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NoPhysicsGenius
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I am having difficulty solving Problem 105 of Chapter 4 from Physics for Scientists and Engineers, 4th edition by Paul A. Tipler:
Though the problem doesn't say so explicity, we may assume that the pulley is massless and frictionless.
For those who are unfamiliar with what an Atwood's machine looks like, try browsing the following link, which shows how to solve this problem in the absence of acceleration:
Calling the two masses in the Atwood's machine [itex]m_1[/itex] and [itex]m_2[/itex], calling their respective accelerations (which are not identical) [itex]a_1[/itex] and [itex]a_2[/itex], and assuming that [itex]m_2 > m_1[/itex], I can easily obtain the following two equations for the tension, [itex]T[/itex]:
[tex]F_{ym1} = m_1a_1 = T - m_1g[/tex]
[tex]\Rightarrow T = m_1a_1 + m_1g[/tex]
[tex]F_{ym2} = m_2a_2 = m_2g - T[/tex]
[tex]\Rightarrow T = m_2g - m_2a_2[/tex]
However, I haven't the slightest clue what the third equation is, though it obviously involves the acceleration [itex]a[/itex].
Can someone please help?
The answers for the two accelerations and the tension as given in the back of the book are:
[tex]a_1 = \frac{m_2 - m_1}{m_1 + m_2}g + \frac{2m_2}{m_1 + m_2}a[/tex]
[tex]a_2 = \frac{m_1 - m_2}{m_1 + m_2}g + \frac{2m_1}{m_1 + m_2}a[/tex]
[tex]T = \frac{2m_1m_2}{m_1 + m_2}(g + a)[/tex]
I should also note that when I take the answer for [itex]a_1[/itex], plug it into my first equation for [itex]T[/itex], and simplify, I do get the book's answer for the tension [itex]T[/itex]. Therefore, I am confident that the two equations I have derived are indeed correct.
The pulley in an Atwood's machine is given an upward acceleration [tex]\overrightarrow{a}[/tex]. Find the acceleration of each mass and the tension in the string that connects them.
Though the problem doesn't say so explicity, we may assume that the pulley is massless and frictionless.
For those who are unfamiliar with what an Atwood's machine looks like, try browsing the following link, which shows how to solve this problem in the absence of acceleration:
Calling the two masses in the Atwood's machine [itex]m_1[/itex] and [itex]m_2[/itex], calling their respective accelerations (which are not identical) [itex]a_1[/itex] and [itex]a_2[/itex], and assuming that [itex]m_2 > m_1[/itex], I can easily obtain the following two equations for the tension, [itex]T[/itex]:
[tex]F_{ym1} = m_1a_1 = T - m_1g[/tex]
[tex]\Rightarrow T = m_1a_1 + m_1g[/tex]
[tex]F_{ym2} = m_2a_2 = m_2g - T[/tex]
[tex]\Rightarrow T = m_2g - m_2a_2[/tex]
However, I haven't the slightest clue what the third equation is, though it obviously involves the acceleration [itex]a[/itex].
Can someone please help?
The answers for the two accelerations and the tension as given in the back of the book are:
[tex]a_1 = \frac{m_2 - m_1}{m_1 + m_2}g + \frac{2m_2}{m_1 + m_2}a[/tex]
[tex]a_2 = \frac{m_1 - m_2}{m_1 + m_2}g + \frac{2m_1}{m_1 + m_2}a[/tex]
[tex]T = \frac{2m_1m_2}{m_1 + m_2}(g + a)[/tex]
I should also note that when I take the answer for [itex]a_1[/itex], plug it into my first equation for [itex]T[/itex], and simplify, I do get the book's answer for the tension [itex]T[/itex]. Therefore, I am confident that the two equations I have derived are indeed correct.