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Tension on a massive rope

  • #1
66
5

Homework Statement


A system, consisting of a wide rope of mass 0.10 kg between two blocks each of mass 0.10 kg, is lifted by an applied force F = 9.0 N (Fig. 4 below). (a) Find the acceleration of the system. Find the tension at (b) the top of the rope, and (c) the bottom of one-fifth of the rope. Take g = 10 m/s2.
upload_2018-10-14_19-22-19.png



Homework Equations


Fnet=ma

The Attempt at a Solution


a) Fnet=ma
F- 3mg = ma
9- 3(0.1)(10) = 0.1a
6=0.1 a
a= 60 m/s^2

b) T- 2mg = ma
T-2(0.10)(10) = 0.10(60) (it's right to use the mass of rope by acceleration right?)
T-2=66
Ttop= 68N
is it right?

c) T- (4/5x m)- 10xm= (4/5)m x a
T- 0.08(10) - (0.10)(10)= 0.08(60) is this equation right
i don't if i got (the bottom of one-fifth of the rope) correctly or not. i considered the tension to equal to that part of the rope below 1/5 from the mass so that m of rope now is 4/5m
and also i have the lower mass (m). so is it right??
 

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Answers and Replies

  • #2
PeroK
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a) Fnet=ma
F- 3mg = ma
9- 3(0.1)(10) = 0.1a
6=0.1 a
a= 60 m/s^2
Doesn't that seem a bit high to you?

b) T- 2mg = ma
T-2(0.10)(10) = 0.10(60) (it's right to use the mass of rope by acceleration right?)
T-2=66
Ttop= 68N
is it right?
You have a ##9N## force and a tension of ##68N## Does that makes sense?
 
  • #3
66
5
Doesn't that seem a bit high to you?



You have a ##9N## force and a tension of ##68N## Does that makes sense?
it does, that's why I am asking here. i get number that are too big
 
  • #4
PeroK
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it does, that's why I am asking here. i get number that are too big
What is the mass of your system?

What is the net external force on your system?

What is the acceleration of your system?
 
  • #5
66
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What is the mass of your system?
the total mass is 3(0.10) the two blocks and the rope
fnet is 9N
the acceleration... do you mean it should be F- 3(mg)=3ma????
 
  • #6
PeroK
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the total mass is 3(0.10) the two blocks and the rope
If you mean: The total mass is ##0.3kg##, then yes.

fnet is 9N
What about gravity?

the acceleration... do you mean it should be F- 3(mg)=3ma????
I didn't mean anything. I just asked a question!
 
  • #7
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What about gravity?
the gravity force is down and each mass including the rope will have a weight downward.
I didn't mean anything. I just asked a question!
Oh yeah, i am just trying to reach the solution fast :)
 
  • #8
PeroK
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the gravity force is down and each mass including the rope will have a weight downward.
Exactly, so you need to take gravity into account when calculating the external force on your system.
 
  • #9
66
5
i JUst checked again my solution and i must have put 3m in a)
F-3mg= (3M)a
right? so a=20

and regarding the b) it would be
T-2mg=2ma
right?

and for c)
it would be
T-(1/5x0.10)g - 0.10g = (1/5x0.10+ 0.10) (20)
right?
 
  • #10
PeroK
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i JUst checked again my solution and i must have put 3m in a)
F-3mg= (3M)a
right? so a=20
20 what? Units are important.

and regarding the b) it would be
T-2mg=2ma
right?
That's the idea. It's called Newton's 2nd Law!
 
  • #11
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20 what? Units are important.
a= 20 m/s^2

so it's right and i am done with it right?
 
  • #12
PeroK
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a= 20 m/s^2

so it's right and i am done with it right?
You just have to calcuate the tension.
 

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