Finding acceleration given mass and tension

  • Thread starter Rionic
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  • #1
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If there is a 70N tension in a rope attached to a 4.0 kg bucket, what will be the vertical acceleration of the bucket?

I'm not really sure how to do this. I can find tension but not acceleration. I think I'm missing something simple. I'm also confused about the direction because it doesn't state whether it is going up or down.

For the formula I got a=Ft-g and m+mg-Ft=a. Idk either are right
 

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  • #2
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For the formula I got a=Ft-g and m+mg-Ft=a. Idk either are right
Where do those formulas come from? Their units do not match.

Did you draw a diagram with all relevant forces in it? That should always be the first step if you are unsure how to proceed.
 
  • #3
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Where do those formulas come from? Their units do not match.

Did you draw a diagram with all relevant forces in it? That should always be the first step if you are unsure how to proceed.
I drew an FBD with Fg downwards and Ft upwards
 
  • #4
jbriggs444
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I drew an FBD with Fg downwards and Ft upwards
If those are the only two forces on the bucket, what is the net force on the bucket?
 
  • #5
Mister T
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For the formula I got a=Ft-g
What did your equation look like before you got this equation from it? It looks like you started out with the right equation but then made mistakes rearranging it.

You know, for example, that Ft = 70 N and g = 9.8 m/s2, but subtracting them doesn't make sense because they have different units.
 
  • #6
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What did your equation look like before you got this equation from it? It looks like you started out with the right equation but then made mistakes rearranging it.

You know, for example, that Ft = 70 N and g = 9.8 m/s2, but subtracting them doesn't make sense because they have different units.
Fnet=Ft-Fg then I turned fnet into ma and Fg into mg so it looked like
Ma=Ft-mg then rearranged it to a=Ft-g by dividing m on both sides
 
  • #7
jbriggs444
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Fnet=Ft-Fg then I turned fnet into ma and Fg into mg so it looked like
Ma=Ft-mg then rearranged it to a=Ft-g by dividing m on both sides
That should be a lower case m on the left ##ma = F_t - mg##
Now, when you divide the right hand side by m, what do you get again?
 
  • #8
Mister T
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Ma=Ft-mg then rearranged it to a=Ft-g by dividing m on both sides
You mean,

##ma=F_t-mg##,

right?

Then dividing the left side by ##m## does give you ##a##, but you got the wrong result when you divided the right side by ##m##.
 
  • #9
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Who says that the rope is pulling upward on the bucket? How do we know that the rope isn't pulling downward on the bucket?

Chet
 
  • #10
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You mean,

##ma=F_t-mg##,

right?

Then dividing the left side by ##m## does give you ##a##, but you got the wrong result when you divided the right side by ##m##.
Doesn't the m on the right side cancel?
And maybe this question doesn't require a direction because it doesn't tell me if it's going upwards or downwards
 
  • #11
Mister T
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Doesn't the m on the right side cancel?
Plug the numbers in and find out! Tell us the magnitude of each force, subtract them, and see if you get a net force that's equal to the mass times the acceleration.
 
  • #12
jbriggs444
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Doesn't the m on the right side cancel?
There are two terms on the right hand side.
 
  • #13
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There are two terms on the right hand side.
Ohh so it's a=Ft/m - g?
 
  • #14
Mister T
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Yes!
 

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