Ch 13 problems¶

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

13.2¶

hand written solution

In [2]:
WA,WB = 25e3,32e3   # lb
mu = 0.5
Q = 5*dtr           # rad
a = gfps*( sin(Q) - mu*WA/(WA+WB)*cos(Q))
FC = WB *(sin(Q)-a/gfps)
print('acceleration = {:.2f} ft/s2'.format(a))
print('coupling force = {:.2f} kip'.format(FC/1e3))
acceleration = -4.23 ft/s2
coupling force = 6.99 kip

13.3¶

hand written solution

In [3]:
mu,m = 0.3,50
P,t  = 200,3
Q = 30*dtr
N = m*g - P*sin(Q)
a = (P*cos(Q) - mu*N)/m
v = a*t
s = 0.5*a*t**2
s,v
Out[3]:
(5.044957268119899, 3.363304845413266)

13.6¶

hand written solution

In [4]:
v0 = 4       # ft/s
f = 0.2*10   # lb
alpha = 8    # lb
t1 = 2       # s
W = 10       # lb
if 0:
    t1 = 3       
    f = 0.1*13
    W = 13       # lb
v1 = v0 + gfps/W * ( alpha/3 * t1**3 - f*t1 )
print('velocity = {:.2f} ft/s'.format(v1))
velocity = 59.81 ft/s

13.12¶

hand written solution

13.19¶

hand written solution

In [5]:
W = 40   # lb
d = 20   # ft
vA = 10  # ft/s
mu = 0.2
Q = 30*dtr # rad
h = 4    # ft
if 0:
    W = 44   # lb
    vA = 14  # ft/s
    mu = 0.25
a = gfps*(sin(Q)-mu*cos(Q))
print('acceleration = {:.2f} ft/s2'.format(a))
vB = sqrt(vA**2 + 2*a*d)
t0 = (vB-vA)/a  # time from A to B
print('vB = {:.2f} ft/s'.format(vB))
t1 = (-vB*sin(Q)+sqrt((vB*sin(Q))**2+2*gfps*h))/gfps
t2 = (-vB*sin(Q)-sqrt((vB*sin(Q))**2+2*gfps*h))/gfps
print('times are {:.3f} s (+) and {:.3f} s (-)'.format(t1,t2))
R = vB*cos(Q)*t1
print('time from A to C = {:.3f} s'.format(t0+t1))
print('R = {:.2f} ft'.format(R))
sqrt(14 + 2*9.128*20)
acceleration = 10.52 ft/s2
vB = 22.82 ft/s
times are 0.257 s (+) and -0.966 s (-)
time from A to C = 1.476 s
R = 5.08 ft
Out[5]:
19.471004082994796

13.21¶

hand written solution

Book solution is incorrect again!

In [6]:
vA = 2.5  # m/s
Dx = 3    # m
mu = 0.3
if 0:
    mu = 0.45
q_crit = arctan(mu)
print('angle at which acceleration is zero:',q_crit/dtr)
alpha = vA**2/(2*g*Dx)
A,B,C = mu**2+1, -2*mu*alpha, alpha**2-1   # quadratic equation for cos(Q)
print('roots are {:.4f} (+) and {:.4f} (-)'.format(   (-B+sqrt(B**2-4*A*C))/(2*A),
                                                      (-B-sqrt(B**2-4*A*C))/(2*A)  ))
Q = arccos((-B+sqrt(B**2-4*A*C))/(2*A))
print('minimum angle is {:.4f} deg'.format(Q/dtr))
angle at which acceleration is zero: 16.69924423399362
roots are 0.9821 (+) and -0.9236 (-)
minimum angle is 10.8618 deg

Alternate (book) method:

In [7]:
a = -vA**2 / (2*Dx)
print('acceleration is {:.4f} '.format(a))
# book's error here
if 0: a = -a  # do what book did
A,B,C = mu**2+1, 2*mu*(a/g),(a/g)**2-1   # quadratic equation for cos(Q)
print('roots are {:.4f} (+) and {:.4f} (-)'.format(   (-B+sqrt(B**2-4*A*C))/(2*A),
                                                      (-B-sqrt(B**2-4*A*C))/(2*A)  ))
Q = arccos((-B+sqrt(B**2-4*A*C))/(2*A))
print('min angle is {:.4f} deg'.format(Q/dtr))
acceleration is -1.0417 
roots are 0.9821 (+) and -0.9236 (-)
min angle is 10.8618 deg

13.22¶

hand written solution

In [8]:
mA,mB = 50,15   # kg
t1 = 2          # s
sBdd = (mB-3*mA)/(mB+9*mA) * g
vB = sBdd*t1
print('acceleration of B is {:.2f} m/s2'.format(sBdd))
print('velocity of B is {:.2f} m/s'.format(vB))
acceleration of B is -2.85 m/s2
velocity of B is -5.70 m/s

13.23¶

In [9]:
F = 150   # N
m = 50    # kg
a = 4*F/m - g
v = sqrt(2*a*3)
print('speed is {:.4f} m/s'.format(v))
speed is 3.6249 m/s

13.25¶

hand written solution

In [10]:
vA = 2   # m/s
mu = 0.2 
Ds = 5   # m
Q = 30*dtr  # rad
h = 2.5  # m
a = g*( sin(Q)-mu*cos(Q))
print('acceleration = {:.3f} m/s2'.format(a))
vC = sqrt( vA**2 + 2*a*Ds)
print('vC = {:.3f} m/s'.format(vC))
A,B,C = .5*g, vC*sin(Q), -h
tB = ( -B + sqrt(B**2-4*A*C) )/(2*A)
print('flight time = {:.3f} s'.format(tB))
R = tB *vC*cos(Q)
print('range = {:.3f} m'.format(R))
acceleration = 3.206 m/s2
vC = 6.005 m/s
flight time = 0.471 s
range = 2.448 m

13.27¶

hand written solution

In [11]:
mu = 0.8
v0 = 4      # m/s
Q = 30*dtr  # rad
a = g*( mu*cos(Q)-sin(Q))
t = v0/a
print('time to stop is {:.2f} s'.format(t))
time to stop is 2.11 s

13.45¶

hand written solution

In [12]:
m = 55  # kg
A = 120  # N.s^(-3/2)
t1 = 6  # s
mus,muk = 0.4, 0.29
t0 = ( mus*m*g/A)**(2/3)
print('time to start moving is {:.4f} s'.format(t0))
v1 = muk*g*(t0-t1) + 2*A/(5*m)*(((t1)**(5/2)-(t0)**(5/2)))
print('velocity at time t1 is {:.4f} m/s'.format(v1))
time to start moving is 1.4789 s
velocity at time t1 is 61.7752 m/s

13.49¶

hand written solution

In [13]:
WA,WB = 8,15  # lb
P = 12        # lb
Q = 15*dtr    # rad
mA,mB = WA/gfps, WB/gfps
T = tan(Q)
sBdd = ( WB*T - P)/( mA/T + mB*T )
print( 'acceleration of B = {:.3f} ft/s2'.format(-sBdd))
acceleration of B = 7.586 ft/s2

F13.10¶

In [14]:
rho = 500  # ft
Q = 30*dtr
mu = 0.2
v = sqrt(rho*gfps*(sin(Q)+mu*cos(Q))/(cos(Q)-mu*sin(Q)))
print('v = {:.2f} ft/s'.format(v))
v = 118.95 ft/s

F13.13¶

hand written solution

In [15]:
l,d = 8,1.5
phi = 45*dtr
r = d + l*sin(phi)
Qd = sqrt(g/r*tan(phi))
print('angular speed = {:.3f} rad/s'.format(Qd))
angular speed = 1.171 rad/s

13.54¶

In [16]:
ma,mb = 15,2
rho = 1.5
v = sqrt(ma*g*rho/mb)
v
Out[16]:
10.505355776935877
In [17]:
ma,mb = 30,5
rho = 2
v = sqrt(ma*g*rho/mb)
v
Out[17]:
10.849884792015075

13.57¶

$\displaystyle ks = m a_c = m \frac{v^2}{s}$

In [18]:
m,k,s = 0.75, 200, 100e-3  # kg,N/m,m
v = s*sqrt(k/m)
print('speed = ',v)
print('N = ',m*g)
speed =  1.6329931618554523
N =  7.3575

13.58¶

hand written solution

In [19]:
mu = 0.2
L = 0.25   # m
v = sqrt(  g*4*L/5 *( 3-4*mu )/(4+3*mu) )
print( 'min speed is {:.3f} m/s'.format(v))
min speed is 0.969 m/s

13.75¶

hand written solution (pdf)

13.86¶

hand written solution

In [20]:
A,B,C,D = 2,10,1.5,6
t,m = 2,5
ar = -(A*t+B)*(2*C*t-D)**2
an = (A*t+B)*2*C + 2*A*(2*C*t-D)
a = sqrt(ar**2+an**2)
m*a
Out[20]:
210.0

13.90¶

hand written solution

In [21]:
m = 40   # kg
R = 1.5  # m
v1 = 2    # m/s
z0 = 2   # m
#d = sqrt(z0**2 +(2*pi*R)**2)
#w = 2*pi*v/d
p = arctan(z0/(2*pi*R))   # angle of slide
w1 = v1*cos(p)/R
Fr = -m*R*w1**2
N1 = m*g*cos(p)
FQ = N1*sin(p)
Fz = N1*cos(p)
print('Fr = {:.3f} N'.format(Fr))
print('FQ = {:.3f} N'.format(FQ))
print('Fz = {:.3f} N'.format(Fz))
p/dtr
Fr = -102.070 N
FQ = 79.682 N
Fz = 375.491 N
Out[21]:
11.980813567686223

13.92¶

hand written solution (pdf)

In [22]:
w = 4      # rad/s
alpha = 3  # rad/s2
Q0 = pi    # rad
A = 2      # m
m = 0.5    # kg
ar = A/Q0**2*(2*w**2/Q0 - alpha) - A*w**2/Q0
aQ = A*alpha/Q0 - 2*A*w**2/Q0**2
print('ar = {:.3f} m/s2'.format(ar))
print('aQ = {:.3f} m/s2'.format(aQ))
cospsi = 1/sqrt(1+Q0**2)
sinpsi = Q0*cospsi
NP = -m*ar/sinpsi
NA = m*aQ + NP*cospsi
print('slotted path force = {:.3f} N'.format(NP))
print('arm force = {:.3f} N'.format(NA))
ar = -8.730 m/s2
aQ = -4.575 m/s2
slotted path force = 4.581 N
arm force = -0.898 N

13.93¶

hand written solution

In [23]:
Q = 45*dtr    # rad
l = 4         # ft
omega = 1.5   # rad/s
W = 4         # lb
#
rd = l*omega*tan(Q)/cos(Q)
rdd = (l*omega**2 + 2*l*omega**2*tan(Q))/cos(Q)
r = l/cos(Q)
ar = rdd - r*omega**2
aQ = 2*rd*omega
F = W*( aQ/gfps + cos(Q) + ar/gfps*tan(Q) + sin(Q)*tan(Q) )
N = W*(ar/gfps+sin(Q))/cos(Q)
print(f'F = {F:.2f} lb')
F = 11.98 lb

13.109¶

hand written solution

In [24]:
w = 4    # rad/s
R = 0.8  # m
Q0 = 45*dtr  # rad
m = 0.5  # kg
a = 4*R*w**2
ar = -a*cos(Q0)
aQ = -a*sin(Q0)
print ('a = {:.2f} ur + {:.2f} uQ m/s2'.format(ar,aQ))
NR = 0 
NC = m*a
print('force of rod: {:.2f} N'.format(NR))
print('force of circle: {:.2f} N'.format(NC))
a = -36.20 ur + -36.20 uQ m/s2
force of rod: 0.00 N
force of circle: 25.60 N

R13-8¶

hand written solution

In [25]:
r,rd,rdd = 0.5,3,1
Qd,Qdd = 6,2
m = 4 
f = -m*(rdd-r*Qd**2)
phi = arctan(g/(r*Qdd+2*rd*Qd))
N = m*g/sin(phi)
print('friction = {:.3f} N'.format(f))
print('normal   = {:.3f} N'.format(N))
friction = 68.000 N
normal   = 153.114 N