How Do I Calculate the Acceleration of a System with 3 Masses and 2 Pulleys?

[gob_b]

Homework Statement

Homework Equations

Fnet=ma
Fres=uFn
Fg=mg

The Attempt at a Solution

ive been stuck on this for 2 days.. i tried splitting the problem into 2 weights and then considering the last mass but NOTHING. i tried SO MANY attempts NOTHING WORKS and its starting to make me angry. the block on the flat surface has a res of 250N and the fapp of the other 2 masses combined is only 130N. HOW DOES IT MOVE? PLEASE help me !

 

[frog210293]

I will leave this to someone who has more of an idea of what they are doing, the only thing i can see is that you have wrote down the angle as 30 degrees on the diagram when it says 35 degrees in the box underneath. Other then that I'm at as much of a loss as you are.

 

[gob_b]

oh no sorry about that, i only used 30 because i wanted to do the numbers in my head rather than do sin35. please pay no attention to my writing

 

[SammyS]

Draw a Free Body Diagram for each block.

 

[gob_b]

ive done all that on a separate paper, i came to a conclusion that there is no acceleration but the question fully asks to solve for a... I am so lost i really don't know what to do

 

[hotvette]

If you draw a FBD for each mass you should end up with 3 equations and 3 unknowns. The key considerations are (1) the acceleration of each mass is the same, and (2) the tension between m₁ and m₂ is different than the tension between m₂ and m₃.

If you tell us the F=ma equation you got for each mass we can help guide you.

 

[gob_b]

i can't get my head around the whole thing..

m1 =
fnet = ma
100 + (-T1) = 10a

m2 =
fnet = ma
50sin35 + (T?) + (-T2) = 5a

m3 =
fres = uFn
fres = 150
fnet = ma
T2 + (-Fres) = 30a

please someone help i have no idea

 

[hotvette]

You pretty much have it. Looks like you are using g = 10 which is ok I guess. For equation 2, the T is T₁ because the tension between m₁ and m₂ is equal and opposite. Also, you might want to double check your geometry, I'm not sure sin35 is right. In summary, you have 3 equations you can solve for T₁, T₂ and a by substitution:

1. 100 - T₁ = 10a
2. 50sin35 + T₁ - T₂ = 5a --- verify that sin35 is correct
3. T2 - 150 = 30a

 

[gob_b]

thank you so much for verifying my equations good sir at least now i don't have to doubt my self if they are right or wrong.

but yes the geometry is correct because 50sin35 will give you the component of gravity that is parallel to the surface and cos would give you the perpendicular

im going to work this out now and see how it goes and ill reply. thanks again :)

 

[gob_b]

according to my work, a = 0 because the combined applied force of m1,m2 is less than the res of m3.

can anyone verify?

 

[diazona]

I get the same result, that there will be no acceleration. Seems a bit odd, but there's nothing in the question that prevents that from being the right answer.

 

[vela]

I think you need to assume the system is already in motion otherwise the coefficient of kinetic friction wouldn't apply. The acceleration will depend on which direction the system is moving.

 

[diazona]

I think you need to assume the system is already in motion otherwise the coefficient of kinetic friction wouldn't apply. The acceleration will depend on which direction the system is moving.

Hm, interesting, I hadn't thought about it that way. Though that does seem like an unusual assumption to make, and besides there's no information about the direction of motion.

I was thinking you'd have to assume that the coefficient of static friction is equal to or greater than the coefficient of kinetic friction, which as far as I know is a reasonable assumption in normal situations.

 

[gob_b]

the system only moves to the left because there isn't anything applying it to move in the right direction. can anyone build on that

agree?

 

[SammyS]

the system only moves to the left because there isn't anything applying it to move in the right direction. can anyone build on that

agree?

If you assume the system is initially in motion, that motion could have m₃ moving to the right, or moving to the left. As vela points out, you will get a different acceleration in each case. (Different in more than simply the sign.)

In either case, the system will eventually come to rest, and stay at rest.

 

[gob_b]

god there is so many possibilities...

can anyone come up with an ultimate correct answer?

 

[gob_b]

heres how i imagine it.

picture the worst case scenario where there is NO incline in this problem, but instead both m1,m2 are hanging straight downward. therefore the down force would be 100N+50N= 150N this force would equal the fres force of m3 and therefore no movement. now if we add this incline back, the app force of m1+m2 is now obviously < 150N.

does this not suggest that a is 0 here? but that would mean fnet = 0?

 

[vela]

With the given information, the problem is ambiguous. You need to make additional assumptions to come up with answers, so just state what those assumptions are and give the corresponding result. There are effectively only three possibilities: the system is already moving to the left; the system is already moving to the right; and the system is at rest (in which case you need to make an assumption about static friction).

 

[gob_b]

for the sake of the question, assume its going to the left.
can anyone come up with an answer?

 

[diazona]

for the sake of the question, assume its going to the left.
can anyone come up with an answer?

OK, if you're going to assume it's going to the left, you should be able to solve the problem. Give it a try and if you have problems, post what you do. (After all, it is your homework )

 

[gob_b]

i got 0.13 for a.

and T1 tension was 110N which is greater than 100N the gravity force acting down for m1.

is this even possible?

 

[SammyS]

i got 0.13 for a.

and T1 tension was 110N which is greater than 100N the gravity force acting down for m1.

is this even possible?

Yes, it's possible, even expected, for T₁ to be greater than 100N. You start with m₁ moving downward, but slowing down, so its acceleration is upward.

T₁ - m₁g = m₁a > 0

so, T₁ > m₁g .

 

[gob_b]

thank you sammy,

so my a was a negative small decimal. and t1 is > 100 like i said.

to my knowledge a negative A means that it did not move correct?
therefore t1 must e > 100 to keep it from moving down

am i right?

 

[Grifter]

thank you sammy,

so my a was a negative small decimal. and t1 is > 100 like i said.

to my knowledge a negative A means that it did not move correct?
therefore t1 must e > 100 to keep it from moving down

am i right?

since you only were given the coef of kinetic friction i would assume the that the system was already in motion. If your accel was negative then that means the system for that instant was slowing down. It would be better if the question gave a little more information.