106 



THE INCLINED PLANE. 



B', 



\ 



\ 

 C 



FIG. 55- 



LM represent a plane inclined to the horizontal line LN. Let A 

 represent a ball weighing 20 Kg. The 

 problem is to find what force acting in the 

 direction LM will hold it in equilibrium. 

 The weight of the body A is a downward 

 force of 20 Kg., which may be graphically 

 represented ( 81) by the vertical line AC, 

 20 mm. in length. Any other convenient 

 unit of length might be used, but the 

 scale of 1 mm. to the Kg. being adopted, 

 it must be maintained throughout the 

 problem. The force represented by AC 

 is resolved into two components repre- 



sented by AD, perpendicular to LM, and by AB, parallel to it. The 

 former component measures the pressure to be resisted by the plane ; 

 the latter component measures the force with which the ball is 

 drawn towards L. This second component is to be balanced by the 

 equal and opposite force AB', the equilibrant of AB, It may be 

 proved geometrically that * 



AB i AC : : MN : ML. (Olney's Geometry, Art. 341.) 

 Careful construction and measurement will give the same result. 

 But AB, or rather its equal AB' , represents the power ; AC repre- 

 sents the weight ; MN represents the height ; and ML, the length 

 of. the plane. Therefore, 



P : W : : h : I, or, P = the y part of W. 



I/ 



2O2. Law for the First Case. In the figure 

 above, ML is twice the length of MN, and AC is twice the 

 length of AB or AB'. This indi- 

 cates that a force of 10 Kg. acting in 

 the direction LM would hold the 

 ball in equilibrium. This result may 

 be easily verified by experiment. 

 We may therefore establish the fol- 

 lowing law : 'JV/ien a given power 

 acts parallel to tfie plane, it will 

 support a weight as many times as great as itself as 

 the length of the plane is times as great as Us verti- 

 cal height. 



20 Kg. 



10 Kg. 



p IG 56> 



