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

[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).

 

[J-dizzal]

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).

i don't see how to get a from these 3 equations

 

[Kinta]

Show me the very first line you have when you try to add the three equations.

 

[J-dizzal]

Show me the very first line you have when you try to add the three equations.

well do i sum all 3 equations or sum B and C and subtract A?

 

[J-dizzal]

if i sum all three everything cancels out except the force from gravity for B and C which is = -578.2

 

[Kinta]

well do i sum all 3 equations or sum B and C and subtract A?

Sum all three equations.

if i sum all three everything cancels out except the force from gravity for B and C which is = -578.2

And what do you have on the left side of that equation?

 

[J-dizzal]

Sum all three equations.

And what do you have on the left side of that equation?

ma

 

[J-dizzal]

ma

three ma's

 

[Kinta]

three ma's

So long as you know the three $m$'s, it looks to me like you have an equation with only one unknown in it.

 

[J-dizzal]

oh i see now the masses just plug into the masses on the left. a = -6.3

 

[Kinta]

oh i see now the masses just plug into the masses on the left. a = -6.3

Looks good to me. Now you can get the desired tension for part (a) and the displacement of block A for part (b) relatively easily.

 

[J-dizzal]

Looks good to me. Now you can get the desired tension for part (a) and the displacement of block A for part (b) relatively easily.

realatively meaning easy for you.

 

[Kinta]

realatively meaning easy for you.

I mean, relative to the rest of the problem, getting the two desired quantities requires less work. :)

 

[J-dizzal]

I mean, relative to the rest of the problem, getting the two desired quantities requires less work. :)

T from B to C = (14kg)(-6.285 m/s/s)=87.99N why is this wrong?

 

[Kinta]

T from B to C = (14kg)(-6.285 m/s/s)=87.99N why is this wrong?

Because this statement directly contradicts one of your previous, correct equations regarding the net force on block C:

mCa=TC to B+mCg

 

[J-dizzal]

Because this statement directly contradicts one of your previous, correct equations regarding the net force on block C:

im getting -861.025 for tension between B and C

 

[J-dizzal]

T_{B to A} + T_{C to B} - T_{B to C} - M_Bg - M_Cg = 861.025
i know this is wrong but don't see why

 

[Kinta]

T_{B to A} + T_{C to B} - T_{B to C} - M_Bg - M_Cg = 861.025

You're making it too complicated and, in doing so, missing something. Try just using the equation you got when you applied Newton's 2nd Law to block C. Now that you have a, you should only have one unknown (and a particularly desirable one at that) in that equation.

 

[J-dizzal]

ok i see, -49.21 but my sign is wrong. i used 14(-(-6.3))-137.2=-49.21, but evidently it should be positive.

 

[J-dizzal]

ok i see, -49.21 but my sign is wrong. i used 14(-(-6.3))-137.2=-49.21, but evidently it should be positive.

T_{C to B}= ma= m_C(-a)

 

[J-dizzal]

T_{C to B}= ma= m_C(-a)

i must be wrong to call a negative for that equation prematurely