RESISTANCE OF MATERIALS 



73 



If the power acts parallel to the plane, during virtual dis- 

 placement r, 



then : 



PXr=FXr=QX r sin a. 

 whence : 



P-Qsin a (fig. 88). 



This assumes that P & F are 

 in the same plane. But P can 



Pio. 88. 



N 



have any direction making an angle 

 9 with the perpendicular, a direction 

 (fig. 89) such that the resultant 

 of P and Q will be perpendicular 

 to the inclined plane and maintain 

 the load there. As Q is already in 

 the plane of that resultant (in AOB) 

 the power must therefore be found 

 in the principal section AOB. As 

 to the value of P, it can easily be 

 obtained by taking GP 1 equal and 

 parallel to P in the same direction, 

 and by projecting P 1 on the direction 

 of the component F of the load. Then P 2 = P l sin 9 = P sin 9 

 (alternate internal angles) ( x ). And as F = Q sin a, the equi- 

 librium is expressed : 



Fin 8!. 



P sin 9 = Q sin a ; whence P = Q 



sin a 

 sin 9 



ViV 



For a given load the power increases as the sine of the angle 

 of inclination of the plane, or as the cosine of its own inclination 

 to that plane, because sin 9 = cos (3. We know that if p = 0, cos 

 P = 1, which will lead to the expression P = Q cos a. 



Consider now, the force of sliding friction, $, remembering that 

 the co-efficient of friction is/= 

 tang 9 ( 39), 9 being the angle 

 of friction, that angle being given 

 by the perpendicular N and the 

 resultant R of the pressure and 

 of the friction. In practice, 9 

 is obtained by inclining the plane 

 until the sliding surface SS 1 just 

 commences to slide. The value 

 of the angle of inclination a 

 will then be the angle of fric- 

 tion 9. Thus if a power P is 



( x ) Instead of sin $, cos P t GP 2 or the angle formed by the direction of the 

 power on that of the inclined plane could be taken. 



