Determine maximum mechanical advantage:
For a lifting machine 15 N effort is needed to raise a load of 70 N with an efficiency equal to 60%, and 25 N effort is needed to lift a load of 1300 N. Find out the law of machine. What shall be the effort required to raise a load of 1000 N? Determine maximum mechanical advantage and the maximum efficiency.
Solution
Given :
Effort P_{1} = 15 N, and load W_{1} = 700 N
Effort P_{2} = 25 N, and load W_{2} = 1300 N
Let m be the gradient and C be the intercept on y-axis. The law of machine can be expressed as :
P = m W + C ----------- (a)
Substituting for P and W in Eq. (a), we obtain:
15 = m 700 + C --------- (b)
and 25 = m 1300 + C ------------ (c)
Subtracting Eq. (b) from Eq. (c),
10 = 600 m, or m = 10/600 = 1 /60
Substituting m in Eq. (b)
15 = (1/60) × 700 + C
or C = 15 - (70/6) = 3.33
Substituting values of m and C, we obtain the law of machine as follows :
P = (W /60) + 3.33 ----- (d) Ans
When load W = 1000 N
P = (1000 /60 )+ 3.33 = 19.99 N
Maximum Mechanical Advantage
Max. M. A. = 1/ m = 1/ (1/60) =60 Ans.
Maximum Mechanical Efficiency
Max. η =1/(m × VR) = 60 Ans.
For 15 N effort to lift a load equal to 700 N,
M. A. = 700/15 = 46.66
Therefore, 0.6 = 46.66 /VR
or, V. R. = 46.66/0.6 = 77.77
Therefore, Max. η = 1 / (m × VR)
= 60/77.77 = 0.77 or 77% Ans.