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

[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

 

[J-dizzal]

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

so only by inspection of the direction of your tension would you know the sign of the tension force?

 

[J-dizzal]

r(t) = r₀+v₀+(1/2)at²
this is the only equation i get for the motion equations for the exam. I am trying to derive an equation for distance box B moves in the time 0.3s

 

[J-dizzal]

r(t) = r₀+v₀+(1/2)at²
this is the only equation i get for the motion equations for the exam. I am trying to derive an equation for distance box B moves in the time 0.3s

ok r0 and v0 cancel. plug in the rest. thanks for help Kinta!

 

[Kinta]

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.

so only by inspection of the direction of your tension would you know the sign of the tension force?

I think the sign of the tension force that you obtain for the cord between two blocks will depend on which equation you use to solve for it. Regardless, the tension everywhere in the cord between two blocks is equal in magnitude (meaning disregarding the sign).

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

This is incorrect. In writing $T_{CtoB} = m_Ca$, you're saying that tension is the only force acting on block C. The acceleration a is the net acceleration of each of the blocks, period. It is the result of ALL forces involved ,not just the tension and not just gravity, all of them.

 

[Kinta]

ok r0 and v0 cancel. plug in the rest. thanks for help Kinta!

My pleasure! :)

 

[J-dizzal]

I think the sign of the tension force that you obtain for the cord between two blocks will depend on which equation you use to solve for it. Regardless, the tension everywhere in the cord between two blocks is equal in magnitude (meaning disregarding the sign).

This is incorrect. In writing $T_{CtoB} = m_Ca$, you're saying that tension is the only force acting on block C. The acceleration a is the net acceleration of each of the blocks, period. It is the result of ALL forces involved ,not just the tension and not just gravity, all of them.

ok i get it.

 

[J-dizzal]

ok i get it.

ok next homework problem, coming soon...hehe