Gent2324
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#1
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#1
https://imgur.com/a/lJE8OI9

force for P where left is positive: T - (mu)R = 3(0.5)
T - (mu)3g = 3(0.5)

force for Q where downwards is positive: mu x 3g is there because of the friction from the table
2g - T - (mu)(3g) = 2x0.5

equation 1 - equation 2:
2T - 2g = 0.5
I know this is wrong because when i solve again to find mu its not 0.54, where am i going wrong? thanks
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RDKGames
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#2
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(Original post by Gent2324)
https://imgur.com/a/lJE8OI9

force for P where left is positive: T - (mu)R = 3(0.5)
T - (mu)3g = 3(0.5)

force for Q where downwards is positive: mu x 3g is there because of the friction from the table
2g - T - (mu)(3g) = 2x0.5

equation 1 - equation 2:
2T - 2g = 0.5
I know this is wrong because when i solve again to find mu its not 0.54, where am i going wrong? thanks
Q isn't on the table so I am not sure why you're including a frictional force on it coming from the table.
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Gent2324
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(Original post by RDKGames)
Q isn't on the table so I am not sure why you're including a frictional force on it coming from the table.
but isnt the frictional force of the other particle pulling particle Q upwards? adding to the tension?
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Gent2324
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#4
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i also tried doing 2g-t = 1 and subbing that value of T into the other equation gives the wrong value (using mu as 0.54)
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RDKGames
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(Original post by Gent2324)
but isnt the frictional force of the other particle pulling particle Q upwards? adding to the tension?
No. The frictional force only acts on P, and hence it affects what T is. And so you don't need to include it because its effect is drilled into T already.
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RDKGames
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(Original post by Gent2324)
i also tried doing 2g-t = 1 and subbing that value of T into the other equation gives the wrong value (using mu as 0.54)
Their value for mu is a misprint. You should work it out to be \mu \approx 0.58.
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Gent2324
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(Original post by RDKGames)
Their value for mu is a misprint. You should work it out to be \mu \approx 0.58.
i got 0.58 originally then i thought i must be doing something wrong...

(Original post by RDKGames)
No. The frictional force only acts on P, and hence it affects what T is. And so you don't need to include it because its effect is drilled into T already.
since thats the case, with my 2 equations are there actually 2 different T's? where T2 = T1 + (Mu)R?
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RDKGames
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(Original post by Gent2324)
i got 0.58 originally then i thought i must be doing something wrong...

since thats the case, with my 2 equations are there actually 2 different T's? where T2 = T1 + (Mu)R?
No.. there is only one tension.
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Gent2324
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#9
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(Original post by RDKGames)
No.. there is only one tension.
im confused because the frictional force on P impacts the motion of Q but its not included in the equation.
when P is going across the force bringing it to the left is the tension subtract the frictional force. so how can the tension include the frictional force from P when talking about the motion on Q when in the equation for P it is not included since you have to subtract the frictional force?
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RDKGames
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(Original post by Gent2324)
im confused because the frictional force on P impacts the motion of Q but its not included in the equation.
when P is going across the force bringing it to the left is the tension subtract the frictional force. so how can the tension include the frictional force from P when talking about the motion on Q when in the equation for P it is not included since you have to subtract the frictional force?
Perhaps you misunderstood what I meant.

Isolate the particle P for now. We have a tension force T and an opposing force of friction. This tension must be some particular value. If we make the surface smooth, so there is no friction, then the tension must decrease if the particle is to accelerate at 0.5 m/s^2 still.

So clearly, the tension adapts in response to us varying friction if we impose the condition of acceleration at 0.5 m/s^2.

Hence, if we consider the particle Q now, it's upwards force is already changing we as we change the friction. This is the effect I am talking about.


But at the end of the day, you literally just need to use the fact that tension in the string is the same throughout, and on particle Q there are ONLY two forces; tension and weight. Don't even try to justify a term for friction, unless the air is made out rubber or something!
Last edited by RDKGames; 1 year ago
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Gent2324
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#11
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(Original post by RDKGames)
Perhaps you misunderstood what I meant.

Isolate the particle P for now. We have a tension force T and an opposing force of friction. This tension must be some particular value. If we make the surface smooth, so there is no friction, then the tension must decrease if the particle is to accelerate at 0.5 m/s^2 still.

So clearly, the tension adapts in response to us varying friction if we impose the condition of acceleration at 0.5 m/s^2.

Hence, if we consider the particle Q now, it's upwards force is already changing we as we change the friction. This is the effect I am talking about.


But at the end of the day, you literally just need to use the fact that tension in the string is the same throughout, and on particle Q there are ONLY two forces; tension and weight. Don't even try to justify a term for friction, unless the air is made out rubber or something!
ah i see, so the tension on Q already implies any friction or whatever is happening from the other end of the string
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