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

[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).

 

[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.