350 PHILOSOPHICAL TRANSACTIONS. [aNNO 1729- 



wards at B, while the same force of 8 stone draws it directly down at d towards 

 F. But as CD is only equal to half of cb, the force at d compared with that 

 at B, loses half its action, and therefore can only take off the force of 4 stone 

 from the push upwards at b ; and consequently the weight w at a will prepon- 

 derate, unless an additional weight of 4 stone be added at b. Therefore a 

 man, &c. pushing upwards under the beam between b and d becomes lighter. 

 Which was also to be demonstrated. 



Scholium 1. — Hence, knowing the absolute force of the man that pushes 

 upwards, that is, the whole oblique force, the place of the point of trusion d, 

 and the angle made by the direction of the force with a perpendicular to the 

 beam at the same point, we may have a general rule to know what force is 

 added to the end of the beam b, in any inclination of the direction of the force 

 or place of the point d. 



First, find the perpendicular force by the following analogy : 



As the radius : to the right sine of the angle of inclination : : so is the 

 oblique force: to the perpendicular force. 



Then the perpendicular force multiplied into the length of the brachium bc, 

 minus the said force multiplied into the distance dc, will give the value of the 

 additional force at b, or of the weight required to restore the equilibrium 

 at A. 



Or, to express it in the algebraical way. Let of express the oblique force, 

 pf the perpendicular force, and x the force required, or value of the additional 

 weight at a to restore the equilibrium. Then de : df :: of : pf, hence pf 

 X BC — Jb/" X DC = T. 



The same rule will serve for the second case, if the quantity found be made 

 negative, and the additional weight suspended at b. Or having found the 

 value of the perpendicular force, the equation will stand thus, — /j/ X bc -f- 

 pf X DC = — .r, and consequently the additional weight must be added at b ; 

 because — ar at a is the same as + x at b. 



Scholium 1. — Hence it follows also, that if, in the first case, the point of 

 trusion be taken at c, the force at b, required, will be the whole perpendi- 

 cular force; because cd is equal to nothing : and if the point d be taken be- 

 yond c towards a; the perpendicular force pushing upwards, at that point, 

 multiplied into dc, must be added to the same force multiplied into bc, that is 

 /j/ X bc -h // X DC = X. 



The machine Dr. D. made use of to prove this experimentally, was as fol- 

 lows, represented in fig. 6. The brass balance ab is 12 inches long, moveable 

 on the centre c, with a perpendicular piece fib hanging at the end b, and 

 moveable about a pin at b, and stopped at its lower end b, by the upright plate 



