Engineering Mechanics

Chapter 11 : Simple Lifting Machines „„„„„ 203

Solution. Given: Dia of load drum = 160 mm or radius (r) = 100/2 = 80 mm; Length of
handle (l) = 360 mm; No. of teeth on pinions (T 2 ) = 20 and (T 4 ) = 30 and no. of teeth on spus
wheels (T 1 ) = 75 and (T 3 ) = 90.

When P = 90 N, W = 1800 N when P = 135 N, W = 3150 N

(a) Law of the machine,

Substituting the values of P and W in the law of the machine, i.e., P = m W + C

90 = (m × 1800) + C ...(i)

and 135 = (m × 3150) + C ...(ii)

Subtracting equation (i) from equation (ii),

45 = m × 1350

or

45 1

1350 30

m==

Now substituting this value of m in equation (i),

1

90 1800 60

30

=× +=+CC

∴ C = 90 – 60 = 30

and now substituting the value fo m and C in the law of the machine,

1

30

30

PW=+^ Ans.

(b) Effort to lift a load of 4500 N

Substituting the value of W equal to 4500 N in the law of the machine,

1

4500 30 180 N

30

P

⎛⎞

=× +=⎜⎟

⎝⎠

Ans.

(c) Efficiency of the machine in the above case

We know that velocity ratio

13

24

360 75 90

V.R. 50.6

80 20 30

l TT

rT T

⎛⎞× ⎛⎞

==×=⎜⎟⎜⎟

⎝⎠× ⎝⎠

and

4500

M.A. 25

180

W

P

== =

∴ Efficiency,

M.A. 25

0.494 49.4 %

V.R. 50.6

η= = = = Ans.

(d) Maximum efficiency of the machine

We also know that maximum efficiency of the machine,

max

11

0.593 59.3%

V.R.^1

50.6

30

m

η= = = =

× ×

Ans.