Find out the effort required to run this machine:
The efficiency of a machine is 75% while an effort of 20 N is required to lift a load of 160 N. Compute the velocity ratio and frictional force of the machine in terms of effort and load. What is the law of machine if frictional resistance of the machine is constant? Find out the effort required to run this machine at a load of 200 N.
Solution
Given :
Efficiency (η) = 75% Effort (P) = 20 N
Load lifted (W) = 160 N
Mechanical Advantage (M A) = 160 /20 = 8
Velocity Ratio (V R) = MA/ η = 8.0/ 0.75
= 10.67 Ans.
Frictional force in terms of load (FW) = P × V R - W
or FW = 20 × 10.67 - 160
= 53.4 N
Frictional force in terms of effort (F _{P} ) = P - (W/ V R)
or F_{P} = 20 - (160 /10.67) = 5 N
The effort needed to overcome friction is 5 N, which is constant. In the law of machine, the factor C is constant which is equal to effort required when load lifted is zero.
Therefore, C = FP = 5 N
Therefore, law of the machine can be written as :
P = m W + 5
For P = 20 N and W = 160 N
20 = 160 m + 5
or m = (20 - 5) /160=15/160 = 3 /32
Substituting for m, the law of machine is given by :
P = (3/32) W + 5 Ans.
For load W = 200 N
P = (3 /32) × 200 + 5 = 23.75 N Ans.