How Do Pulley Systems Affect Block Speeds with Friction and Spring Forces?

[obnoxiousris]

Homework Statement

a block of mass m2=4kg hangs by a light string that passes over a massless, frictionless pulley and is connected to a block of mass, m1=6kg that rests on a rough shelf. the coeff of kinetic friction is uk=0.2. the 6 kg block is pushed against a spring, but not attached to it. the spring has a force constant of 180nm^-1 and is compressed to 30cm.

find the speed of the blocks after the spring is released and the 4 kg block has moved a distance of 40cm. (the connecting rope has zero mass and remains taut the whole time.)

a diagram can be found here http://anulib.anu.edu.au/anuonly/exams/2006/PHYS1101_06_1.pdf

Homework Equations

F=ma?
Ffriction=u*normal force
F spring=kx

The Attempt at a Solution

i started by working out the forces on the 2 masses first.
at rest
on the hanging 4kg mass, there are weight force pulling it down, and tension force pulling it up. the two forces are equal and opposite to each other. on the rest 6kg mass, there is tension force pulling it toward the pulley, (which is the same as the weight force of the 4kg mass??) and friction pointing backwards, these two forces must be equal and opposite.

when the spring is released
on the 6kg mass, there is still the same amount of tension force pulling it towards the pulley, and now also the spring is also pushing at the same direction. since the mass is moving now, there must be friction pointing in the direction away form the pulley. from this, the net force on the mass can be found and acceleration can be known.
on the 4kg mass, there is still weight, and since the tension in the string has to be the same, does it mean now the downward force is the sum of weight force and the spring force?

also i watched a video on khan academy http://www.khanacademy.org/video/a-more-complicated-friction-inclined-plane-problem?playlist=Physics and he calculated the acceleration of the system, does this mean the 2 objects are accelerating at the same rate?

can some body please help? I am really confused D=

 

[tiny-tim]

welcome to pf!

hi obnoxiousris! welcome to pf!

(have a mu: µ and try using the X₂ icon just above the Reply box )

no, the accelerations are not the same, you'd need to do a bit of trig

but anyway can't you just use the work energy theorem ?​

 

[obnoxiousris]

thanks!

um, to use the energy theorem, i still need to know the forces on the masses right?
i just don't know whether my force analysis is right...

 

[tiny-tim]

hi obnoxiousris!

(just got up :zzz: …)

um, to use the energy theorem, i still need to know the forces on the masses right?

you don't need the tension as such, because that's an internal force, but unfortunately you do need to find the vertical component of the tension (where the rope is slanting), to calculate the normal force so that you can find the https://www.physicsforums.com/library.php?do=view_item&itemid=75" by the friction (i didn't notice that at first! )

(and you don't need the spring force, because you can use the spring https://www.physicsforums.com/library.php?do=view_item&itemid=269" )

i just don't know whether my force analysis is right...

let's see …

F=ma?
Ffriction=u*normal force
F spring=kx

The Attempt at a Solution

i started by working out the forces on the 2 masses first.
at rest on the hanging 4kg mass, there are weight force pulling it down, and tension force pulling it up. the two forces are equal and opposite to each other. on the rest 6kg mass, there is tension force pulling it toward the pulley, (which is the same as the weight force of the 4kg mass??) and friction pointing backwards, these two forces must be equal and opposite.

nooo …

the forces are not equal and opposite if there's acceleration (in a particular direction) …

that's the whole point of F = ma !

 

[obnoxiousris]

but i analysed the rest first,
'm1=6kg that rests on a rough shelf.'
so since the spring hasnt been released yet, the whole system is at rest, and there can be no unbalanced force right?

 

[tiny-tim]

but i analysed the rest first,
'm1=6kg that rests on a rough shelf.'
so since the spring hasnt been released yet, the whole system is at rest, and there can be no unbalanced force right?

sorry, i missed the words "at rest"

i don't understand why you've done that …

the acceleration and the tension keeps changing, so what is the advantage of finding the initial tension?

call the acceleration of the hanging mass "a", and the tension "T", and carry on from there (to find the normal force) ​

on the 4kg mass, there is still weight, and since the tension in the string has to be the same, does it mean now the downward force is the sum of weight force and the spring force?

even if we ignore the acceleration, the tension and the spring force aren't parallel

 

[obnoxiousris]

the string was taut the whole time, so doesn't this mean the tension in the string must be the same in both parts? originally the tension was by the weight, but since now the spring is also pushing on it, wouldn't it be the weight plus the spring? and that's the same for both parts?

 

[tiny-tim]

the string was taut the whole time, so doesn't this mean the tension in the string must be the same in both parts? originally the tension was by the weight, but since now the spring is also pushing on it, wouldn't it be the weight plus the spring? and that's the same for both parts?

yes, the tension is the same in both parts of the string

but it isn't the weight "plus" (or "minus") the spring force, since they're not parallel, so you need vector addition (and of course you also need to subtract the mass times acceleration)

 

[obnoxiousris]

ok, i guess that's where i got confused, um, so how do i do it vectorically? they are perpendicular to each other

 

[tiny-tim]

ok, i guess that's where i got confused, um, so how do i do it vectorically? they are perpendicular to each other

perpendicular?

i assumed it was an acute angle, but i can't access your picture without an anu password

can you post a picture?​

 

[obnoxiousris]

http://www.turboimagehost.com/p/7639040/untitled.JPG.html
oh sorry i ddnt realize that. here is the photo

 

[tiny-tim]

ohhh! it's simpler than i thought

i think i'd better start again …

i started by working out the forces on the 2 masses first.
at rest
on the hanging 4kg mass, there are weight force pulling it down, and tension force pulling it up. the two forces are equal and opposite to each other. on the rest 6kg mass, there is tension force pulling it toward the pulley, (which is the same as the weight force of the 4kg mass??) and friction pointing backwards, these two forces must be equal and opposite.

at rest, yes, the tension equals the friction

when the spring is released
on the 6kg mass, there is still the same amount of tension force pulling it towards the pulley, and now also the spring is also pushing at the same direction. since the mass is moving now, there must be friction pointing in the direction away form the pulley. from this, the net force on the mass can be found and acceleration can be known.
on the 4kg mass, there is still weight, and since the tension in the string has to be the same, does it mean now the downward force is the sum of weight force and the spring force?

no, you now need to include the acceleration (times the mass) in the equation, F_total = ma

the string was taut the whole time, so doesn't this mean the tension in the string must be the same in both parts?

yes, the tension is the same in both parts of the string

… does this mean the 2 objects are accelerating at the same rate?

yes, the displacements speeds and accelerations are always the same

but you should try using the work energy theorem (that'll avoid calculating the tension)

um, to use the energy theorem, i still need to know the forces on the masses right?
i just don't know whether my force analysis is right...

you need to know the external forces (ie the weight and the friction), but not the internal force (the tension), and not the spring force (because that's already included in the spring potential energy)

 

[obnoxiousris]

yes! me and few friends used energy and worked it out! (after much debate)
our eqn:
spring energy+ gravitational potential energy- work done by friction= kinetic energy of the system after
so this means the velocity is the same for both masses?

thank you for your patience! i learned a lot!

 

[tiny-tim]

hi obnoxiousris!

spring energy+ gravitational potential energy- work done by friction= kinetic energy of the system after
so this means the velocity is the same for both masses?

yes!

(though for clarity i'd start "decrease in spring potential energy + decrease in gravitational potential energy - work done by friction" )