Question

On a machine efforts of 100 N and 160 N are required to lift the loads of 3000 N and 9000 N respectively. Find the law of the machine.

(a) P = 1/100W + 60

(b) P = 1/100W + 70

(c) P = 1/100W + 80

(d) P = 1/100W + 90

I got this question in an international level competition.

Question is from Velocities in a Slider Crank Mechanism and Motion of a Link topic in division Velocity in Mechanisms of Machine Kinematics

Answer 1

Correct option is (c) P = 1/100W + 80

For explanation I would say: Let the law of machine be P = mW + C

where P = effort applied, W = load lifted and m and C being constants.

when P = 100 N    W = 3000 N

when P = 160 N    W = 9000 N

Putting these values in the law of machine.

100 = 3000m + C                                        …………(i)

160 = 9000m + C                                        …………(ii)

Subtracting (i) and (ii), we get

60 = 6000 m

or, m = 1/100

Putting this value in equation (i), we get

100 = 3000 x 1/100 + C

C = 70

Hence, the machine follows the laws

P = 1/100W +70.