Applications of Friction
4. Depth of thread. It is the distance between the top and bottom surfaces of a thread (also known as crest and root of thread)
9.7. EFFICIENCY OF A SCREW JACK
We have seen in Art. 9.6 that the effort (P) required at the mean radius of a jack to lift the load (W),
P = W tan (α + φ) ...(i)
where α = Helix angle, and
μ = Coefficient of friction between the screw and the nut,
= tan φ ...(where φ = Angle of friction)
If there would have been no friction between the screw and the nut, then φ will be zero. In such a case, the value of effort (P0) necessary to raise the same load, will be given by the equation :
P0 = W tan α ...[substituting φ = 0 in equation (i)]
∴ Ideal effort 0 tan tan
Efficiency ( ) =
It shows that the efficiency of a screw jack is independent of the weight lifted or effort ap-plied. The above equation for the efficiency of a screw jack may also be written as:
sin
Now for the efficiency to be maximum, the term (1 – η) should be the least. Or in other words, the value of sin (2α + φ) should be the greatest. This is only possible, when
2α + φ = 90° or 2α = 90° – φ
∴ α =45 –° 2φ
It shows that the maximum efficiency of a screw jack is also independent of the weight lifted or effort applied.
Example 9.13. A load of 2.5 kN is to be raised by a screw jack with mean diameter of 75 mm and pitch of 12 mm. Find the efficiency of the screw jack, if the coefficient of friction between the screw and nut is 0.075.
Solution. Given: Load (W) = 2.5 kN ; Mean diameter of the screw (d) = 75 mm; Pitch of the screen (p) = 12 mm and coefficient of friction between the screw and nut (μ) = 0.075 = tan φ.
We know that tan 12 0.051
75 p
α = d = =
π π ×
and efficiency of the screw jack,
Chapter 9 :
Applications of Friction 169
Example 9.14. A screw jack has a square thread of 75 mm mean diameter and 15 mm pitch. The load on the jack revolves with the screws. The coefficient of friction at the screw thread is 0.05. (i) Find the tangential force to be applied to the jack at 360 mm radius, so as to lift a load of 6 kN weight. (ii) State whether the jack is selflocking. If it is, find the torque necessary to lower the load. If not, find the torque which must be applied to keep the load from descending.
Solution. Given: Mean diameter of square thread (d) = 75 mm or mean radius (r) = 37.5 mm; Pitch (p) = 15 mm; Coefficient of friction (μ) = 0.05 = tan φ; Radius of effort arm = 360 mm and load lifted = 6 kN = 6000 N.
(i) Tangential force to be applied at the jack.
Let P1 = Tangential force to be applied at 36 cm radius to lift the load, and α = Helix angle.
We know that tan 15 0.064
75 p
α = d = =
π π ×
and tangential force required at the mean radius to lift the load, tan tan
Now the effort applied at a radius of 36 cm may be found out from the relation P1 × 360 = P × r = 686.2 × 37.5 = 25 732
∴ P1=25 732360 =71.48 N Ans.
(ii) Self-locking of the screw jack
We know that efficiency of the screw jack,
tan tan
Since efficiency of the jack is more than 50%, therefore, it is not *self-locking. Ans.
Torque, which must be applied to keep the load from descending
We know that the force which must be applied at the mean radius to keep the load from descending (i.e. to prevent the load from descending).
2 tan ( – ) tan – tan
∴ Torque, which must be applied to keep the load from descending
= P2 × r = 83.73 × 37.5 = 3140 N-mm Ans.
* For details, please refer to Art. 10.14.
EXERCISE 9.3
1. A square threaded screw jack of mean diameter 25 mm and a pitch of 6 mm is used to lift the load of 1500 N. Find the force required at the mean circumference if the coefficient of friction between the screw and nut is 0.02. [Ans. 144.2 N]
2. A square threaded screw jack of mean diameter 50 mm has 3° angle of inclination of the thread and coefficient of friction 0.06. Find the effort required at the end of handle 450 mm long (i) to raise a load of 20 kN, and (ii) to lower the same load.
[Ans. 126 N; 8.4 N]
QUESTIONS
1. What is a wedge? State its uses and the method of solving the problems on wedge friction.
2. What is a screw jack? Explain the principle, on which it works.
3. Establish a relation between the effort and load, when a square threaded screw is used for lifting purposes, taking friction into account.
4. Derive a relation for the efficiency of a screw jack, taking friction into account.
5. In a screw jack, the helix angle is α and the angle of friction is φ. Show that its efficiency is maximum, when
2α = 90° – φ.
OBJECTIVE TYPE QUESTIONS
1. If a ladder is not in equilibrium against a smooth vertical wall, then it can be made in equilibrium by
(a) increasing the angle of inclination.
(b) decreasing the angle of inclination.
(c) increasing the length of the ladder.
(d) decreasing the length of the ladder.
2. The efficiency of a screw jack may be increased by (a) increasing its pitch.
(b) decreasing its pitch.
(c) increasing the load to be lifted.
(d) decreasing the load to be lifted.
3. The efficiency of a screw jack is maximum when the helix angle is equal to (a) 45
2
° + φ (b) 45 – 2
° φ (c) 30
φ + °2 (d) – 30 2
φ °
ANSWERS
1. (a) 2. (a) 3. (b)
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Chapter 10 :
Principles of Lifting Machines 171
In olden times, thousands of slaves had to be arranged, whenever a heavy load had to be lifted or dragged. Even today, in the absence of a suitable device, many people have to be arranged to lift a motor car so that its tyres can be changed. In order to overcome such difficulties, a few simple machines were invented, which could save the man power i.e., a single man can do the same work as many could do, though at a lesser speed.
Before entering into the details of simple machines and their working, following terms should be clearly understood at this stage :