Chapter 10:Principles of Lifting Machines

Example 10.1, Page no.173

In [1]:
#variable declaration
W=1000     #Weight in N
P=25       #Effort in N
x=0.1      #Distance through which the weight is moved in m
y=8        #Distance through which effort is moved in m

#calculation
MA=W/P
VR=y/x
Eta=MA/VR

#Result
print"Mechanical advantage of the machine, M.A.=",int(MA)
print"Velocity ratio of the machine, V.R.=",int(VR)
print"Efficiency of the machine, Eta=",int(round(Eta*100,1)),"%"

Mechanical advantage of the machine, M.A.= 40
Velocity ratio of the machine, V.R.= 80
Efficiency of the machine, Eta= 50 %

Example 10.2, Page no.174

In [3]:
#variable declaration
VR=30     #Velocity ratio
W=1500.0  #Load in N
P=125     #Effort in N

#calculation
MA=W/P
Eta=MA/VR

#Result
print"Efficiency, Eta=",int(round(Eta*100,1)),"%"
print"Since efficiency of the machine is less than 50%, therefore the machine is non-reversible"

Efficiency, Eta= 40 %
Since efficiency of the machine is less than 50%, therefore the machine is non-reversible

Example 10.4, Page no.176

In [2]:
#variable declaration
P=100.0   #Effort in N
W=840.0   #Load in N
VR=10     #vecocity ratio

#calculation
#To calculate efficiency of the machine
MA=W/P
Eta=(MA/VR)*100
#To calculate friction of the machine
F_effort=P-(W/VR) #In terms of effort
F_load=(P*VR)-W   #In terms of load

#Result
print"Efficiency of the machine, Eta=",int(Eta),"%"
print"Friction of the machine in terms of effort, F_effort=",int(F_effort),"N"
print"Friction of the machine in terms of load, F_load=",int(F_load),"N"

Efficiency of the machine, Eta= 84 %
Friction of the machine in terms of effort, F_effort= 16 N
Friction of the machine in terms of load, F_load= 160 N

Example 10.9, Page no.181

In [12]:
#variable declaration
m=0.02   #On dividing W/50 we get 0.02W where m=0.02
W=600    #Load in N
VR=100   #velocity ratio

#calculation
Max_MA=1/m
Max_Pos_Efficiency=1/(m*VR)
P=m*W+8  #Given
F_effort=P-(W/VR)
MA=W/P
Eta=MA/VR

#Result
print"Maximum possible mechanical advantage, Max M.A.=",int(Max_MA)
print"Maximum possible efficiency=",int(round(Max_Pos_Efficiency*100,1)),"%"
print"Effort required to overcome the machine friction, F_effort=",int(F_effort),"N"
print"Efficiency of the machine, Eta=",int(round(Eta*100,1)),"%"

Maximum possible mechanical advantage, Max M.A.= 50
Maximum possible efficiency= 50 %
Effort required to overcome the machine friction, F_effort= 14 N
Efficiency of the machine, Eta= 30 %
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