Question about the Normal Force

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  • #1
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Two buckets of nails are hung one above the other and are pulled up to a roof by a rope. Each bucket has a mass of 5.0kg. The tension in the rope connecting the buckets is 60 N. Calculate the acceleration of the buckets.

So here's what I did:

mt= ma + mb (m which is equal to mass)
mt= 5.0kg + 5.0kg
mt= 10.0 kg (total mass is 10.0kg)

Fnet= Fn + Fg
ma= Fn + mg

if i isolate the a it would be:

a= (Fn/m) + g

How do I find the Normal force to solve the question?
 

Answers and Replies

  • #2
haruspex
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Clue: forget about the upper bucket and just think about the forces and acceleration of the lower bucket.
 
  • #3
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Clue: forget about the upper bucket and just think about the forces and acceleration of the lower bucket.
still dont get it please elaborate it
 
  • #4
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Make a force equation of the lower bucket. Which are the forces acting on it? Due to the string(tension)? due to earth? how is the net force on it related to its acceleration, then?
 
  • #5
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forcers that are acting on it are. Fn and Fg. I dont know what to do with the tension its a big question for me. Net force = ma
 
  • #6
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Fn in your post #5, IS the tension in the string...You should probably call it FT

Net force [tex]F_{net} = ma[/tex]

But, what is [tex]F_{net}[/tex] in terms of Fn and Fg??

remember, this is only for the lower bucket.
 
  • #7
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Oh thanks for the tension thing.
I think the Fnet is=T+mg ?
 
  • #8
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Oh thanks for the tension thing.
I think the Fnet is=T+mg ?

But tension is acting upwards, mg is acting downwards, and Fnet, I'll leave to you. :wink:
 
  • #9
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Oh ok now i got it, is this right? Fnet= T-mg
 
  • #10
881
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Oh ok now i got it, is this right? Fnet= T-mg

Yep. :approve:

Now use the relation with acceleration.
 
  • #11
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a= Fnet/m Ok i used the 5 kg and I got the answer. But I have one quick question how come I didnt use the total mass?

Thank You man ! :)

But I could also do this

Fnet=T-mg
ma=T-mg
when I isolate acceleration -->> a=Ft-(mg)/m Instead of a=Ft/m ( both correct)

Thank You so much man :smile:
 
Last edited:
  • #12
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a= Fnet/m right? Ok i used the 5 kg and I got the answer.

Great! :biggrin:

But I have one quick question how come I didnt use the total mass?

Why would you use something when you don't need it? :wink: Its simply not necessary to use the total mass for the situation you are given.

But I could also do this

Fnet=T-mg
ma=T-mg
when I isolate acceleration -->> a=Ft-(mg)/m I still got the right answer

That basically is using a = Fnet/m :wink: You just jumbled with terms to get there, in this answer.

PS : Post such questions in homework section, next time!
 
Last edited:
  • #13
haruspex
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But I have one quick question how come I didnt use the total mass?
The total mass is distributed across two buckets. If you want an equation involving the total mass then you'll need to treat the two buckets as a unit. So then you want the net force on that unit. That will be sum of the two gravitational forces, downwards, against the tension in the string above the top bucket acting upwards. But you are not told the tension in that part of the string, so it doesn't get you far.
You can view the question as an exercise in figuring out which pieces of information are useful and which aren't.
 
  • #14
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The total mass is distributed across two buckets. If you want an equation involving the total mass then you'll need to treat the two buckets as a unit. So then you want the net force on that unit. That will be sum of the two gravitational forces, downwards, against the tension in the string above the top bucket acting upwards. But you are not told the tension in that part of the string, so it doesn't get you far.
You can view the question as an exercise in figuring out which pieces of information are useful and which aren't.
Thank You so much :smile:
 

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