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 P₁ = 15 N, and load W₁ = 700 N

 Effort P₂ = 25 N, and load W₂ = 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.