196 



MOTION UNDER CONSTANT FORCES 





If a string having two equal weights attached to its ends is placed 

 over a smooth vertical pulley, so that the weights hang freely, it is 

 obvious that there will be equilibrium. If the weights are unequal, 

 equilibrium cannot exist. In Atwood's machine the difference 

 between the weights is made small, so that the 

 motion is slow and is therefore easily measured. 

 Let m lt m 2 be the masses of the weights, 

 of which the former will be supposed to be the 

 greater. When set free, let us suppose that 

 the former descends with an acceleration /. 

 Regarding the string as inextensible, the second 

 mass must ascend with an acceleration /. 



The string will be treated as weightless, so 

 that the mass- of any element of it may be dis- 

 regarded. The second law of motion accordingly shows that the 

 resultant force acting on any element must vanish. Thus the forces 

 acting on the string must be in equilibrium (even although the 

 string is not at rest), and it follows, as in 54, that the tension 

 must be the same at all points, say T. 



. The forces acting on either mass consist of its weight acting 

 downwards and the tension of the string acting upwards. Thus 

 the resultant downward forces on the two masses are respectively 

 m^g T and m z g T. The equations of motion for the two masses 

 are accordingly m ^ g _ T = m j^ 



m *9 T = m 2 f. 



If we eliminate T, we obtain 



mi -m ^ 



/ ~ ' 



giving the acceleration. On eliminating /, we obtain as the value 



of the tension o m 



(50) 



ra 



Clearly, if m l is nearly equal to ra 2 , the acceleration will be small. For 

 instance, if the weights are 100 and 101 grammes, we find that 

 /= ^jg = feet per second per second. 



