m1a = T-m1g (1) m2a = m2g -T (2) Solve (1) for a gives a = (T/m1) - g So now tak

Question

m1a = T-m1g (1)
m2a = m2g -T (2)

Solve (1) for a gives
a = (T/m1) - g

So now take 1 and insert it in 2.

so m2 (T/m1 -g) = m2g - T I think I'm okay up to here.

My first thought would be to multiply both sides by m1 to get it out of the fraction.

m2(T - g)= m2m1g- Tm1

Now divide m2 to get it to the right

T - g = m2m1g - Tm1/m2

Now add g to both sides to get T alone

T = (m2m1g - Tm1/m2) + g

Can I multiply both sides of the right by m2 to get it out of the denominator?

That would give 2m2m1g - Tm1 + g

which still is not right.

The correct form is
T = (2m1m2g)/m2 + m1

Thanks for the help!

Explanation / Answer

m1a = T-m1g T = m1(a+g) (1) m2a = m2g -T a = g-(T/m2) from (1) T = m1(g - T/m2 + g ) T = 2gm1 -(Tm1/m2) T + (Tm1/m2) = 2gm1 (Tm2 + Tm1)/m2 = 2gm1 T(m1+m2) = 2gm1m2 T = 2gm1m2/(m1+m2)

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