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Frictional Force

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An object of mass 5kg is on a horizontal surface with a coefficient of kinetic friction .45. The object is connected by a string of negligible mass through a frictionless pulley to an object with a mass of 3kg (hanging vertically off the edge of a table, sort of situation..sorry, this is all paraphrased). Find the acceleration of the two objects.

kinetic friction on object 1:

f = N * u = 9.81m/s^2 * 5kg * .45 = 22.073N


Fgrav on object 2:

F = a*m = g*m = 9.81m/s^2 *3kg = 29.43N


Fgrav gets transferred through tension to object one, and the net force in the x direction is:

Fnet = Fgrav -f = 29.43 - 22.073 = 7.36N

----------------------------------------------------------


I can't figure out how to get this into acceleration though..little help?
 
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  • #2
G01
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An object of mass 5kg is on a horizontal surface with a coefficient of kinetic friction .45. The object is connected by a string of negligible mass through a frictionless pulley to an object with a mass of 3kg (hanging vertically off the edge of a table, sort of situation..sorry, this is all paraphrased). Find the acceleration of the two objects.

kinetic friction on object 1:

f = N * u = 9.81m/s^2 * 5kg * .45 = 22.073N


Fgrav on object 2:

F = a*m = g*m = 9.81m/s^2 *3kg = 29.43N


Fgrav gets transferred through tension to object one, and the net force in the x direction is:

Fnet = Fgrav -f = 29.43 - 22.073 = 7.36N

----------------------------------------------------------


I can't figure out how to get this into acceleration though..little help?

Use Newton's second law:

[tex]F_{net}=Ma=F_g-f[/tex]
 
  • #3
tiny-tim
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hi anonymity! :smile:

(try using the X2 and X2 iocns just above the Reply box :wink:)
F = a*m = g*m
nooo … the acceleration of mass 2 is less than g, it's the same as the acceleration of mass 1 (in a different direction, of course)

you need to do full Ftotal = ma on each of mass 1 and on mass 2 (separately), and then eliminate T …

what do you get? :smile:
 
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Use Newton's second law:

[tex]F_{net}=Ma=F_g-f[/tex]
Hey, thanks for the response.

That is what I did (this was for an online problem set), but it was asking for one answer for both of them..I squared the two, added them, and took the square root because I didnt know what to do =x (it was multiple choice and I saw an answer very close to that which I found, so I gambled it)

It seemed wrong, and it was! lol

I had a = 7.36 / 5 and a = 7.36 / 3

Tiny: is that what you meant?

edit:

i'll clarify how i got that in case there is confusion.

Fnet, m1, x = m2g - fk = T => T = 29.43 - 22.073 = 7.36

Ma = T
a = T / M

Is that wrong?

I don't know what you mean by full, that is every force acting on it in the x direction, and vice versa (for m2 in the y direction). They each only motion in one dimension. Plus, the tension is the same across the whole length of the rope, so solving for it once (as above) would be enough..solving for it again through the m2 in the y direction would just yield the same result, wouldnt it?
 
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tiny-tim
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hi anonymity! :wink:
I had a = 7.36 / 5 and a = 7.36 / 3

Tiny: Is that what you meant?
(you were just guessing, weren't you?)

no, i meant call the tension T, and do Ftotal = ma twice, once for each block (including T in each Ftotal), and then eliminate T

(G01's :smile: method avoids using T, but i'm not sure the professor would let you do that … you don't seem to be familiar with it :redface:)
 
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hi anonymity! :wink:


(you were just guessing, weren't you?)

no, i meant call the tension T, and do Ftotal = ma twice, once for each block (including T in each Ftotal), and then eliminate T

(G01's :smile: method avoids using T, but i'm not sure the professor would let you do that … you don't seem to be familiar with it :redface:)
deleted..

can you just tell me what i'm doing wrong =|

It seems to me that I did exactly what G01 recommended, and that regardless of the method, you should find the same acceleration. You are saying the accelerations are the same (which makes sense), but I dont see how you could get a single acceleration from what I did and what G01 suggested.
 
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  • #7
tiny-tim
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hmm … you edited after my last reply …
edit:

i'll clarify how i got that in case there is confusion.

Fnet, m1, x = m2g - fk = T => T = 29.43 - 22.073 = 7.36

Ma = T
a = T / M

Is that wrong?
Fnet, m1, x = m2g - fk = T => T = 29.43 - 22.073 = 7.36

is this supposed to be F = ma on the upper block?

then there's no "a" in it, and you have 3 forces even though there are only two forcesw on the upper block :confused:

or were you trying to do F = ma on both blocks combined?
I don't know what you mean by full, that is every force acting on it in the x direction, and vice versa (for m2 in the y direction). They each only motion in one dimension. Plus, the tension is the same across the whole length of the rope, so solving for it once (as above) would be enough..solving for it again through the m2 in the y direction would just yield the same result, wouldnt it?
no, that would give you 2 equations for 2 unknowns …*eliminate one unknown (T) and you find the other (a)
 
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G01
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Hey, thanks for the response.

That is what I did (this was for an online problem set), but it was asking for one answer for both of them..I squared the two, added them, and took the square root because I didnt know what to do =x (it was multiple choice and I saw an answer very close to that which I found, so I gambled it)

It seemed wrong, and it was! lol

I had a = 7.36 / 5 and a = 7.36 / 3

Tiny: is that what you meant?

edit:

i'll clarify how i got that in case there is confusion.

Fnet, m1, x = m2g - fk = T => T = 29.43 - 22.073 = 7.36

Ma = T
a = T / M

Is that wrong?

I don't know what you mean by full, that is every force acting on it in the x direction, and vice versa (for m2 in the y direction). They each only motion in one dimension. Plus, the tension is the same across the whole length of the rope, so solving for it once (as above) would be enough..solving for it again through the m2 in the y direction would just yield the same result, wouldnt it?
You are dividing by the wrong mass. Using my approach, the mass multiplying a is the mass of the entire system, not just the one block.

However, you seem really confused about this problem. So, I agree with tiny tim that you should solve this going about it the standard way. Draw free body diagrams for each block and write Newton's second law equations for each block. You will end up with two unknowns, a and T, and two equations. You should then be able to solve for a.

Going about it this way may help clear up some confusion about the method I posted above.
(I essentially skip a step and combine both newton's second law equations into one right from the start.)

Like this?
Sorry. I can't read anything in that image.
 

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