It is found that the efficiency of a simple machine increases with load but not in a linear fashion.
As can be seen in Figure 1.81, the efficiency eventually levels off at a limiting value. This is found to depend on the constant a, in the law of the machine, and its velocity ratio.
efficiency¼mechanical advantage velocity ratio
¼MA
VR ¼ W
E VR
Now, from the law of the machine, E¼ aW þ b
¼ W
ðaW þ bÞ VR
Dividing numerator and denominator by W gives
¼ 1
aþ b W
VR
h = 1
a VR+ b
WVR ð1:66Þ
Examination of this expression shows that as the load W, increases, the second term in the denominator becomes smaller and smaller. When the load becomes very large, this term tends to zero and the limiting value of efficiency is
h = 1
a VR ð1:67Þ
In this limiting condition the mechanical advantage levels off to a value given by
¼MA VR MA¼ VR MA¼1
a ð1:68Þ
The limiting value of mechanical advantage is thus the reciprocal of the gradient of the effort v. load graph.
Limiting value of efficiency Limiting value of MA
Mechanical advantage, Efficiency,η % MA
Load, W Load, W
(a) Graph of efficiency v. load (b) Graph of mechanical advantage v. load Figure 1.81 Variation of efficiency and mechanical advantage with load
Key point
The mechanical advantage and the efficiency of a simple machine increase with load but eventually level off at a limiting value.
Overhauling
A simple machine is said to overhaul if when the effort is removed, the load falls under the effects of gravity. If a machine does not overhaul, then friction alone must be sufficient to support the load. If you have changed a wheel on a car you probably used a screw jack to raise the wheel off the road surface. Friction in the screw thread is then sufficient to support the car. This means, of course that there must be quite a lot of friction present which results in a low value of efficiency. This however is the price that often has to be paid for safety in lifting devices such as car jacks and engine hoists.
Frictional resistance can be considered as an additional load which the effort must overcome.
total
¼effort to overcome
þeffort to overcome effort friction load, F actual load, W
Now in an ideal machine where there is no friction, the mechanical advantage is equal to the velocity ratio and
VR¼ MA ¼ load
If friction force, F is being considered as a separate load, the total effort can be written as
total effort¼ F VRþ W
VR E= F+ W
VR ð1:69Þ
Now, the efficiency is given by
¼MA
VR ¼ W
E VR Substituting for E gives
¼ W
When the effort is removed there is only the frictional resistance F, to oppose the load W. To stop it overhauling the friction force must be equal or greater than the load. In the limit when F¼ W, the efficiency will be
h = W
2W = 0:5 or 50% ð1:71Þ
It follows that a machine will overhaul if its efficiency is greater than 50%.
Key point
A simple machine will over-haul if the load falls under the effects of gravity when the effort is removed. Over-hauling is likely to occur if the efficiency of a machine is greater than 50%.
Example 1.25
The law of the gear winch shown in Figure 1.82 is E ¼ 0.0105 W þ 5.5 and it is required to raise a load of 150 kg. Determine, (a) its velocity ratio, (b) the effort required at the operating handle, (c) the mechanical advantage and efficiency when raising this load, (d) the limiting mechanical advantage and efficiency and (e) state whether the winch is likely to overhaul when raising the 150 kg load.
Operating handle
(a) Finding velocity ratio:
VR ¼2r
(b) Finding effort required:
E ¼ 0.0105 W þ 5.5 ¼ (0.0105 150 9.81) þ 5.5 E = 21.0 N
(c) Finding mechanical advantage:
MA ¼W
(d) Finding limiting mechanical advantage:
limiting MA ¼1
a ðwhere a ¼ 0:0105; from the law of the machine) limiting MA= 1
(e) When raising the 150 kg load the efficiency is greater than 50%
and so the winch will overhaul under these conditions.
Test your knowledge 1.11 1. How does the
mechanical advantage of a simple machine vary with the load raised?
2. How does the frictional resistance in a simple machine vary with the load raised?
3. How is the efficiency of a simple machine defined?
4. What information does the law of a machine contain?
5. What is meant by overhauling and how can it be predicted?
Activity 1.11
A screw jack has a single start thread of pitch 6 mm and an operating handle of radius 450 mm. The following readings of load and effort were taken during a test on the jack.
Load (kN) 0 1 2 3 4 5 6 7 8 9 10
Effort (N) 2.2 6.6 11.8 17.0 22.2 27.2 32.2 36.8 41.4 46.9 52.4 (a) Plot a graph of effort against load and from it, determine the
law of the machine.
(b) Plot a graph of mechanical advantage against load.
(c) Plot a graph of efficiency against load and state whether the machine is likely to overhaul.
(d) Calculate the theoretical limiting values of mechanical advantage and efficiency, and state whether your graphs tend towards these values.
(e) Calculate the work input and the work done against friction when raising the 10 kN load through a height of 50 mm.
Problems 1.9
1. A screw jack has a single start thread of pitch 10 mm and is used to raise a load of 900 kg. The required effort is 75 N, applied at the end of an operating handle 600 mm long. For these operating conditions, determine (a) the mechanical advantage, (b) the velo-city ratio, (c) the efficiency of the machine.
[118, 377, 31.3%]
2. The top pulleys of a Weston differential pulley block have diameters of 210 mm and 190 mm. Determine the effort required to raise a load of 150 kg if the efficiency of the system is 35%. What is the work done in overcoming friction when the load is raised through a height of 2.5 m?
[200 N, 6.83 kJ]
3. A differential wheel and axle has a wheel diameter of 300 mm and axle diameters of 100 mm and 75 mm. The effort required to raise a load of 50 kg is 55 N and the effort required to raise a load of 200 kg is 180 N. Determine (a) the velocity ratio, (b) the law of the machine, (c) the efficiency of the machine when raising a load of 100 kg.
[24, E¼ 0.085 W þ 13.3, 42.3%]
4.
A D
100 mm
B C
450 mm
Teeth on A = 20 Teeth on B = 100 Teeth on C = 25 Teeth on D = 110
Figure 1.83
The crab winch shown in Figure 1.83 has a law of the form E¼ 0.01 W þ 7.5. Determine (a) its velocity ratio, (b) its mech-anical advantage and efficiency when raising a load of 50 kg, (c) the limiting values of its mechanical advantage and efficiency.
[198, 20%, 100, 50.5%]