(^180) „„„„„ A Textbook of Engineering Mechanics
Effort required to run the machine at no load
Substituting the value of W = 0 in the law of the machine (for no load condition),
P = 4.8 N Ans.
Effort required to run the machine at a load of 900 N
Substituting the value of W = 900 N in the law of machine,
1
900 4.8 54.8 N
18
P=× + =^ Ans.
10.17. MAXIMUM MECHANICAL ADVANTAGE OF A LIFTING MACHINE
We know that mechanical advantage of a lifting machine,
M.A.
W
P

For maximum mechanical advantage, substituting the value of P = mW + C in the above
equation,
11
Max. M.A.
W
mW C C m
m
W

• 
  
  
  
  • ... NeglectingC
    W
    ⎛⎞
    ⎜⎟
    ⎝⎠
    10.18. MAXIMUM EFFICIENCY OF A LIFTING MACHINE
    We know that efficiency of a lifting machine,
    Mechanical advantage
    Velocity ratio V.R. V.R.
    W
    P W
    P
    η= = =
    ×
    For *maximum efficiency, substituting the value of P = mW + C in the above equation,
    11
    Max.
    ()V.R V.R.
    V.R.
    W
    mW C C m
    m
    W
    η= = =
    +× ⎛⎞ ×
    ⎜⎟+×
    ⎝⎠
    ... NeglectingC
    W
    ⎛⎞
    ⎜⎟
    ⎝⎠
    Example 10.8. The law of a machine is given by the relation :
    P = 0.04 W + 7.5
    where (P) is the effort required to lift a load (W), both expressed in newtons. What is the mechanical
    advantage and efficiency of the machine, when the load is 2 kN and velocity ratio is 40? What is the
    maximum efficiency of the machine?
    If (F) is the effort lost in friction, find the relation between F and W. Also find the value of F,
    when W is 2 kN.
    Solution. Given: Law of machine P = 0.04 W + 7.5 ; Load (W) = 2 kN = 2000 N and velocity
    ratio (V.R.) = 40.
  
  
  
  

• 
  

  We know that efficiency of a lifting machine,
  M.A.
  V.R.
  η=
  A little consideration will show that the efficiency will be maximum, when the mechanical advantage
  will be maximum.
  or
  Max. Max. M.A.
  V.R.
  η=
  1
  m V.R.

  
  

  × ...
  Max. M.A.^1
  m
  ⎛⎞=
  ⎜⎟⎝⎠Q