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

[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