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1. Two particles A and B, of masses 0.4kg and 0.3kg respectively, are connected by a light inextensible string.
The particle A is placed near the bottom of a smooth plane inclined at 30 degrees to the horizontal.
The string passes over a small smooth light pulley which is fixed at the top of the inclined plane and B hangs freely.
The system is released from rest, with each portion of the string taut and in the same vertical plane as a lane of greatest slope of the inclined plane.

Calculate:

a) the common acceleration, in ms^-2, of the 2 particles

b) the tension, in N, in the string

Given that A has not reached the pulley, find:

c) the time taken for B to fall 6.3m from rest

d) the speed that B then acquireid
2. a) 0.3g - 0.4g sin30 = 0.7 a => a = g/7 m/s²

b) 0.3g - T = 0.3g / 7 => T = 2.52 N

c) 6.3 = 1/2 (g/7) t² => t = 3 s

d) v = 3 * g/7 = 4.2 m/s

edit: oops, forgot about the g!
3. (Original post by elpaw)
a) 0.3g - 0.4g sin30 = 0.7 a => a = 1/7 m/s²
...
Err, shouldn't that be,

a) 0.3g - 0.4g*sin30 = 0.7*a
0.3g - 0.2g = 0.7*a
0.7*a = 0.1g
a=g/7
a=1.4 m/s
======

For the rest I get,

b)
0.3g - T = 0.3*a
T = 0.3*9.81 - 0.3*1.4
T = 2.52 N
======

c)
s=ut + ½at²
6.3 = 0 + ½*1.4*t²
6.3 = 0.7*t²
t² = 6.3/0.7 = 9
t = 3 s
====

d)
v² = 2as
v² = 2*1.4*6.3
v² = 17.64
v = 4.2 m/s
=======
4. yeah, a = 1.4m/s^2
5. yeah, sorry, i dropped a g in the acceleration (e.c.f.)
6. (Original post by Fermat)
Err, shouldn't that be,

a) 0.3g - 0.4g*sin30 = 0.7*a
0.3g - 0.2g = 0.7*a
0.7*a = 0.1g
a=g/7
a=1.4 m/s
OK Thanks. But shouldn't that 'sin' be 'cos'? And if not, why?

Also, I would've written:

Newton II of A: T - 0.4g cos 30 = 0.4a

Newton II of B: 2.94 - T = 0.3a

Is that wrong? If it's right, where would I go from there please?

Thanks
7. Looking at the diagram, the weight of A is resolved into "horizontal" and perpindicular forces, Where "horizontal" means parallel to the slope.
The "horizontal component of A is 0.4gsinø.

Doing NewtonII at A and B.

T - 0.4gsinø = 0.4.a (at A)
0.3g - T = 0.3.a (at B)

0.3g - 0.4gsinø = 0.7.a

etc.

So, the first of the two eqns you wrote should have sin30° instead of cos30°.
0.4gcos30 is the normal reaction of the plane against the mass A. (See diagram)
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