I42 CENTRIFUGAL FORCE CH. IX 



The fractions |, \, \, show the proportions of the 



overturning- force to the holding down force. 



In a coach weighing 2300 pounds, 



loaded with 1 700 pounds on top, the 



total weight will be 4000 pounds, 



which acting at the end of the arm 



AB will give a holding down force 



r of 10,000 pounds. The centrifugal 



Fig. 71. r & 



force will act at the centre of gravity, 

 5.5 feet above the ground, and at different speeds 

 the forces will be as follows : — 



FOR THE LOADED COACH. 



At 7 miles an hour, 263.6 X 5-5 — '45° pounds. * 

 10 ,, „ 534.0 X 5-5 = 2 937 „ ^ 



15 ,, ,, 1202.0 X 5.5 = 661 1 



In the case of an empty coach, the centrifugal 

 force, acting at a height of four feet, required to 

 balance the weight of the coach, 2300 pounds, with 

 a leverage of 2.5 feet, will be 1437 (2300 x 2.5 = 

 5750, and 1437 x 4 = 575°)- This is the force 

 due to a speed of 21.6 miles an hour, on a 

 curve of 50 feet radius, at which speed the wheels 

 on the inside of the curve would be lifted from the 

 ground and the coach overturned. 



_.. -.2 



= 1437 or v l — 1006.6 



32.2 X 5° 

 v— 31.73 feet per second or 21.6 miles an hour. 



