52 



FORCES ACTING ON A SINGLE PARTICLE 



3. A particle is tied to an elastic string, the other end of which is fixed at a 

 point in a rough inclined plane. Find the region of the plane within which the 

 particle can rest. 



The forces acting on the particle are 



(a) its weight, say TF, vertically down ; 



(6) the tension of the string ; 



(c) the reaction with the rough plane. 



Let the natural length of the string be Z, the modulus of elasticity X ; then 



when the actual length of the 

 string is r, where r is greater 



than Z, the tension is 



Let a be the inclination of 

 the plane, and let /* be the 

 coefficient of friction between 

 the plane and the particle. Let 

 the reaction with the plane be 

 resolved into a normal component 

 R, and a component F along the 

 plane. The condition that the 

 particle can remain at rest is 

 that F shall be less than pR. 

 Resolving at right angles to the plane, the only forces which have compo- 

 nents in this direction are found to be the weight of the particle and its reaction 



with the plane. Thus 



R W cos a = 0. 



Consider the equilibrium of the particle when at some point P, distant r ( > I) 

 from 0. The components in the inclined plane, of the forces acting on it, are 

 (a) W sin a down the line of greatest slope through P ; 



(6) the tension along PO ; 



(c) the f rictional component of the reaction, which we have called F. 



Let OP make an angle 6 with the line of greatest slope in the plane. Then, 

 since the resultant of the first two forces must be of magnitude F, we must have 



2(r- 



FIG. 26 



W sin a cos 0, 



giving the magnitude of the f rictional force required to maintain equilibrium. 

 If the particle is on the point of motion, F = /J.R = /tTFcos a, so that 





= 0. 



(a) 



Since r, 6 are polar coordinates of the point P, equation (a) is the polar equa- 

 tion of the boundary of the region within which the particle can remain at rest. 



