Understanding Tension, Weight, and Forces: A Scientific Inquiry

AI Thread Summary
The discussion centers on the equation T = m(g - a) in the context of forces acting on mass m. One participant questions whether it should be T = m(g + a), suggesting a misunderstanding of the signs associated with gravitational and acceleration forces. Clarification is provided that T = m(g - a) is correct if g and a are oriented oppositely. Another participant argues that the orientation of acceleration should consistently align with tension and gravity, implying that absolute values should be uniformly applied. The conversation highlights the importance of consistent vector orientation in force equations.
soljaragz
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There is a problem I am looking at and in its explanations for the answer it says

"...We know the sum of forces acting on m is T-mg which is equal to ma. Therefore, T=m(g-a)..."

um...Shouldn't T=m(g+a)?
 
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soljaragz said:
There is a problem I am looking at and in its explanations for the answer it says

"...We know the sum of forces acting on m is T-mg which is equal to ma. Therefore, T=m(g-a)..."

um...Shouldn't T=m(g+a)?
Yes, provided the signs of g and a are opposite. This is explicit in T = m(g-a), where g and a are the magnitudes of the vectors \vec{g} \text{ and } \vec{a}.

AM
 
Doesn't make sense to me. a should be oriented so that + is in the direction of the tension and - is in the direction of gravity. It shouldn't be an absolute value. Anyway if they want to use it as an absolute value in the second part they should have been doing that in the first part.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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