To find the mechanical advantage, velocity ratio and efficiency of a simple screw jack

Experiment No. 5 Objective : To find the mechanical advantage, velocity ratio and efficiency of a simple screw jack and plot the graph between : (i) (ii)

Efficiency v/s Load Effort v/s Load.

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Apparatus :

Screw jack apparatus, slotted weights, string, outside caliper etc.

Apparatus Description

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SCREW JACK A screw jack is simply a ramp wrapped around an axle, with the axle rotated by a handle used as a lever.

A screw jack consists of the following parts: 1. Screw threads: It consists of threads that may be square or V in shape. However square threads are more efficient than V threads and are used for power transmission. 2. Head: The screw has a head on its upper end on which the load W rests. 3. Nut: In the case of a screw jack, the threads of the screw will slide around upon the fixed threads of the nut fixed in the frame and which generally forms a part of the body of the screw jack.

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Screw Jack Apparatus

Theory

Screw jack is used to raise heavy loads. The apparatus works on a simple principle of screw and nut. The axial distance between the corresponding threads is known as PITCH. Let this pitch be p & d is the diameter of the flanged table on which the load w is to be placed and lifted. Let the table turns through one revolution. Load rises in one revolution = length of the screw jack = l l = 2×pitch

Or l = 2×p Effort moved in one revolution = πD Velocity ratio = distance moved by effort / distance moved by load = πD/l = πD/2*p Total effort in the two hangers including the weight of the hangers=P

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Mechanical advantage = W/P %η = M.A. /V.R. *100

Mechanical advantage (MA) is the factor by which a mechanism multiplies the force put into it. It is the ratio of the force exerted by a machine (the output) to the force exerted on the machine, usually by an operator (the input). The theoretical mechanical advantage of a system is the ratio of the force that performs the useful work to the force applied, assuming there is no friction in the system. In practice, the actual mechanical advantage will be less than the theoretical value by an amount determined by the amount of friction. i.e., M.A. =

Load Lifted W = Effort Applied P

Procedure :

2. 3.

4. 5. 6.

Wrap the string round the circumference of the flanged table and pass it over the pulley. Similarly, wrap another string over the circumference of the flanged table and take it over the second pulley. The free ends of both the strings be tied to the two hangers where the weights are to be hanged, Measure the pitch of the thread with the help of the Vernier Caliper. Place the load W on the screw head and some weight on hangers so that the load W is just lifted. The effort P is equal to the sum of the weights hanged in the two hangers. Increase the loads and find the corresponding efforts applied for the consecutive readings. Calculate mechanical advantage, velocity ratio and efficiency in each case. Plot the graph between efficiency v/s load, effort v/s load.

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1.

Observations : (i) Circumference of table (ii) Load of screw = 2 x p (iii) Velocity ratio = π D

= 39.88 cm. = 12 mm. = 33.23 L

1 2 3 4

Load kg (W) 5.875+0.969=6.844 5.875+1.927=7.802 5.875+2.427=8.302 5.875+2.929=8.804

Effort P Kg= P1 + P2 0.570 0.650 0.680 0.720

12.00 12.00 12.20 12.22

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S.No.

M.A. = %η = ( M.A./ W/P V.R.) * 100

Calculations :

Diameter of load pulley = 127 mm. Pitch of screw jack = 6 mm. Weight of screw jack system = 5.875 K.g.

Velocity Ratio (V.R.) = π D = 3.14 x 127 2p 2x6 (V.R.) = 398.78 = 33.23 12

36.11 36.11 36.71 36.79

1.

Mechanical Advantage (M.A.) = W = 6.844 P 0.570 M.A. = 12.00

%ή = M.A. x 100 V.R.

%ή = 12.00 x 100 = 36.11

2.

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33.23

Mechanical Advantage (M.A.) = W = 7.802 P 0.650 M.A. = 12.00

%ή = M.A. x 100 V.R.

%ή = 11.44 x 100 = 36.11 33.23

3.

Mechanical Advantage (M.A.) = W = 8.302 P 0.680 M.A. = 12.20

%ή = M.A. x 100

V.R.

%ή = 12.20 x 100 = 36.71 33.23

Mechanical Advantage (M.A.) = W = 8.804 P 0.720 M.A. = 12.22

%ή = M.A. x 100 V.R.

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4.

%ή = 12.22 x 100 = 36.79 33.23

Graphical analysis

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Graph 1

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Graph 2

Result: The Mechanical Advantage of the given simple screwjack = 12.10. Velocity Ratio = 33.23 Efficiency = 36.43%

Conclusion: A. On analysis of Effort v/s Load graph, we observe that that a minimum initial effort has to be applied even when the load is zero, and then the effort varies linearly with load with a low value of slope of the straight line being formed. Hence, we conclude that a large amount of load can be lifted by applying comparatively much lesser effort.

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B. On analysis of Efficiency v/s Load, we conclude that the efficiency of an apparatus remains constant even if the load is varied, and hence the Mechanical Advantage is also constant.

precautions :

1. Use both the pulleys to find the values of effort P to avoid the side thrust. 2. The load and effort should move slowly. 3. Add weights in hangers gently. 4. Lubricate the screw to decrease friction. 5. The string should not overlap. 6. There should be no knot in the string. 7. See that both the pans should move downwards.

Sources of Error 1. The friction offered by pulleys is bound to introduce some error in the readings. No matter what amount of lubrication is done, pulleys can never be made completely frictionless. 2. Friction due to the screws of the screw jack may also introduce error. 3. If the string is not inextensible or if it overlaps or if there is a knot in the string, then it will result in error. 4. Weights in the pan may be added forcefully. This should not happen, as this will introduce the impulse causing an error.

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References:

• Write ups provided in the Applied Mechanics Lab. • www.wikipedia.org/wiki/jackscrew • “Applied Mechanics-Statics and Strengths of Materials” by U.C.Jindal • Encyclopedia : Encarta