What am I missing here? Mechanics

  • Thread starter Abukadu
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  • #1
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Homework Statement



Hello guys :)
I'm a physics student from Israel and I couldnt find a decent physics forum to help me answer some hard to solve exercises (btw, forgive my english if I fail to express myself well).

The exercise:
(hopefully this works..)
http://img143.imageshack.us/img143/8707/88816184zp7.jpg [Broken]
http://g.imageshack.us/img143/88816184zp7.jpg/1/ [Broken]

The mass of object A is 2kg
The mass of object B is also 2kg
The mass of object C is yet unknown
The friction coefficient between object A and B is μ=0.808
(no friction between A and the rail)


1) I need to find the maximum value for the mass of object C so that all 3 object will accelerate in the same acceleration (answer:Mc=3.5kg)

2) Now it is known that the mass of object C=4kg. They are asking for the acceleration of object A,B and C (answers: the acceleration of objects A and C is 2.67 m/s^2, acceleration of object B is 2 m/s^2)

2. The attempt at a solution

1) What I tried to do is solve the problem by using 3 equations:

object B-
T=fk (fk=friction) => fk=μ*N=0.808 * mgCos30=14N
so: T=14N

objects A+B-
T-MgSin30=MgSin30*a when M=4kg and T=14
so that: a=-0.3 m/s^2

object C-
Mc*g - T = Mc*a when Mc=The unknown mass of object C and a=-0.3
so that: Mc=1.5kg

But, the right answer by the book is 3.5kg

2) Tried to solve but with no luck.

Can someone please help me understand what is it that I'm doing wrong? I'm in a point where I'm completely helples, I have no clue..
 
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Answers and Replies

  • #2
tiny-tim
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Welcome to PF!

object B-
T=fk (fk=friction) => fk=μ*N=0.808 * mgCos30=14N
so: T=14N

Hello Abukadu! Welcome to PF! :smile:

If T is the tension, then your equation is wrong …

there is no tension on B …

the only forces on B are the friction the normal force and the weight.

Try again! :smile:
 
  • #3
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Besides that my other 2 equations are right?

Hmm.. I'm trying to imagine how object B behaves in this situation.
I know that in this case the normal force is there for me to find the friction (since i have μ, and I know that the normal force is equal to mgCos30=2*10*sin30=17.32). Now, is object B accelerating? how can i create an equation for object B when there is no force pulling him to the right?
 
  • #4
tiny-tim
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Now, is object B accelerating? how can i create an equation for object B when there is no force pulling him to the right?

B is accelerating at the same rate as A and C.

Call it a.

And there is a force on B pulling it to the right, it's … ?

Hint: use good ol' Newton's second law for B to find a.

Then use Newton's second law for A to find T.

Then … :smile:
 
  • #5
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Object B-

-Fk = Mb*a ?

Feeling kinda dumb right now with this guessing game.. I'm in a point where I'm trying things I know arent true just for the chance I'll get to the right answer :uhh:
 
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  • #6
tiny-tim
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Object B-

-Fk = Mb*a

?

Yes :smile: (but why the minus sign?)
 
  • #7
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Yes :smile: (but why the minus sign?)

I figured it's a minus sign because it is opposing the direction of the acceleration.. Anyway, I tryed that option before :) and yet no luck..

Object B-
Fk=Mb*a

Fk=μ * N=0.808 * mgCos30=0.808 * 2*10*Cos30=13.99

therefor:
13.99=2a
a=6.99 m/s^2

Object A-
T-Mg=Ma now i figured the mass is 4kg because object B is on top so:
T-40=4*6.99 -> T=67.98

Object C-
Mc*g-T=Mc*a -> 10Mc - 67.98 = 6.99Mc -> Mc=22.5kg which is,of course, incorrect!

I'm not sure what is the scientific term to call it, but did I place a wrong variable or something like that?

By that way, Thanks a lot for your help! I'm realy going to adopt this forum as a second home for my physics course and share it with my friends :approve:
You dont have a problem with foreigners, right? :tongue2:
 
  • #8
Since it looks like tiny-tim logged off (maybe he had to a teach class! I just finished!), I'm looking at it now. I like your above equations for A & C. But you're missing one force in B that acts parallel to the slope. There's the friction force, but also part of the weight wants to oppose forward motion up the slope... include that.

P.S. One of the reasons I like PF is that I can help students in all different parts of the world (at all hours of the day!). Welcome! :smile: It was a big boost to my procrastination when I wrote my dissertation! :rofl:
 
  • #9
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Since it looks like tiny-tim logged off (maybe he had to a teach class! I just finished!), I'm looking at it now. I like your above equations for A & C. But you're missing one force in B that acts parallel to the slope. There's the friction force, but also part of the weight wants to oppose forward motion up the slope... include that.

P.S. One of the reasons I like PF is that I can help students in all different parts of the world (at all hours of the day!). Welcome! :smile: It was a big boost to my procrastination when I wrote my dissertation! :rofl:

Okay so you mean I should do something like this:
mgSin30 + Fk = ma -> 2*10*sin30 + 13.99 = 2a -> a=12 m/s^2 ?
because I still dont get the right answer.. :(

BTW
"It was a big boost to my procrastination when I wrote my dissertation! "
Even if I was a native speaker I dont think I could understand that sentence :D
 
  • #10
I think there's a sign error wrong in your diagram for B.

Let's examine the concept of static friction a bit (I think maybe I misled you a bit originally, probably hopping in too quick...)

Static friction is present to KEEP the block from sliding relative to the other block... as Tiny-tim is hinting: "And there is a force on B pulling it to the right, it's … ?"

It's the static friction! The static friction has a maximum value (given by the normal force and the coefficent). This value can be reached by: some of it being used to keep the block from sliding down due to weight. The rest of it can be used to keep the block from sliding relative to the other block as the other block accelerates beneath it... i.e. to give your acceleration.

I think set it up like this for B: ma + Fg sin theta <= Fk (with <= being less than or equal.. you want the "equal part to give the maximum value for the acceleration)
 
  • #11
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Okay I started all over again..

C:

∑F_y = T-m_C*g = m_C*a
∑F_x = 0

B:

∑F_y = N_2 - m_B*gcos(a) = 0
∑F_x = f_s = m_B*a

f_s:

f_s = μN_2

A:

∑F_x = T - m_A*gsin(a) - f_s = m_A*a
∑F_y = N1 - N2 - m_Agcos(a) = 0

And still.. the results turns out to be 2.23kg, and not 3.5kg.
Can you please try to solve the question and tell me if you get the same answer?
 
  • #12
I have some intermediate answers.... what did you get for your acceleration?
 
  • #13
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6.997 m/s^2
 
  • #14
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Homework Statement



Hello guys :)
I'm a physics student from Israel and I couldnt find a decent physics forum to help me answer some hard to solve exercises (btw, forgive my english if I fail to express myself well).

The exercise:
(hopefully this works..)
http://img143.imageshack.us/img143/8707/88816184zp7.jpg [Broken]
http://g.imageshack.us/img143/88816184zp7.jpg/1/ [Broken]

The mass of object A is 2kg
The mass of object B is also 2kg
The mass of object C is yet unknown
The friction coefficient between object A and B is μ=0.808
(no friction between A and the rail)


1) I need to find the maximum value for the mass of object C so that all 3 object will accelerate in the same acceleration (answer:Mc=3.5kg)

2) Now it is known that the mass of object C=4kg. They are asking for the acceleration of object A,B and C (answers: the acceleration of objects A and C is 2.67 m/s^2, acceleration of object B is 2 m/s^2)

2. The attempt at a solution

1) What I tried to do is solve the problem by using 3 equations:

object B-
T=fk (fk=friction) => fk=μ*N=0.808 * mgCos30=14N
so: T=14N

objects A+B-
T-MgSin30=MgSin30*a when M=4kg and T=14
so that: a=-0.3 m/s^2

object C-
Mc*g - T = Mc*a when Mc=The unknown mass of object C and a=-0.3
so that: Mc=1.5kg

But, the right answer by the book is 3.5kg

2) Tried to solve but with no luck.

Can someone please help me understand what is it that I'm doing wrong? I'm in a point where I'm completely helples, I have no clue..

basically, what you need to understand here is that, the frictional force between the blocks A and B is slowing down A, but accelerating B. Also there is no difference in the MAGNITUDE of accelerations, so acceleration of the system = (Driving force/ Inertia)
 
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  • #15
you still have the "x" part of b wrong -- you didn't include the weight parallel to the slope.
 
  • #16
Note: since I have to go, I'll let you know: the a I get is 2 m/s^2. I think it then all works out.
 
  • #17
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B-
∑F_x; f_s - mg*sin30= m_B*a
that way I get the acceleration of 2 m/s^2
but still the final result is Mc=2.3 kg and not 3.5kg ..
 
  • #18
32
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C:

∑F_y = T-m_C*g = m_C*a
∑F_x = 0

B:

∑F_y = N_2 - m_B*gcos(a) = 0
∑F_x = f_s - mgSin30 = m_B*a

f_s:

f_s = μN_2

A:

∑F_x = T - m_A*gsin(a) - f_s = m_A*a
∑F_y = N1 - N2 - m_Agcos(a) = 0

is it possible that after i find a=2 m/s^2 and I'm moving on to object's A ∑F_x equation, object A has a different f_s than the one I calculated for object B?
it turns out that in order to get the right answer, object A's f_s has to be twice has big. maybe its some sort of a coincidence, I dont know..
 
  • #19
tiny-tim
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Hi physics girl! Hi Abukadu! :smile:

Yup … it's definitely 2 m/s2, and, like physics girl, I get 3.5kg.

Hint: you should get a tension of 28N. :smile:
 
  • #20
tiny-tim
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sorry!

Hi Abukadu! :smile:

You beat me by a minute!
is it possible that after i find a=2 m/s^2 and I'm moving on to object's A ∑F_x equation, object A has a different f_s than the one I calculated for object B?
it turns out that in order to get the right answer, object A's f_s has to be twice has big. maybe its some sort of a coincidence, I dont know..

ah … I think I misled you :redface: when I said:
Then use Newton's second law for A to find T.

To make the friction an internal force (so that you can ignore it), you need to use Newton's second law for A-and-B combined … that's much simpler than using it on A separately! :wink:
 
  • #21
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The equations above are right?
I do get a tension of 28N, and yet when I put all the variables in object C's equation I get a mass of 2.333 kg.

And this is only the second part of the question :(
 
  • #22
tiny-tim
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I do get a tension of 28N, and yet when I put all the variables in object C's equation I get a mass of 2.333 kg.

hmm … let's see :rolleyes: … 2.333 = 28/12, and 3.5 = 28/8 …

i think you added the acceleration instead of subtracting it! :redface:
 
  • #23
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Toda Lael (thank god) I finally solved it :cool:
I used the mass of both A+B together and got a tension of 42N which led to the right answer.

Now I have the last part of the question where they tell me that Object C's mass is 4kg and I need to find the acceleration of all three objects.
Now the friction is not static, how does it influence the system? I know that object B's acceleration is probably different than Object A and C (which I think is the same) but I cant realy explaion why, maybe you can help me understand?

Another thing is that someone told me that there is no Mc that can make Object A accelerate up the slope while object B accelerates down the slope. He tried to explain it by saying that while theres a kinetic friction on object B directed up the slope there is a force that accelerates it upwards. Can you please try to clarify it a bit more?
 
  • #24
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hmm … let's see :rolleyes: … 2.333 = 28/12, and 3.5 = 28/8 …

i think you added the acceleration instead of subtracting it! :redface:

but isnt the equation T-m_C*g = m_C*a ?
so that 28 = 10m_C + 2m_C ?
 
  • #25
tiny-tim
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but isnt the equation T-m_C*g = m_C*a ?
so that 28 = 10m_C + 2m_C ?

Nope … the acceleration is down

m_C*g - T = m_C*a
so that 28 = 10m_C - 2m_C :smile:
 

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