How to solve for acceleration in a block on incline system with a pulley?

[vs55]

Homework Statement

a)which way does this system go?
b)write force equations
c)find acceleration of the 10kg mass
THE ROPE IS SUPPOSED TO BE ATTACHED TO THE MASS NOT THE RAMP IN THE PICTURE

Homework Equations

m1 is the 10kg mass..m2 is the 8kg mass
u = coefficient of friction

The Attempt at a Solution

The system goes to the right because:
Tension>m1gsin-um1gcos

Tension = m2g-ma
Tension-m1gsin-um1gcos = ma
sub in for tension to get
m2g-m2a-m1gsin-umgcos = m1a
78.4 - 8a - 10(9.8)(sin30)-(.05)(10)(9.8)cos30 = 10a
solving for a gives me 1.397
the REAL answer is supposed to be .464
what did i do wrong?♦

 

[Doc Al]

I will assume that you meant to show the cord attached to the mass m1, not to the ramp.

Tension = m2g-ma
Tension-m1gsin-um1gcos = ma

Two problems:
(1) The rope (and thus the tension) pulls twice on m2.
(2) The accelerations of m1 and m2 are not the same. (Figure out how are they related.)

 

[Dweirdo]

I Attached a file which could help you get to the relations as Al above me said.
Have fun and good luck.

 

[vs55]

yes sorry the rope is supposed to be attatched to the mass not ramp

so is T1 at the ramp = 2T1 at the hanging mass?

 

[tiny-tim]

Welcome to PF!

Hi vs55! Welcome to PF!

yes sorry the rope is supposed to be attatched to the mass not ramp

so is T1 at the ramp = 2T1 at the hanging mass?

Doc Al is offline, so I'll jump in and say …

no, the tension is the same all the way along the same rope.

Use good ol' Netwon's second law twice, once on the 8kg mass, and once on the 10 kg mass, with the same T.

 

[vs55]

i'm still getting the wrong answer...the tension is messing me up

 

[Dweirdo]

Hi vs55! Welcome to PF! Doc Al is offline, so I'll jump in and say …

no, the tension is the same all the way along the same rope.

Use good ol' Netwon's second law twice, once on the 8kg mass, and once on the 10 kg mass, with the same T.

Are You absolutely sure??
I'm pretty sure Doc Al is right, the tension in the rope attached to mass m2 is twice as the tension in the other rope,it's not the same rope,thus not the same acceleration.
O.P:
the answer i got is 0.47 approx.

 

[Doc Al]

i'm still getting the wrong answer...the tension is messing me up

Please show the revised versions of your force equations.

Are You absolutely sure??
I'm pretty sure Doc Al is right, the tension in the rope attached to mass m2 is twice as the tension in the other rope,it's not the same rope,thus not the same acceleration.

Of course Doc Al is right, but so is tiny-tim. I didn't say that the tension in the rope attached to m2 was twice the tension elsewhere in the rope. (As tiny-tim says, the tension is the same throughout the rope.) What I did say was that the rope pulls twice on m2, thus the total force exerted by the rope on m2 is twice the tension.

 

[Dweirdo]

Please show the revised versions of your force equations.Of course Doc Al is right, but so is tiny-tim. I didn't say that the tension in the rope attached to m2 was twice the tension elsewhere in the rope. (As tiny-tim says, the tension is the same throughout the rope.) What I did say was that the rope pulls twice on m2, thus the total force exerted by the rope on m2 is twice the tension.

but it's not the same rope!
the rope attaching M2 with the pulley is not the same rope attaching M1 with the wall or w/e.
The rope pulls twice on M2 thus the tension in the rope attaching m2 with the pulley is 2T.which is basically the same as saying that 2T pulls M2.
Pretty sure about it...Correct me again if I'm wrong...

 

[tiny-tim]

but it's not the same rope!
the rope attaching M2 with the pulley is not the same rope attaching M1 with the wall or w/e.

Well, it's your rope, so I'm reluctant to contradict you …

but your diagram, in post #1, shows only one rope, going from the block up the slope to the pulley, down to the other pulley, and up to the ceiling …

(and, I assume, the upper pulley is fixed, while the lower pulley can move … is that correct?)

if there are two ropes, where on the diagram does the first end and the second begin?

 

[Lok]

The mass tied to the pulley exerts half it's weight as force upon the block. So Fp=mp*g/2 (just how it is)
The block exerts only Fb=mb*sin30*g-Ffriction
I favor the block out of sympathy. That is the system might pe pulled by the block.

 

[Doc Al]

The mass tied to the pulley exerts half it's weight as force upon the block.

That would be true if it were in equilibrium, but it's not.

 

[Doc Al]

but it's not the same rope!
the rope attaching M2 with the pulley is not the same rope attaching M1 with the wall or w/e.

There's only one continuous rope in this problem, with a single tension throughout.

The rope pulls twice on M2

Yes!

thus the tension in the rope attaching m2 with the pulley is 2T.

No, the tension is the same (T), but the net force exerted by the rope on m2 is 2T, since it pulls twice.

which is basically the same as saying that 2T pulls M2.

I think you're being a bit sloppy with your terminology here. It's important to realize that there's a single, continuous rope with one tension throughout.

 

[Dweirdo]

Well, it's your rope, so I'm reluctant to contradict you …

but your diagram, in post #1, shows only one rope, going from the block up the slope to the pulley, down to the other pulley, and up to the ceiling …

(and, I assume, the upper pulley is fixed, while the lower pulley can move … is that correct?)

if there are two ropes, where on the diagram does the first end and the second begin?

First,I'm not the original poster, I'm just trying to help as You do.
I'm attaching a sketch of the problem , highlighted rope 1 and rope 2, this is system is hard to build with only 1 rope...
what do You think?

 

[Doc Al]

I'm attaching a sketch of the problem , highlighted rope 1 and rope 2, this is system is hard to build with only 1 rope...
what do You think?

It looks like one rope to me, with each end dyed a different color! Of course you can make "one rope" by tying two different ropes together--but as far as the physics is concerned, that's the same as "one rope". It will have a single tension throughout (given the usual approximations used in these problems).

 

[Dweirdo]

It looks like one rope to me, with each end dyed a different color! Of course you can make "one rope" by tying two different ropes together--but as far as the physics is concerned, that's the same as "one rope". It will have a single tension throughout (given the usual approximations used in these problems).

Common,I like these kind of Arguments\Discussions,they some times lead to...
But man, I'm starting to feel I'm a total idiot, in Every book that i have learned from, they divide it to 2 ropes ,the rope that holds the pulley(same rope all around the pulley) and a rope attached to the center of the pulley, tension in the rope attached to the center of the pulley is twice than the rope around the pulley.
Try to make A System a like with 1 rope , You'll never reach what's in the sketch , no way!
Maybe You don't understand what I'm trying to say( I really do hope so) these are same kind of ropes but definitely not the same rope! same rope=continues rope,this is what I mean...
here is another link http://www.juniorengineering.usu.edu/lessons/machines/images/pulley2to1.jpg"
lots of books show the same thing, and some books instead of a rope attaching the center of the pulley with the mass, they use a hook...
Am I the only one Seeing 2 ropes?"Of course you can make "one rope" by tying two different ropes together--but as far as the physics is concerned, that's the same as "one rope"
hmm? take a mass less rope, tie another rope to it hold the first rope from it's sides ,connect the another rope with a body, the same tension will be in each of the ropes?
than why won't both of them crack?

 

[tiny-tim]

… these are same kind of ropes but definitely not the same rope! same rope=continues rope,this is what I mean...
here is another link http://www.juniorengineering.usu.edu/lessons/machines/images/pulley2to1.jpg"
lots of books show the same thing, and some books instead of a rope attaching the center of the pulley with the mass, they use a hook...
Am I the only one Seeing 2 ropes?

Hi Dweirdo!

Your new diagram does only show one rope … it goes from you (that is you, isn't it? ) over the top pulley, under the bottom pulley, and back up to where it is attached to the centre of the top pulley.

In the original diagram, the only rope attached to the centre of a pulley is is the one with the 8 kg mass on the end.

hmm? take a mass less rope, tie another rope to it hold the first rope from it's sides ,connect the another rope with a body, the same tension will be in each of the ropes?
than why won't both of them crack?

uhh?

join two different ropes, and the tension (if the rope is not stretching) will still be the same all the way along … use Newton's second law on any section of the rope … the force at one end must be the same as at the other end.

 

[Doc Al]

But man, I'm starting to feel I'm a total idiot, in Every book that i have learned from, they divide it to 2 ropes ,the rope that holds the pulley(same rope all around the pulley) and a rope attached to the center of the pulley, tension in the rope attached to the center of the pulley is twice than the rope around the pulley.
Try to make A System a like with 1 rope , You'll never reach what's in the sketch , no way!

I just realized that you are talking about a second rope! You meant the rope that attaches m2 to its pulley. Sorry about that! (I guess I didn't look at your diagram too carefully... I didn't see any purple and assumed it was my eyes going. I didn't think of looking at that small piece of rope, since I routinely ignore it. ) Of course, that might not even be a rope--it could be a metal rod, a bracket, or whatever.

I generally ignore such things and treat the mass and its pulley as a single object. Saves time! If you include that second tension, you need a third equation relating it to the other tension: T₂ = 2T₁. But yes, you are correct: the tension in that second rope does equal twice the tension in the other rope.

Of course, tiny-tim and I were talking about the rope that connects from ceiling to m1. It pulls up twice on the m2/pulley subsystem (which I just lump together).

 

[Dweirdo]

I just realized that you are talking about a second rope! You meant the rope that attaches m2 to its pulley. Sorry about that! (I guess I didn't look at your diagram too carefully... I didn't see any purple and assumed it was my eyes going. I didn't think of looking at that small piece of rope, since I routinely ignore it. ) Of course, that might not even be a rope--it could be a metal rod, a bracket, or whatever.

I generally ignore such things and treat the mass and its pulley as a single object. Saves time! If you include that second tension, you need a third equation relating it to the other tension: T₂ = 2T₁. But yes, you are correct: the tension in that second rope does equal twice the tension in the other rope.

Of course, tiny-tim and I were talking about the rope that connects from ceiling to m1. It pulls up twice on the m2/pulley subsystem (which I just lump together).

At least I'm not as idiot as i thought xD
"you need a third equation relating it to the other tension"
Yes, exactly like You do for accelerations... well now I'm calm XD
Have a nice day.

 

[vs55]

uhhh so are there 2 ropes or 1 rope??
if the system is not moving what is the equations for m2?
is it 2T1 + T2 - m2g=0?

 

[Dweirdo]

Look , I'll try to settle it down for You.
You can look as if there were 2 ropes or 1 rope, it doesn't matter the thing that matters for you is that the tension pulling the pulley which is attached with m2, is 2T, when T is the tension in the rope...cause you can see that the pulley is covered by the rope thus is being pulled as if from both sides.
look at a picture i attached in page 1, that should help.
anyway:
2T1 + T2 - m2g=0?
is wrong, if You look at the problem as if another rope is pulling m2, You'd say T2-m2=0(if it's not moving) while T2=2T1 is that clear?
the equation of motions:
2T-m2g=m2*(a2)
XXXXXXX=m1*(a1)
the XXXXX means You should try alone... and You should get the relation of the accelerations (hint: while m1 goes Distance X, mass m2 raises height of X/2)
Good luck!

 

[vs55]

hmm ok thanks i kind of get it now
but the work that i posted in the first post, is it right? except for force in mass2( 2T1-m2a = 0) ofcourse
so would my FBD show T1 to the right for mass m1 and 2T1 upwards for mass m2?