Tension on a Massive Rope: How to Calculate Tension at Different Points

[SakuRERE]

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.

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??

 

[PeroK]

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?

 

[SakuRERE]

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

 

[PeroK]

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?

 

[SakuRERE]

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?

 

[PeroK]

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!

 

[SakuRERE]

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 :)

 

[PeroK]

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.

 

[SakuRERE]

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?

 

[PeroK]

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!

 

[SakuRERE]

20 what? Units are important.

a= 20 m/s^2

so it's right and i am done with it right?

 

[PeroK]

a= 20 m/s^2

so it's right and i am done with it right?

You just have to calcuate the tension.