Two blocks with masses M_1 and M_2 hang one under the other. For this problem, take the positive direction to be upward, and use g for the magnitude of the acceleration due to gravity. (Figure 1)

the blocks are now accelerating upward (due to the tension in the strings) with acceleration of magnitude a

Find T_1, the tension in the upper rope.
Express your answer in terms of some or all of the variables M_1, M_2, a, and g.

Question

Two blocks with masses M_1 and M_2 hang one under the other. For this problem, take the positive direction to be upward, and use g for the magnitude of the acceleration due to gravity. (Figure 1)

the blocks are now accelerating upward (due to the tension in the strings) with acceleration of magnitude a

Find T_1, the tension in the upper rope.
Express your answer in terms of some or all of the variables M_1, M_2, a, and g.

anonymous · Answer

figure 1 : http://session.masteringphysics.com/problemAsset/1010969/28/MLD_2l_1.jpg $$T _{2}=M _{2}a + M _{2}g$$

anonymous · Answer

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Why is A not the same as B if the string is massless?

anonymous · Answer

$T_2$ is simply $m_2g$. $T_1$ is greater than $T_2$ because $T_1$ holds more blocks. $T_1=m_1g+m_2g $

anonymous · Answer

the correct answer involves the variable a. I forgot to mention the blocks are now accelerating upward (due to the tension in the strings) with acceleration of magnitude a

anonymous · Answer

then yeah, T2 = m2g+m2a

anonymous · Answer

So I'm trying to find T_1 while T_2 = M2g + M2a

anonymous · Answer

T1 = m1g +m1a +m2g +m2a

anonymous · Answer

Could you explain how you got that?

anonymous · Answer

like... with equations?

anonymous · Answer

Yeah I don't understand why T_1 can't equal M2a+M2g+M1a

anonymous · Answer

T1 -m1g -T2 = m1a
T2 -m2g = m2a

anonymous · Answer

sketch an FBD

anonymous · Answer

Ah I see now thank you.

anonymous · Answer

two forces act downward on the mass m1...  m1g and T2
T1 acts upwards...
sum of forces = mass*acceleration

anonymous · Answer

sure:)