M1 Pulleys/Tension Question: Where have I gone wrong?
The answer should be 12/7g, but I can't seem to find where I went wrong..
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Original post by creativebuzz
The answer should be 12/7g, but I can't seem to find where I went wrong..
it appears that the question is missing
Original post by TeeEm
it appears that the question is missing
Apologies. I am referring to part (a):
The 3kg mass is the heavier (going down), so for that mass, taking upwards as positive, the acceleration is negative.
Original post by creativebuzz
Apologies. I am referring to part (a):
I think the other 2 helpers have answered your question
Original post by creativebuzz
The answer should be 12/7g, but I can't seem to find where I went wrong..
If you define up as positive, then in this case the acceleration should be negative (and therefore the RHS).
Original post by creativebuzz
The answer should be 12/7g, but I can't seem to find where I went wrong..
Hey, in Mechanics it is very important that you state the direction which you are taking to be positive; if you are taking up to be positive, then your acceleration should be negative.
Hope this helps
Original post by ThatPerson
If you define up as positive, then in this case the acceleration should be negative (and therefore the RHS).
Ah thank you! Would you mind explain why the acceleration would be negative/decelerating?
Original post by HenryHiddler
Hey, in Mechanics it is very important that you state the direction which you are taking to be positive; if you are taking up to be positive, then your acceleration should be negative.
Hope this helps
Hope this helps
Ah thanks! Would you mind explaining why the acceleration would be negative?
Original post by creativebuzz
Ah thanks! Would you mind explaining why the acceleration would be negative?
Well g, acceleration due to the Earth's gravitational field, acts towards the centre of the Earth. And also, acceleration is a vector, as it is derived from velocity, not speed - thus it has a direction. Therefore, taking down to be positive, g is positive - but taking up to be positive means that g is negative.
Hope this helps
Original post by creativebuzz
Ah thank you! Would you mind explain why the acceleration would be negative/decelerating?
As m<3 the mass on the LHS of the pulley is larger, so when the system is released, the LHS moves downwards and the RHS will move upwards.
Original post by ThatPerson
As m<3 the mass on the LHS of the pulley is larger, so when the system is released, the LHS moves downwards and the RHS will move upwards.
Ah that makes sense, thank you!
How would you do part b of that question? I tried resolving upwards to give me T - 3g = m(3/7g) but I wasn't sure if the tension in both strings were equal because if they were then that would give me -29.4 when I should get 1.2
Original post by creativebuzz
Ah that makes sense, thank you!
How would you do part b of that question? I tried resolving upwards to give me T - 3g = m(3/7g) but I wasn't sure if the tension in both strings were equal because if they were then that would give me -29.4 when I should get 1.2
How would you do part b of that question? I tried resolving upwards to give me T - 3g = m(3/7g) but I wasn't sure if the tension in both strings were equal because if they were then that would give me -29.4 when I should get 1.2
As the string is inextensible the tension in both sides is equal. Form another equation for the RHS of the pulley and then solve simultaneously for m (or just use your value for T and rearrange for m)
(edited 9 years ago)
Original post by creativebuzz
Ah that makes sense, thank you!
How would you do part b of that question? I tried resolving upwards to give me T - 3g = m(3/7g) but I wasn't sure if the tension in both strings were equal because if they were then that would give me -29.4 when I should get 1.2
How would you do part b of that question? I tried resolving upwards to give me T - 3g = m(3/7g) but I wasn't sure if the tension in both strings were equal because if they were then that would give me -29.4 when I should get 1.2
Check your arithmetic.
The tension is equal in both strings and you should have T - mg = ma as your equation of motion. If you plug in the value of a given and the result you found for T then you should get m = 12/10 = 1.2 as you say.
Original post by davros
Check your arithmetic.
The tension is equal in both strings and you should have T - mg = ma as your equation of motion. If you plug in the value of a given and the result you found for T then you should get m = 12/10 = 1.2 as you say.
The tension is equal in both strings and you should have T - mg = ma as your equation of motion. If you plug in the value of a given and the result you found for T then you should get m = 12/10 = 1.2 as you say.
Ah I see where I went wrong. Thank you.
Would you mind seeing where I went wrong in these two questions based on C4 integration:
Original post by ThatPerson
As the string is inextensible the tension in both sides is equal.
It's the fact that the pulley is smooth that implies the tension is the same on either side.Original post by ghostwalker It's the fact that the pulley is smooth that implies the tension is the same on either side.
It's the fact that the pulley is smooth that implies the tension is the same on either side.Ah, oops. (Was thinking of acceleration for some reason).
Original post by ThatPerson
As m<3 the mass on the LHS of the pulley is larger, so when the system is released, the LHS moves downwards and the RHS will move upwards.
What difference does it make to accelerate if I decide to make:
(a) upwards direction positive
(d) downwards direction positive
Original post by creativebuzz
What difference does it make to accelerate if I decide to make:
(a) upwards direction positive
(d) downwards direction positive
(a) upwards direction positive
(d) downwards direction positive
It shouldn't make a difference as long as you are applying your chosen convention consistently in the equations you form. (And there will be a sign difference in your accelerations, but they'll be referring to the same direction).
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