screw on its axis. From this is obtained the following:

RULE. The force multiplied by the circumference of the circle through which the force arm moves equals the weight multiplied by the lead of the screw.

From this rule is obtained the statement:

shown in Fig. 23, the differential screw. This is made with two screws of different pitches, or leads of threads, either both right or both left hand threads.

It will be seen from Fig. 23 that one turn of the large screw lifts the weight only the difference between the leads of the large and small screws; then by the General Law,

L

The screw can be compounded like the other elements of machines, as W

Note. The nut N, Fig. 22, is the part that must be used with a screw to make it effective.

Note.

Owing to the amount of friction in the differential screw its practical use is limited.

The wedge is a pair of inclined planes placed back to back. It is used in two ways; by being driven with a blow of the hammer, and by pressure which usually acts parallel to the base of the planes. The difficulty in calculating the effectiveness of the first kind is to determine the force of the hammer blow, otherwise, the statement and rule is the same for either kind.

The rule is thus the same as for the inclined plane where force acts parallel to the base of incline.

back end is 4 in., the length is 20 in. What weight can

be raised by its use?

Note.

MISCELLANEOUS PROBLEMS

In the following problems friction of moving parts will not be considered.

1. An iron ball weighing 398 lbs. rests on a surface which is inclined 16° 45′ to the horizontal. What force, acting at an angle of 14° 30′ to the incline, is required to hold the ball in position?

2. A weight of 3,500 lbs. is to be drawn up an incline 640 ft. long, 85 ft. above the horizontal. What force acting parallel to incline will be required to keep the load on incline?

3. A cylinder of cast iron 24 in. dia., 30 in. long is to be rolled up an incline of 18° 15′. What force, acting at 8° 15′ to the incline, will be required to hold the cylinder from rolling down?

4. What weight will be raised with a screw of in. lead when 100 lbs. of force is applied at the end of a lever 18 in. from the center of screw?

5. When the lead of a screw is in., R is 20 in. and a weight of 12,000 lbs. is to be raised, what is the force required? 6. In Fig. 22, a screw of in. lead is turned with a bar 11⁄2 in. long, with 1 lb. of force on end of R; what weight can be raised?

7. A sliding wedge, Fig. 24, is used to raise the knife on a shearing press that weighs 100 lbs., the wedge is to move 18 in. and is 3 in. thick at back end. What force will be required?

8. A truck loaded with an engine weighing 6 tons is to be drawn up an incline 12 ft. long and 5 ft. above the hori

zontal. What force will be required when the pull is parallel to the incline?

9. What weight can be drawn up an incline 10 ft. long and 4 ft. high with a pull of 300 lbs.?

10. Two men each pulling 125 lbs. can pull what weight up an incline 8 ft. long and 6 ft. high?

11. What force will be required on a single thread screw having 3 threads per inch, with a bar 18 in. long from center of screw to raise a block of granite that weighs 5 tons?

12. Two jack screws are to be used to raise a block 10 ft. long, weighing 10,000 lbs. One is a third more powerful than the other. Make sketch showing the position of screws to give the proportionate load on each.

13. How many jack screws within. lead and having R 16 in. long, will be required to raise a building weighing 50 tons, if the pull on each lever is 50 lbs.?

14. How many jacks will be required with screws of in. lead, and 12 in. levers, with 25 lbs. pull on each lever to raise the building of problem 13?

15. A wedge 8 in. long, 14 in. thick at end will require how many lbs. of a hammer blow to drive it into a log that has a resistance of 2,000 lbs. against splitting?

16. When the screws of a differential are 8 and 12 pitch single threads respectively, with

a pull of 5 lbs., what length of lever will be required to raise a weight of 5,000 lbs.?

25

P

36"

W

-121"

Fig.26.

17. What length of bar will be required to raise a building of 100 tons weight with 10 lbs. pull each on 100, in. lead jack screws? 18. A cylinder 25 in. dia. weighs 5,000 lbs., Fig. 26. What force at P will hold the cylinder on incline?

19. If the cylinder of problem 18 weighs 6,575 lbs., the lever is 36 in. from P to fulcrum, 12 in. from weight to incline, and incline is 10 ft. long, 4 ft. high, what force on the end of lever P will prevent cylinder from rolling back?

SCREW THREADS

-P

-60°

A

The formula for finding tap drill sizes is based on the depth from point to bottom of thread, as follows: Fig. 27 shows outline of the sharp, or V thread. The depth A of a thread of 1 inch pitch is found by trigonometry and is equal to .8660 inch. If then the thread is taken on both sides of a cylinder the double depth for 1 inch pitch=.8660×2=1.732.

Fig.27.

1.732

For any other pitch of thread the double depth=

N

where N is the number of the threads per inch of the

Fig. 28 shows the outline of the United States Standard

(U. S. S.) thread. Here oneeighth of the total depth of the sharp V is flattened on the points, and the same amount filled in at the bottom of the V, thus making the flattened parts of the U. S. S. thread each equal

Fig.28.

-8

to one-eigth of the pitch so that the double depth for a