Friction. 1 43 



friction, the machine must remain in equilibrium so long as the 

 direction of the resultant, supposed to be AD, does not make with 

 the surface NDM (that is, with the tangent at the point where 

 AD meets this surface,) an angle less than the angle of friction. 

 This will be evident by imagining the force in question decom- 

 posed into two others, one perpendicular to the tangent at D ; 

 and the other in the direction of this tangent. 



This being done, since AD is the direction of the resultant, 

 we shall have 48 



q : p : : sin GAD : sin DAC, 



: : sin (GAF -f- FAD) : sin (FAC FAD,} 



Now, (1.) If we let fall upon AD the perpendicular FE, in the 

 right-angled triangle FED, the angle FDE is the complement of 

 the angle EDO which AD makes at D with the surface NDM, and 

 is accordingly supposed tojknown ; so that if we call /the angle 

 of friction, FDE will be the complement of/, and if we call d the 

 distance FD, or the radius of the axle, we shall have 



, Tng.3C. 



the radius of the tables being equal to 1. (2.) As the directions 

 of p and q are supposed to be known, as well as the dimensions 

 of the machine, the angles GAF, FAC, are supposed to be known 

 as also the distance AF. Thus, in the right-angled triangle FAE 

 in which AF, FE, are known, it will be easy to calculate the 

 angle FAE-, calling this angle e, and the angles GAF, FAC, a, b, Trig. so. 

 respectively, we shall have 48. 



q : p : : sin (a -f e) : sin (6 e) ; 



p sin (a -4- e] , . . , , 



consequently q = ~ - rr ; this is the value of the power 

 sin ( o 6j 



in the case of friction. 



If the friction is nothing, the angle /of the friction is 90; 

 that is, the resultant must be perpendicular to the surface of the 

 axis, and consequently pass through the centre F. We have 

 accordingly, cos/ = 0, and hence FE = 0, and e = ; therefore 



p sin a 



