m/s
R = 50   # kg/s
m = 10e3+500  # kg
a = u/m*R
print('initial acceleration is {:.4f} m/s2'.format(a))
initial acceleration is 0.0476 m/s2

Review Problem 15.5

In [35]:
m,M = 0.2,15
v1,v2 = 900,300
mu = 0.2
vw = m/M*(v1-v2)
print('speed of block = {:.2f} m/s'.format(vw))
t = vw/(mu*g)
print('time to stop = {:.2f} s'.format(t))
speed of block = 8.00 m/s
time to stop = 4.08 s

In [36]:
v0 = 2
m = 10
t = 2
R = 0.75
F = 60 
Q = arctan(4/3)
A,B = 8,5
v = v0 + (1/3*A*t**3 + B*t + R*F*t*cos(Q))/(m*R)
print('speed = {:.2f} m/s'.format(v))
speed = 13.38 m/s