Tension Equation Question - Algebra Related

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The discussion centers on the tension equation T - mg = ma for an upward moving object. The user questions the disappearance of the mass variable when solving for tension, T. It is clarified that T can be expressed as T = m(g + a), which retains both mass terms when expanded. The user acknowledges the explanation and recognizes the importance of factoring in algebra. Understanding this concept is essential for solving similar physics problems.
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Just a quick question regarding the following equation for tension of an upward moving object.

T - mg = ma

According to my textbook (for this question at least), solving for T gives us...

T = m (g + a)

Looking back at the initial equation we have an m on both sides of the equal sign. But when bringing - mg to the other side and thus solving for T we then only have one m. Where did the second m go?

Thanks!
 
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T = m(a+g) = mg + ma. See. There are two ms .
 
That makes perfect sense! Thank you! :)

Forgive my ignorance, I've got a lot to brush up on with math and algebra, that is called factoring...right?
 
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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