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M1 question..i need help please!!!

Two particles A and B, of mass mkg and 3 kg respectively, are connected by a light inextensible string. The particle A is held resting on a smooth fixed plane inclined at 30° to the horizontal. The string passes over a smooth pulley P fixed at the top of the plane. The portion AP of the string lies along a line of greatest slope of the plane and B hangs freely from the pulley, as shown in the figure m1.doc. The system is released from rest with B at a height of 0.25 m above horizontal ground. Immediately after release, B descends with an acceleration of 2/5 g. Given that A does not reach P, calculate

a) the tension in the string while B is descending,

b) the value of m.


i dont understand how to calculate part b. so far i have done :

For particle A going upwards : F=ma
but i dont know what to minus from T
T- ......= m x 2/5 g

for a) i got T=17.64N -->18 (2sf) which is correct

Reply 1

You need to minus the component of the weight of the particle acting down the slope.

Reply 2

I got the tension as 17.143N so then i didT - 10m = 10m(30/7)17.143= 10m(30/7) x 10m17.143= 52.86mm=0.32i mean thats probs wrong but i got there?