1835.] 



Prop on an inclined plane. 



337 



For, by meclianics, the power exerted is to the weight 

 sustained, as the perpendicular distances upon the di- 

 rections in which the forces act, inversely. 

 : Hence P:W::BC:AC orPx AC=Wx BC, and since 

 AC and W are constant, P will be the greatest whea 

 B C is the least, that is when perpendicular to the line 

 AB, the plane of descent. 



For if not, let CD be the position of the prop, andbe* 

 cause DEC is a right angle, BC is manifestly less than 

 DC, (being the hypothenuse) and the same may be 

 shown of any other direction whatever. 



The construction may be as follov/s, upon AC de- 

 scribe the semicircle ABC cutting AB in B, join BG 

 which is the position of the prop for the gallery on a de- 

 scent or an ascent. 



It is required to determine the position of a pro^ ED 

 to support a body ACB that shall be least liable to break. 

 (By the Rev. G. Barker.) 



Solution 2d. —Conceive the whole mass to be con- 

 centrated into the centre of gravity g, then since every 

 body tends towards the Earth's centre, the whole weight 

 is acting in direction g D. Let this be resolved into two 

 GH perpendicular to A B and D H 



parallel to A B 



