296 APPLIED MECHANICS 
15. A body weighing 322 lb. is lifted by a force F lb. which alters. When 
the body has risen through the distance x feet, the force in Ib. for the several 
values of x is as follows (or would be if the body rose as far) :— 
x 0 1 2 3 4 55 7 9 11 |12°5| 14 | 17 | 20 
F | 540 | 540 | 540 | 530 | 500 | 460 | 310 | 220 | 190 | 190 | 190 | 190 | 190 
’ 
Using squared paper, find the velocity in each position and the time taken 
by the body to get to each position counting from x=0, the velocity then being _ 
5 feet per second. |. 7 (By 
16. A tram-car, weighing 15 tons, suddenly has the electric current cut off. 
At that instant the speed of the car was !6 miles per hour. Reckoning time 
from that instant, the following velocities, V (miles per hour),and times, ¢ (seconds), 
were noted. V=16, f=0. V=14, ¢=9:3. V=12, ¢=21.) V=10) ¢=386, 
Calculate the average value of the retarding force, and find the average velocity 
from t=0 tot=35. Also find the distance travelled between these times. . — ~ 
If the law of resistance be F (lb.)=a+6V+cV%, where V is in miles per hour 
as before, indicate the method by which values of a, b, and ¢ could’ be found 
from the above observations. Also calculate the relation between V and T (the ~ 
time taken to come to rest from velocity V) for such tests. What is T when V 
is very large? [B.E.] 
17. During the up stroke of the piston of a Bull engine the effective pressure 
p of the steam on the piston, in lbs. per square inch, varies, as shown in the 
following table, where « is the distance of the piston from the bottom of its 
stroke in feet; the piston will however not rise so high as 10 feet. The weight 
2} 00/05 }1:0} 15] 2:0] 2:5|3-0| 3°5 | 4-0| 4:5] 5:0/ 6-0 7-01 8:0 | 9:0 10:0 
p| 55 | 54 mg he 41 | 34 | 28 | 24 21 | 18 | 16 |12°5 |10°5| 8°5 | 7 | 6 
: 
of the piston, piston-rod, and ‘‘ pitwork” amounts to 22 lbs. per square inch of 
piston, and the frictional resistances are equivalent to 2 lbs. per square inch 
of piston. Draw the velocity curve for the piston on a stroke base, and find the 
Jength of the stroke. Find the time taken to make one up stroke, and draw the 
velocity curve on a time base. 
258. Simple Harmonic Motion.—A (Fig. 459) is a point which is 
moving with a uniform velocity V along the circumference of the circle 
BACD. | a@ is another point which is 
moving backward and forward along the . 
diameter BOC of the same circle in such 
a manner that Aa is always perpendicular 
to BC; in other words, a is the projection 
of Aon BC. Under these circumstances, 
the point a has simple harmonic motion. 
Let A and A, be two positions, near 
to one another, of the point which is moving 
round the circle, and a and a, correspond- 
ing positions of the point which has simple 
harmonic motion. Let v be the velocity of a 
at a, and v, the velocity of a at a, along BOC. Let Oa=2, Oa, =«,, and 
angle BOA= 6. Resolving V, the velocity of A, parallel and perpendicular 
to BOG, it is evident that the component parallel to BOC is equal to », 
the velocity of a, and v=Vsin d= : Vr—a. Also v,= rd 7? — gn 
Fig. 459... 
