In [2]:

```
import math
# Given Data
h = 100.; #lb, horizontal force
W = 300.; #lb, weight of block
us = 0.2; # Coeffiecient of static friction
uk = 0.20; #Co = efficient of kinetic friction
# Calculations and Results
#Applying sumFx = 0 , we get
F = h-3./5*W; #lb, Force along plane
F = -F
#Applying sumFy = 0, we get
N = 4./5*W #lb, Normal force to the plane
print "Force F required to maintain the equillibrium is thus %.0f lb, up and to right"%(F);
# Maximum friction force
Fm = us*N; #lb,Maximum friction force
print " Maximum friction force is %.2f lb is less than that of required to maintain equillibrium\
\n that is %.2f lb So, equillibrium will nat maintain and block wil move down"%(Fm,F);
# Actual value of friction force
Fk = (0.6*300)-(h)-(Fm); #lb, Actual value of friction force
print "Actual value of friction force is %.2f lb directed up and to the right"%(Fk);
```

In [3]:

```
import math
# Given Data
F = 800.; #N Firce in verical direction
us = 0.35; # Coeffiecient of static friction
uk = 0.25; #Co = efficient of kinetic friction
theta = 25.; #degree, angle of inclination
# Calculations and Results
theta = theta*math.pi/180; #rad, Conversion into radian
# Force P start block moving up
# At static equillibrium Tan(Theta_s) = us
theta_s = math.atan(us); #rad
P = F*math.tan(theta+theta_s); #N,Force P to start block moving up
print "Force P to start block moving up is %.0f N"%(P);
# Force P to keep block moving up
# At kinetic equillibrium Tan(Theta_k) = uk
theta_k = math.atan(uk); #rad
P = F*math.tan(theta+theta_k); #N,Force P to keep block moving up
print "Force P to keep block moving up is %.0f N"%(P);
# Force P to prevent block from sliding down
theta_s = math.atan(us); #rad
P = F*math.tan(theta-theta_s); #N,Force P to prevent block from sliding down
print "Force P to prevent block from sliding down is %.0f N"%(P);
```

In [4]:

```
import math
# Given Data
us = 0.25; # Coeffiecient of static friction
# Calculations and Results
#Applying equillibrium equation we get relation in x
print "Apply equillibrium equations. It is theoritical part. ";
x = 12-(.75*2)+1.5 #in, Dismath.tance at which the applied load can be supported
print "Minimum dismath.tance at which the applied load can be supported is %.0f in"%(x);
```

In [5]:

```
import math
# Given Data
F = 400.; #lb, force exerte
us = 0.35; # Coeffiecient of static friction
phi = math.degrees(math.atan(us)); #rad, angle of friction
#print (phi)
theta = 8.; #degree, angle of inclination
theta = theta*math.pi/180; #rad, Conversion into radian
# Calculations and Results
#Umath.sing math.sine rule
#force p to raise block
#free body , block B
R1 = F*math.sin(math.radians(109.3))/(math.sin(math.radians(43.4)))
#free body wedge A
P = R1*math.sin(math.radians(46.6))/(math.sin(math.radians(70.7)))
print " force required to raise block is P = %.0f lb"%(P);
#force to lower block
#free body , block B
R1 = F*math.sin(math.radians(70.7))/(math.sin(math.radians(98.0)))
#free body wedge A
P = R1*math.sin(math.radians(30.6))/(math.sin(math.radians(70.7)))
print " force required to lower block is P = %.0f lb"%(P);
```

In [6]:

```
import math
# Given Data
pitch = 2.; #mm, pitch of screw
d = 10.; #mm, mean diameter of thread
r = d/2; #mm, radius
us = 0.30; # Coeffiecient of static friction
M = 40.; #kN.m , Maximum couple
# Calculations and Results
#Force exerted by clamp
L = 2*pitch; #mm, as screw is double threaded
theta = math.atan(L/(2*math.pi*r)); #rad, angle of inclination
phi = math.atan(us); #rad, angle of friction
Q = M/r*1000; #N, Force applied to block representing screw
Q = Q/1000 #kN, Conversion into kN
W = Q/math.tan(theta+phi); #kN, Magnitude of force exerted on the piece of wood
print "Magnitude of force exerted on the piece of wood is W = %.2f kN "%(W);
#Couple required to loosen clamp
Q = W*math.tan(phi-theta); #kN, Force required to loosen clamp
Couple = Q*r; #N.m, Couple required to loosen clamp
print "Couple required to loosen clamp is %.2f N.m"%(Couple);
```

In [11]:

```
import math
from sympy.mpmath import cot
#from numpy import cot
# Given Data
r = 1. #in in
us = 0.20; # Coeffiecient of static friction between shaft and pully
# Calculations and Results
#Vertical Force required to raise load
rf = r*us; #in, Perpendicular dismath.tance from the center Of pully to line of action
#summing moment about B
P1 = (2.20*500)/1.8 #lb , downward Force required to raise load
print "Force required to raise load is %f lb in downward direction"%(P1);
#Vertcal Force required to hold load
#summing moment about C
P = (1.80*500)/2.20 #lb , downward Force required to hold load
print "Force required to hold load is %.0f lb in downward direction"%(P);
#Horizontal force P to start raimath.sing the load
OE = rf; #mm,
OD = math.sqrt(2)*2; #in, pythagorus theorm
theta = math.asin(OE/OD); #rad,
# from force triangle
P = 500*cot(math.radians(40.9)); #lb, Horizontal force P to start raimath.sing the load
print "Horizontal force P required to start raising the load is %.0f lb"%(P);
```

In [12]:

```
import math
# Given Data
T1 = 150.; #N, Force on free end of hawser
T2 = 7500.; #N, Force on other end of hawser
# Calculations and Results
#a, coefficient of friction
bta = 2*2*math.pi; #rad, angle of contact, 2 turns
#By equation 8.13
us = math.log(T2/T1)/bta; # Co-efficient of static friction
print "Coefficient of static friction between hawser and ballard is us = %0.3f "%(us);
#Number of wraps when tension in hawser = 75 kN
bta = 3*2*math.pi #in rad
#One turn = 2* pi angle, bta corresponds to
ten = T1*math.exp(bta*us)
print "Tension is %f N "%(ten);
```

In [13]:

```
import math
#Given
T2 = 600.; #lb, Tension from side 2
us = 0.25; # Coeffiecient of static friction between pulley and belt
bta = (2*math.pi)/3; #Co = efficient of kinetic friction between pulley and belt
r1 = 8. #in in
#Pulley B
# Calculations
T1 = T2/(math.exp(us*bta)) #N, Tension from side 1
#print (T1)
#Pulley A
#Aumming moment about A
MA = (T2*r1)-(T1*r1); #lb-ft, Couple MA applied to pulley which is equal and opposite to torque
# Results
print "The largest torque which can be exerted by belt on pulley A is MA = %0.0f lb-in"%(MA);
```