# mechanics pullies help

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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

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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#2

(Original post by

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

**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

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(Original post by

Q isn't on the table so I am not sure why you're including a frictional force on it coming from the table.

**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.

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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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#5

(Original post by

but isnt the frictional force of the other particle pulling particle Q upwards? adding to the tension?

**Gent2324**)but isnt the frictional force of the other particle pulling particle Q upwards? adding to the tension?

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#6

(Original post by

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)

**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)

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(Original post by

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.

**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.

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#8

(Original post by

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?

**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?

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(Original post by

No.. there is only one tension.

**RDKGames**)No.. there is only one tension.

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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#10

(Original post by

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?

**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?

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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(Original post by

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!

**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!

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