M1 doubt .. URGENT..

[crazed2009]

hey can smone plz help me with this...

Figure 3 ( attachment) shows a particle X of mass 3 kg on a smooth plane inclined at an angle 30° to the
horizontal, and a particle Y of mass 2 kg on a smooth plane inclined at an angle 60° to the
horizontal. The two particles are connected by a light, inextensible string of length 2.5 metre
passing over a smooth pulley at C which is the highest point of the two planes.
Initially, Y is at a point just below C touching the pulley with the string taut. When the
particles are released from rest they travel along the lines of greatest slope, AC in the case of
X and BC in the case of Y, of their respective planes. A and B are the points where the planes
meet the horizontal ground and AB = 4 metres.

initial acceleration of the system is given by : g/10(2x3^0.5 -3)

    By finding the tension in the string, or otherwise, find the magnitude of the force
exerted on the pulley and the angle that this force makes with the vertical.


thanks,,

[Amr Fouad]

7 minutes

[Amr Fouad]

please post original question..i cudnt figure out whether the multiplied part of acceleration is with the num or denom

[crazed2009]

Q. 7  b)

[Amr Fouad]

Thanks..give me some time..

[Amr Fouad]

it gets really simple now..

first, you have got to find the tension:
T=3gsin30+3a
or
T=2gsin60-2a

then there's a genral rule to find normal force on pulley:

R=T x square root{2(1+cos theta)}

where theta equals the angle between the two ropes, int his case 90 degrees..

Try it..

[vanibharutham]

You have to consider it as pythagoras

or just learn this formula:

THE REACTION ON A PULLEY BY STRING = 2 x TENSION cos ( 90 - theta / 2 ), where theta is angle of slope

[Amr Fouad]

You have to consider it as pythagoras

or just learn this formula:

THE REACTION ON A PULLEY BY STRING = 2 x TENSION cos ( 90 - theta / 2 ), where theta is angle of slope

mine is easier ..and works all the time

[crazed2009]

Thanks a LOT guys... its weird we never took these formulas, nt tht i remeber at least..
but wht about the rest of the question.. the angle btn the force and the vertical???