The stamp hammer is an example of the useful application of the principle of the falling weight. In finding the work accumulated in any moving body, such, for instance, as energy stored up in a flywheel, the work of a railway engine when ascending an incline, the work of a cannon ball, and similar questions, it is necessary to introduce the force of gravity into the calculation, since the law of gravitation necessarily has an effect upon such bodies as are dealt with in these calculations. It is therefore perhaps necessary to briefly review the action of gravity on falling bodies to enable the practical mechanic to understand what g means.

If a body be raised 16 1/12 ft., then allowed to fall freely, it will fall through this space of 16 1/12 ft. in one second, and at the end of that second it will have attained a velocity of 32 1/6 ft. per second. This velocity of 32 1/6 ft per second, which is simply due to the force of gravity, is denoted by g, and the velocity v attained at the end of t seconds will be t x 32 1/6, or v = t x 32 1/6. Therefore the velocity of a body which has fallen for four seconds will be at the end of that time travelling at a velocity = 4 x 32 1/6 = 128 2/3 ft. per second.

The mean velocity-that is, the velocity in the middle of that time-will be 2 x 32 1/6 = 64 1/3 ft. per second, and the space described will be 42 x 16 1/12 = 257 1/3 ft.

## Work Done By A Stamp Hammer

Formula for Falling Bodies when Falling from Rest.

 h = height of fall in feet. V = velocity in feet per second. g = force of gravity = 32.2. t = time of fall in seconds.
 h = g t2 = 1/2 g t2 = V 2 2g
 V = g t = 2h = √2gh. t
 t = V = 2 h = √ 2h 9 V √ g

If the stamp hammer be raised to a definite height, the work expended in raising it will be W x h, and in doing so the force of gravity will have to be overcome. If the hammer be now supported, there will be"potential energy" stored up in it, and when allowed to fall it will attain a certain velocity depending upon the distance fallen. When the hammer reaches the end of its fall, the accumulated work in the hammer will be

W v2 / 2g and

W v2 / 2g = W h

Example.-Suppose a stamp hammer of 500 lb. weight be raised through a height of 579 ft. The work expended in raising this hammer will be

W x A = 500 x 579 = 289500 foot-pounds.

If the hammer be now supported at this height, the potential energy which exists in, or is stored up, will be = 289,500 foot-pounds. When allowed to fall the accumulated work will be

= W v2 / 2g

First find v:

 V = √2 g h = √2x 32 1/6 x 579 = 193 ft. per second;

.. when the hammer reaches the end of its fall it will have attained a velocity = 193 ft. per second;

 • • • W v2 = 500 x 193 x 193 = 289500 foot-pounds, 2 g 2 x 32 1/6

and this is the same result as W x h.

A certain amount of energy is passed into the stamp hammer in raising it up, and when it falls the energy is given out again. There is neither gain nor loss of power. It may be that a great pressure is exerted through a small space, or a less pressure exerted through a greater space, and in both instances the work may be the same.

If a resistance is offered to a 5601b. hammer, falling from a height of 16 ft., during the last one foot of its fall the average pressure acting against the resistance will be 8,9601b., the pressure being much greater at the commencement, and reducing as it reaches the last inch; i.e. (this reasoning applies to the last 12 in. of a fall of 16 ft.), the accumulated energy gradually decreasing to simply that of the weight of the hammer itself as it reaches the end.

But should the hammer be brought to rest in a fraction of 1 ft, then the resistance offered must be proportionally greater.

A stamp hammer 2001b. weight falls 10 ft, and in stamping a piece of metal the hammer is brought to rest in the space of the last 1/2 in. in its fall. What resistance has been offered by the metal article ?

W x h = 200 x 10 = 2000 foot-pounds,

1/2 in. = 1/2 of 1/12 = 1/24 of 1 ft., and since

2000 = R x 1/24 ,

..R= 2000 x 24 / 1 = 48000 lb

Notwithstanding the fact that the work of a stamp hammer may be calculated without considering gravity, we will now consider the case of another kind of hammer, which is assisted by the workman's arm-the hand hammer. The conditions under which the hand hammer is used make it necessary that the law of gravitation shall be introduced into the calculation. Take the case of a fitter striking a blow upon the head of a chisel, or driving a nail into a piece of wood, with a 2 lb. hand hammer.

As another example, consider the case of a fitter driving a key into the boss of a flywheel with a 4 lb. jack hammer. In the first case there are two forces acting upon the hammer, namely, force of gravity and the man's muscular force. The workman raises the hammer; he then drives it home, delivering a blow upon the head of the chisel. The first portion of the distance through which the hammer moves is traversed by a movement of the whole arm from the shoulder.

This is followed up by the workman straightening his arm at the elbow; then, just as he is about to reach the head of the chisel with the hammer to strike the blow, he straightens his wrist, thereby adding impetus to the hammer, which is already rapidly falling, and in this manner a very great velocity is given; probably at the exact moment of impact the actual velocity may be 50 ft. per second.

In the second example, where a blow is delivered upon the head of a steel key by a jack hammer, and the hammer is driven in a horizontal line, the fitter will swing the hammer through a comparatively long distance, and will probably put the weight of the upper half of his body into the blow, thereby considerably increasing the velocity of the hammer, which may be 50 ft. per second, as before. In both these cases of hand hammers the accumulated work or energy stored up in the hammer will be the same as though the hammer had fallen from a sufficient height to attain that velocity which the hammer has at the moment of impact. But here the only information we have to assist us in solving the problem is the weight of the hammer and the assumed velocity at which it is moving, say 50 ft. per second. Since having no particulars as to the height from which a body must fall to attain this velocity, it is necessary to introduce the law of gravitation into the calculation to enable reliable results to be obtained.

Here we have for the 2 lb. hammer accumulated work

 = W v 2 = 2 x 50 x 50 = 78 foot-pounds. 2 g 2 x 32

If the face of the hammer moves the head of the chisel 1/16 in., then

1/16 x 1/12 =1/192 of 1 ft;

 R = 78 x 192 = 14976 1b. 1

If a nail had been driven 1/4 in.,

1/4 x 1/12 = 1/48 of 1 ft.;

 R = 78 x 48 = 3744 lb. 1

In the case of the 4 lb. jack hammer we have accumulated work

= W v2 / 2g

 = 4 x 50 x 50 = 156 foot-pounds. 2 x 32

If the key is driven 1/8 in. by the blow,

1/8 x 1/12 = 1/96 of 1 ft;

 R = 156 x 96 = 14976 lb. resistance. 1

The work done by the jack hammer, namely, 14,976 1b., is approximately the same that would be obtained by a dead load of 14,976 lb. giving a direct pressure.