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VCE Stuff => VCE Science => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Physics => Topic started by: /0 on May 30, 2008, 09:54:02 pm

Title: nice pulley question
Post by: /0 on May 30, 2008, 09:54:02 pm: In the apparatus shown below, two masses (20 kg and 30 kg) are hung over a frictionless pulley. The ropes are weightless. Assume the tensions in each rope are equal. Take $g=10ms^{-2}$ .
Calculate:
a) The acceleration of the masses. Give size and direction.
b) The tension in rope 1.
c) The tension in rope AB.
d) Explain why the tension in rope AB is less than the combined weight of the 20 and 30kg masses.
Title: Re: nice pulley question
Post by: enwiabe on May 30, 2008, 10:24:56 pm: Are we to assume that the pulley is weightless as well? If not, we'll need a radius of gyration and a moment of inertia to work out that question
Title: Re: nice pulley question
Post by: /0 on May 30, 2008, 10:36:42 pm: Nah, dw the pulley is weightless. The only things with mass are the blocks.
Title: Re: nice pulley question
Post by: cara.mel on May 30, 2008, 11:03:43 pm: a) System will slide to left:
sum of forces = 30g - 20g = 10g = 100N
F = ma 100 = 50a a=2m/s

b) Looking at the block on the left
sum of forces = ma
T - 30g = 30*-2
T = 300-60 = 240N

c) To be honest I would have said 500N here :( Otherwise I'd say 240N
Title: Re: nice pulley question
Post by: bigtick on May 30, 2008, 11:47:18 pm: 480N
Title: Re: nice pulley question
Post by: /0 on May 31, 2008, 12:14:10 am: Thanks, for part c) the answers say 480N. I think it's got something to do with the centre of mass of the system, but it would be helpful if I knew the acceleration of the centre of mass.
Title: Re: nice pulley question
Post by: bigtick on May 31, 2008, 10:27:04 am: Consider the system pulley+2 masses.
30 kg accel downwards + 20 kg accel upwards = 10 kg accel downwards
Apply net force = ma to the system
Force of gravity + tension in ropeAB = 10x2
30x10+20x10+0x10-T=10x2
T=480
Title: Re: nice pulley question
Post by: Mao on May 31, 2008, 10:59:23 am: using the result from part b:

$T_1=T_2=240$

given that $F_{net} = 0$

$\implies T_1 + T_2 = T_{AB}$

$\implies 240 + 240 = T_{AB}$

$\implies T_{AB} = 480N$
Title: Re: nice pulley question
Post by: dcc on May 31, 2008, 11:23:40 am: This question tastes like Specialist Mathematics.