€ 
270 APPLIED MECHANTCS 
inch at the highest speed, and 90 Ibs. per square inch at the lowest 
speed. ; 
r 236. Friction of an Axle.—AB (Fig. 413) is a body mounted on 
an axle whose axis is O and radius 7. The body AB is either fixed to 
‘the axle and the axle rotates ina bearing, 
or the axle is fixed and AB rotates on 
the axle. The resultant of, the forces 
which resist the rotation of AB is a force 
Q, acting at a perpendicular distance 
OB=6 from O. An effort P acts at a per- 
pendicular distance OA=a from O, and 
is just able to cause AB to rotate in the 
direction of the arrow e. The lines of 
action of P and Q meet at C, and make 
an angle @ with one another. 
R, the resultant of the pressure of the 
bearing on the axle when the axle rotates 
(lower part of figure), or the resultant of 
the pressure of the axle on the lever when 
the axle is fixed (upper part of figure), must 
be equal and opposite to the resultant of Fic. 413. 
P and Q, and therefore its line of action 
must pass through C, and must make with the normal OD to the 
sliding surfaces an angle equal to ¢; also, the line of action of R will 
evidently lie between O and A. 
Draw OE at right angles to the line of action of R. Then OF is 
evidently equal to r sin ¢. In most cases ¢ is so small an angle that 
sing may be taken equal to tan¢ or » without sensible error. Let 
OE=r sin ¢=s. If a circle be described with centre O and radius s, 
the line of action of R will obviously be tangential to this circle. Hence 
the construction for finding the line of action of R is to draw through 
C a tangent to the circle whose centre is O and radius s. This circle is 
called the friction circle. 
If CF be’made equal to Q, and FH be drawn parallel to CA to meet 
CE at H, the triangle CFH will be the triangle of forces for Q, P, and R, 
and FH will be equal to P and CH equal to R. 
If there were no friction the line of action of R would be CO, and 
then FK would be equal to P and CK equal to R. Hence the efficiency 
of the mechanism is equal to FK/FH. The moment of. the friction is 
R xs or CH x OE. 
Proceeding analytically, R= ,/P?+ Q?+2PQ cos 0. Taking moments 
about O, Pa=Qb+ Rs=Qd+s ,/P? + Q?+2PQ cos 0, a quadratic equa- 
tion which gives 
P= 
ae slab+ 8? cos O48 ,/a?+b?+ 2ab cos 0 —s? sin? 6}. 
The + sign in front of the surd is to be taken when P overcomes Q, and 
the — sign when Q overcomes P. 
When 6=0°, OA and OB are in the same straight line, and P and Q 
b+s 
Fa 
are on opposite sides of O, then P=Q 5 
asxs 
the upper sign to be taken 
VW 
