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M1 Tension problem

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The peg is smooth and fixed. The system is released from rest: find the acceleration of the particle.

I'm really having problems with this (and the whole of Newton's Third Law infact). Will the tension in string 2 act downwards on the 3kg particle? Will the tension in string 1 act upwards on the 1 kg block? I set up the following system of equations with T1 and T2 representing the tension in S1 and S2 respectively

Particle 1: (T₁+T₂)-24.5=2.5a
Particle 2: 29.4+T₂-T₁=3a
Particle 3: 9.8-(T₁+T₂)=a

But I just get a=-4.2 which is incorrect.

Help seriously needed.

Reply 1

Tulian

Can't see you're attatchment.

Anyone?

Bump - I need help with this.

Reply 5

ghostwalker

Original post by Julii92

Will the tension in string 2 act downwards on the 3kg particle?

Yes

Will the tension in string 1 act upwards on the 1 kg block?

No, as S1 is not connected to the 1 kg block.

I set up the following system of equations with T1 and T2 representing the tension in S1 and S2 respectively

Particle 1: (T₁+T₂)-24.5=2.5a
Particle 2: 29.4+T₂-T₁=3a
Particle 3: 9.8-(T₁+T₂)=a

But I just get a=-4.2 which is incorrect.

Help seriously needed.

T2 is not included in particle 1's equation as S2 is not connected to particle 1.

Similarly T1 is not included in particle 3's calculation as S1 is not connected to particle 3.

Note:
There is a quicker method for doing this treating all the masses together, but it's worth going through the exercise above so you understand your original method.

(edited 11 years ago)

Reply 6

Julii92

Original post by ghostwalker

Yes

No, as S1 is not connected to the 1 kg block.

T2 is not included in particle 1's equation as S2 is not connected to particle 1.

Similarly T1 is not included in particle 3's calculation as S1 is not connected to particle 3.

Note:
There is a quicker method for doing this treating all the masses together, but it's worth going through the exercise above so you understand your original method.

Thank you - finally I get a=2.67ms^-2. My textbook gives a=2.26ms^-2, but I hope that's just an error.

How does the quicker method work?
Thanks for your time.

Reply 7

ghostwalker

Original post by Julii92

Thank you - finally I get a=2.67ms^-2. My textbook gives a=2.26ms^-2, but I hope that's just an error.

'Fraid not. Book's right.

I suspect an arithmetic slip. If you made the modifications I mentioned earlier, then just add all the equations, it does work out at 2.26... m/s/s.

Quicker method:

Weight on one side = 4g N,
Weight on other side =2.5g N

So nett force = 4g-2.5g = 1.5g

Total mass = 6.5 kg.

Hence acceleration = Force/mass = 1.5g/6.5 =....