ILLUSTRATIVE EXAMPLES 



71 



(As regards the f rictional force acting between the wheel and the ground, we 

 notice that although motion is about to take place, this motion is not one of 

 slipping between the wheel and the ground, so that the ratio of F to B for any 

 wheel is not the coefficient of friction between the wheel and the ground.) 



The forces just enumerated hold the car in equilibrium. Thus the sum of 

 their components in any direction must vanish and the sum of their moments 

 about any line must vanish. Resolving horizontally and vertically, we obtain 



P = FI + F 2 + F 3 + *V (a) 



W = B! + U 2 -f R s + 4 - (b) 



There is nothing to be gained by taking moments about any line ; as we do not 

 know the line of action of P, we cannot 

 know its moment. 



Next let us consider the equilibrium of 

 a single wheel. The wheel touches the 

 ground and also touches the axle at some 

 point C. (We may think of the axle as a 

 circle of radius very slightly less than 

 that of the inside of the hub of the 

 wheel.) The forces acting on the wheel 

 are accordingly 



(a) its reaction with the ground ; 



(6) its reaction with the axle ; 



(c) its weight, which we shall neglect 

 as being insignificant in comparison with 

 that of the car. 



Neglecting the third force, the two former forces must be equal and opposite. 

 The line of action of each is accordingly the line joining B, the point of contact 

 with the ground, to C the point of contact with the axle. Since slipping is just 

 about to take place at (7, the reaction at C will make an angle e, equal to the 

 angle of friction, with AC the normal at C. Thus the angle BCA is equal to e. 



In the triangle A CB we have AC = 6, AB = a, and the angle ACS = e. 



a b 



Thus 



sin c sin ABC 



Since, however, a force along BC has components 

 pendicular to AB, we have 



tan ABC = 



I and FI along and per- 



Thus 



sin ABC 



Vl_ sin2 ABC 



b sin e 



Va 2 - 



