What are the tensions and movements in this force and motion problem?

[J-dizzal]

Homework Statement

In the figure here three ballot boxes are connected by cords, one of which wraps over a pulley having negligible friction on its axle and negligible mass. The masses are mA = 33 kg, mB = 45 kg, and mC = 14 kg. When the assembly is released from rest, (a) what is the tension in the cord connecting B and C, and (b) how far does A move in the first 0.300 s (assuming it does not reach the pulley)?

http://edugen.wileyplus.com/edugen/courses/crs7165/art/qb/qu/c05/fig_5_E.gif

Homework Equations

F=ma

The Attempt at a Solution

http://s1294.photobucket.com/user/jhoversten/media/be59570a-8c6b-4612-aa51-52d1bd5d3a7d_zpsqc5pc7ce.jpg.html

Im not sure how to solve the simultaneous equations for T or a. my attempt at solving for T was incorrect.[/B]

 

[Kinta]

The first part of the question asks for the tension in the cord between blocks B and C. So, if you've run into trouble, you may want to put some of your initial assumptions under more scrutiny. Do you know for a fact that the tension in the cord is uniform throughout both segments or was that just a gut feeling?

 

[J-dizzal]

The first part of the question asks for the tension in the cord between blocks B and C. So, if you've run into trouble, you may want to put some of your initial assumptions under more scrutiny. Do you know for a fact that the tension in the cord is uniform throughout both segments or was that just a gut feeling?

ah crap. yea the tension in the cord is not the same through out.
I just understood the problem wrong.

 

[Kinta]

ah crap. yea the tension in the cord is not the same through out.
I just understood the problem wrong.

Live and learn :)

 

[J-dizzal]

Live and learn :)

im going to take another stab at this.

 

[J-dizzal]

Live and learn :)

So i need to find a for the whole system because its asking for tension after the system is released right?

 

[Kinta]

So i need to find a for the whole system because its asking for tension after the system is released right?

From what I've done, I am seeing that finding $a$ for the system will be beneficial, yes.

 

[J-dizzal]

From what I've done, I am seeing that finding $a$ for the system will be beneficial, yes.

So what is the difference between the tension between box B and A and the weight of B+A?
T=ma
T=(45+14)(-9.8)=-578.2N that is the same as the weight of box B and A? So then how do i find acceleration for the system?

 

[Kinta]

Have you written down Newton's 2nd law for the 3 masses?

 

[J-dizzal]

Have you written down Newton's 2nd law for the 3 masses?

here what i have so far; F_A=(33kg)(a), F_B=572.8N - T, F_C=mg + (14kg)(-a)

 

[J-dizzal]

F_B= 572.8 + mg + TBtoC
edit: i think mg and + TBtoC are the same = -137.2N
edit again: nevermind they are different T in B=-5.4N

 

[J-dizzal]

Have you written down Newton's 2nd law for the 3 masses?

how would I find the tension between C and B? if F_C= mg + tension from B?

 

[J-dizzal]

Have you written down Newton's 2nd law for the 3 masses?

no

 

[J-dizzal]

im having trouble finding the acceleration on the system because i am not getting the net forces correctly.

 

[J-dizzal]

Have you written down Newton's 2nd law for the 3 masses?

F_A=M_A a
F_B=T_{B to A} + T_{B to C} + mg
F_C=T_{C to B} + mg
how does that look for Newtons 2nd law for each box?

 

[Kinta]

im having trouble finding the acceleration on the system because i am not getting the net forces correctly.

Sorry, I had to step out for a while. When writing Newton's Second Law for a thing, just put $m \vec{a} = \vec{F}^{(net)}$ where the net force would include things like gravity and tension (not the total acceleration, $\vec{a}$ which appears on the left).

 

[Kinta]

F_A=M_A a
F_B=T_{B to A} + T_{B to C} + mg
F_C=T_{C to B} + mg
how does that look for Newtons 2nd law for each box?

This might be right depending on your notation! Are those tensions meant to be vectors or positive scalars?

You may also find it more instructive to have $m_{A,B,C}a$ on the left side instead of $F_{A,B,C}$.
Edit: I jumped the gun a bit. Your equation for the forces on mass A isn't wrong, but go ahead and put the forces acting on it in the place of $F_A$. The previous suggestion does apply to your second two equations though.

 

[J-dizzal]

This might be right depending on your notation! Are those tensions meant to be vectors or positive scalars?

You may also find it more instructive to have $m_{A,B,C}a$ on the left side instead of $F_{A,B,C}$

the tensions are vectors, eg. T_{B to A} = (45kg + 14kg)a -59a

if $m_{A,B,C}a$ is on the left then what is on the right?

 

[Kinta]

the tensions are vectors, eg. T_{B to A} = (45kg + 14kg)a -59a

Just so that I can be sure that you're sure, can you note your coordinate system (which directions are positive and negative) and write the three equations in scalar form?

 

[J-dizzal]

Just so that I can be sure that you're sure, can you note your coordinate system (which directions are positive and negative) and write the three equations in scalar form?

Sure, F_netA=33a, F_netB= -59a + 14a - 441, F_netC= -14a - 137.2
standard coordinate system: up positive, down negative, right positive

 

[J-dizzal]

Just so that I can be sure that you're sure, can you note your coordinate system (which directions are positive and negative) and write the three equations in scalar form?

i got a = -22.24 :(

 

[J-dizzal]

Just so that I can be sure that you're sure, can you note your coordinate system (which directions are positive and negative) and write the three equations in scalar form?

 

[J-dizzal]

my signs are all messed up hold on. and i think Tension from B to A is not (45+14)a its just 45a

 

[Kinta]

Don't worry about calculating numbers yet. Keep everything symbolic (i.e., stick to letters and symbols) until you can just plug everything in for a single equation. When writing the net force on an object of mass $m$, try writing $m\vec{a} = \sum \vec{F}$ where the left side is written exactly as is (except for subscripts if applicable) and the right side is to be replaced with all of the forces acting on the mass. Ex: If you have a mass acted on by only gravity, you'd have something like $m \vec{a} = m \vec{g}$ where $\vec{g}$ points toward the center of mass of the gravitational body. If there was a tension force also acting on the object, it would be included on the right side. You basically want the right side determining the left side.

Edit: After you have this equation in vector form for each mass, you can begin assigning a coordinate system to determine positive/negative signs for the various quantities.

 

[J-dizzal]

Don't worry about calculating numbers yet. Keep everything symbolic (i.e., stick to letters and symbols) until you can just plug everything in for a single equation. When writing the net force on an object of mass $m$, try writing $m\vec{a} = \sum \vec{F}$ where the left side is written exactly as is (except for subscripts if applicable) and the right side is to be replaced with all of the forces acting on the mass. Ex: If you have a mass acted on by only gravity, you'd have something like $m \vec{a} = m \vec{g}$ where $\vec{g}$ points toward the center of mass of the gravitational body. If there was a tension force also acting on the object, it would be included on the right side. You basically want the right side determining the left side.

ok let me try that

 

[J-dizzal]

ok let me try that

im getting ma= m_Aa + T_{B to A} + T_{B to C} + mg + T_{C to B} + mg

 

[J-dizzal]

Don't worry about calculating numbers yet. Keep everything symbolic (i.e., stick to letters and symbols) until you can just plug everything in for a single equation. When writing the net force on an object of mass $m$, try writing $m\vec{a} = \sum \vec{F}$ where the left side is written exactly as is (except for subscripts if applicable) and the right side is to be replaced with all of the forces acting on the mass. Ex: If you have a mass acted on by only gravity, you'd have something like $m \vec{a} = m \vec{g}$ where $\vec{g}$ points toward the center of mass of the gravitational body. If there was a tension force also acting on the object, it would be included on the right side. You basically want the right side determining the left side.

Edit: After you have this equation in vector form for each mass, you can begin assigning a coordinate system to determine positive/negative signs for the various quantities.

im getting a = -7.413

 

[Kinta]

im getting ma= m_Aa + T_{B to A} + T_{B to C} + mg + T_{C to B} + mg

Those forces aren't all acting on the same mass. I wrote, "when writing the net force on an object of mass $m$..." So, you should definitely have three equations, one for each mass.

 

[J-dizzal]

Those forces aren't all acting on the same mass. I wrote, "when writing the net force on an object of mass $m$..." So, you should definitely have three equations, one for each mass.

ok i have 3 equations, how do i solve for a if i cannot set them equal to each other?

 

[Kinta]

ok i have 3 equations, how do i solve for a if i cannot set them equal to each other?

Can you show me the three equations?

The problem with where you wrote

im getting ma= m_Aa + T_{B to A} + T_{B to C} + mg + T_{C to B} + mg

was that it looked like you lumped all of the forces from the separate masses onto some joint mass, which isn't practical in this case.

 

[J-dizzal]

Can you show me the three equations?

The problem with where you wrote

was that it looked like you lumped all of the forces from the separate masses onto some joint mass, which isn't practical in this case.

thanks for helping I am having real difficulty with this supposedly easy question.
ma=m_Aa
ma=T_{B to A} + T_{B to C} + m_Bg
ma=T_{C to B} + m_Cg

 

[Kinta]

thanks for helping I am having real difficulty with this supposedly easy question.
ma=m_Aa
ma=T_{B to A} + T_{B to C} + m_Bg
ma=T_{C to B} + m_Cg

This is almost to a point of being useful! Go ahead and label the masses on the left-hand sides of your equations with the appropriate subscripts. Then, instead of having your first equation be redundant, replace the right-hand side of it with all of the interesting forces acting on it (there's only one non-canceled force on it, the tension) like you did with the others.

 

[J-dizzal]

sorry got hung up on something.

A; ma=33a
B; ma=31a -441
C; ma=14a -137.2

 

[Kinta]

sorry got hung up on something.

A; ma=33a
B; ma=31a -441
C; ma=14a -137.2

I just meant for you to put subscripts on the masses on the l.h.s. of your equations like you had on the right and then to replace the r.h.s. of the first equation with $T_{A to B}$. Keep symbols until you absolutely know you're ready to calculate stuff.

 

[J-dizzal]

I just meant for you to put subscripts on the masses on the l.h.s. of your equations like you had on the right and then to replace the r.h.s. of the first equation with $T_{A to B}$. Keep symbols until you absolutely know you're ready to calculate stuff.

m_Aa=T_{A to B}
m_Ba=T_{B to A}+T_{B to C}+m_Bg
m_Ca=T_{C to B}+m_Cg

like that? or write out the masses for each tension multiplied by a?

 

[J-dizzal]

then add everything together?

 

[J-dizzal]

then add everything together?

or can i move each mass on the left side to the right then add everything up to solve for a?

 

[Kinta]

m_Aa=T_{A to B}
m_Ba=T_{B to A}+T_{B to C}+m_Bg
m_Ca=T_{C to B}+m_Cg

like that? or write out the masses for each tension multiplied by a?

What you have looks great!

Because the a that we're talking about is the net acceleration of the whole system, replacing those tensions with the corresponding masses multiplied by a would be incorrect. This is because tension is not the only force present on two of the three masses. To see why this wouldn't be right, imagine replacing the tension in your $m_C$ equation with the $m_Ca$ product you suggest. You would get that $m_Ca = m_Ca + m_Cg$ which is equivalent to writing $m_Cg = 0$ and we know that's not true! You could make the replacement you suggest in the first equation, but that would just be redundant and not helpful.

Now, take a little while to think about the tensions you have in each equation and how they may or may not relate to the other tensions. I think you can solve the rest of the problem from here. Just make sure that if you add something to one side of an equation, you add the same thing (or something equivalent) to the other side to keep the equality true. Try not to make another reply at the first sign of trouble. You should be able to think through it from here :) After a while, if you just can't get it, I'll try to offer a little more help based on what you've done.

 

[J-dizzal]

What you have looks great!

Because the a that we're talking about is the net acceleration of the whole system, replacing those tensions with the corresponding masses multiplied by a would be incorrect. This is because tension is not the only force present on two of the three masses. To see why this wouldn't be right, imagine replacing the tension in your $m_C$ equation with the $m_Ca$ product you suggest. You would get that $m_Ca = m_Ca + m_Cg$ which is equivalent to writing $m_Cg = 0$ and we know that's not true! You could make the replacement you suggest in the first equation, but that would just be redundant and not helpful.

Now, take a little while to think about the tensions you have in each equation and how they may or may not relate to the other tensions. I think you can solve the rest of the problem from here. Just make sure that if you add something to one side of an equation, you add the same thing (or something equivalent) to the other side to keep the equality true. Try not to make another reply at the first sign of trouble. You should be able to think through it from here :) After a while, if you just can't get it, I'll try to offer a little more help based on what you've done.

Im getting stuck I am only able to find Tension between A and B being =441 but that's not even correct. I set msubBa = Tsub A to B.
and a = -17.52

 

[J-dizzal]

Im getting stuck I am only able to find Tension between A and B being =441 but that's not even correct. I set msubBa = Tsub A to B.
and a = -17.52

What you have looks great!

Because the a that we're talking about is the net acceleration of the whole system, replacing those tensions with the corresponding masses multiplied by a would be incorrect. This is because tension is not the only force present on two of the three masses. To see why this wouldn't be right, imagine replacing the tension in your $m_C$ equation with the $m_Ca$ product you suggest. You would get that $m_Ca = m_Ca + m_Cg$ which is equivalent to writing $m_Cg = 0$ and we know that's not true! You could make the replacement you suggest in the first equation, but that would just be redundant and not helpful.

Now, take a little while to think about the tensions you have in each equation and how they may or may not relate to the other tensions. I think you can solve the rest of the problem from here. Just make sure that if you add something to one side of an equation, you add the same thing (or something equivalent) to the other side to keep the equality true. Try not to make another reply at the first sign of trouble. You should be able to think through it from here :) After a while, if you just can't get it, I'll try to offer a little more help based on what you've done.

i don't understand why my tensions between A and B is not equal to the tension from B to A. and same for between B and C. they should be the same?

 

[J-dizzal]

Tensions from B + C should equal the tension on A but that is giving me a=17.52

 

[Kinta]

m_Aa=T_{A to B}
m_Ba=T_{B to A}+T_{B to C}+m_Bg
m_Ca=T_{C to B}+m_Cg

From this point, what exactly did you do to try obtaining a?

 

[J-dizzal]

From this point, what exactly did you do to try obtaining a?

well, i was trying a few different things, i tried summing everything to get a net force which was 572 i think then plugging that into ma = T for box A that didnt work. i tried substitution of Tension at A for ma in equation for box B. i keep getting the same wrong answers though.

 

[J-dizzal]

now i tried vector addition between B and C and got a=12.85. but when i plug that into T i got the wrong value

 

[Kinta]

If you sum all of the equations, you won't be obtaining a "net force". A net force is an overall force acting on a single object, not a collection of separate objects. The right-hand sides of each of your equations are the net or total forces acting on the corresponding mass. Can you show me what your summation of equations looks like (keeping symbols because we're still not ready to put in any numbers)?

 

[J-dizzal]

If you sum all of the equations, you won't be obtaining a "net force". A net force is an overall force acting on a single object, not a collection of separate objects. The right-hand sides of each of your equations are the net or total forces acting on the corresponding mass. Can you show me what your summation of equations looks like (keeping symbols because we're still not ready to put in any numbers)?

the summation should be equal to the tension above box B point up?

 

[Kinta]

the summation should be equal to the tension above box B point up?

Why? What's leading you to that conclusion?

 

[J-dizzal]

Why? What's leading you to that conclusion?

sum of force from B and C. then subract A from B+C would give the net force?

 

[J-dizzal]

the problem is i don't know the tension between B and C and between A and B

 

[Kinta]

sum of force from B and C. then subract A from B+C would give the net force?

You shouldn't be looking for any net forces. You already have the three net forces acting on each of the respective masses, (not explicitly because you don't know the tensions yet, but you should see that you don't need the tension forces to find a).