2 3 8

M E c

HANICS,

9. If the Height H of the Obstacle be proportionalto the Radius of the Wheel, that is, if H be as R, andthe Force draw in a Direction parallel to OK ; then

. r v/aRH—H 1 /2RR-RR v^R - *

because - - ~ —

R R “ R

— 1, therefore F — W, or the Force will be proportionalto the Weight of the Wheel.

10. If the Direction of the Force be parallel to thehorizontal Flane, that is, if C m be parallel to N D,then because the Angle m C E is (in that Cafe) equalto the Angle C E H, their Sines will be equal, that is,I7)I = CH = R — H ; therefore the Expression of the

Force (Art. 6.) will

and if the

Height

become F- WX/egH-H'R — H 1

H be given it will be F —

W X •/ 2 R — 1R^i

11. From the Expression F ^ ^ ~

we have this Equation— — ^—_ 2 _ , which

gives the following Analogy F :W :: \/1 R H — H*: S.That is, The Force is to the Weight as the Sine of theAngle ECH (viz. E H) is to the Sine of the Anglem C E, which the Line of Direction makes with the Line

EC.

12. If the Obstacle is capable of beinborne down by the Wheel

greases will be the Force to do this ; for since C E re-presents the whole Force with which the Wheel bearsupon the Obstacle, and this is resolvable into the twoParts C H and H E, of which the former C H beingparallel to E F tends to press it down, it will be ex-prefs’d by R — H, and since H is given, the depressingForce will be as R—1, and therefore will increasewith R, or the Radius of the Wheel.

13. If the Obstacle be such, as that it can neitherbe surmounted nor depressed, but must be driven for-ward, the Force to do that will be expressed by H E =

V' 2 R H

6 depress’d orthe larger the Wheel the