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I got an exercise to solve but I can't understand something and need your help. First of all, I want to apologize for my English and I'll try to explain myself in the best possible way.

Diagram

As shown on the picture, we have two masses connected to each other with a mass-less rope: $$m_2=500\:\mathrm{kg}\quad\text{and}\quad m_1=200\:\mathrm{kg}$$

There is no resistance between the ground and $m_1$ or between $m_1$ and $m_2$.

Now, at time $t=0$ the $m_2$ mass starts to move downwards because of the $m_2g$ force and causes $m_1$ to move to the right. The task is to to calculate the acceleration of the two masses. From the answer in the book, they treat the tension as $2T$ and I can't understand why. I would have thought that if mass $m_2$ is moving downwards because of the $m_2g$ force, then we can set up the force equation for $m_2$ as: $$m_2g-T=ma$$ Why does the answer say $2T$?

I am looking forward to the answers. Thank you.


Here is the given full solution for finding the acceleration. I can't understand why it says 2 times $T$:

enter image description here

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  • $\begingroup$ The rope pulls on the cart ($m_1$) on two points, the attach point and the pulley. $\endgroup$ – Aetol Dec 12 '18 at 15:25
  • $\begingroup$ Yes, i understand but can't understand why 2*T, lets suppose that m2g =T = 500N for example, how its can be thath in 2 connected points there is 2*T = 1000N that effects on M1 ? @Aetol $\endgroup$ – Rasim Muradov Dec 12 '18 at 15:27
  • $\begingroup$ can show the full solution?where he takes 2*T $\endgroup$ – Abdelrhman Fawzy Dec 12 '18 at 15:28
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    $\begingroup$ RasimMuradov and @AbdelrhmanFawzy, I have updated the question with some spelling correction and added the information (the answer sheet solution) that was put as an answer below. And I will vote for deleting the answer (since it is not an answer). Rasim, I hope nothing is corrected wrongly or misunderstood in my edits to the question. Otherwise please edit it to fit or let me know. $\endgroup$ – Steeven Dec 12 '18 at 16:13
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    $\begingroup$ @Steeven I want to thank you for your attention and for fix my spelling mistakes. $\endgroup$ – Rasim Muradov Dec 12 '18 at 16:19
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Newton's 2nd law horizontally on $m_2$: $$\sum F_x=m_2 a_x\quad\Leftrightarrow\quad N=m_2a_x$$

Newton's 2nd law horizontally on $m_1$: $$\sum F_x=m_1 a_x\quad\Leftrightarrow\quad 2T-N=m_1a_x$$

There are 2 times $T$ here, because $m_1$ is being pulled in by the string both at the bottom and at the top. Plug the first equation for $N$ into this:

$$2T-N=m_1a_x\quad\Leftrightarrow\quad 2T-m_2a_x=m_1a_x \quad\Leftrightarrow\quad 2T=(m_1+m_2)a_x$$

And here it is. The doubled $T$ actually comes from $m_1$. $m_2$ is just being mixed into this equation because it touches $m_1$.

You are correct that if we only look at $m_2$ and set up Newton's 2nd law vertically on it, then we don't see a doubled $T$:

$$\sum F_y=-m_2 a_y\quad\Leftrightarrow\quad T-w=-m_2a_y\quad\Leftrightarrow\quad T-m_2g=-m_2a_y$$

Both expressions are correct at the same time.

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  • $\begingroup$ Steeven, there are 2 different accelerations for m1 and m2.. a1>a2 more than two times as you can see on the answer picture that I attached to the post and it's logical because when the mass m2 start move downwards the mass m1 moving right because of the tension and the [delta]X of the treat decrease two times from bottom and top. and its makes [Delta]Y to be 2 times [Delta]X. but i cant understand lets say for example that the force that makes this system to start moving is 500N= T how its possible to say that because 2 points that connected to M1 its 2 times T (1000N) $\endgroup$ – Rasim Muradov Dec 12 '18 at 17:08
  • $\begingroup$ We can see from the picture that T its same T for all over the lenght, Hope i explain my self right $\endgroup$ – Rasim Muradov Dec 12 '18 at 17:12
  • $\begingroup$ I just added a subscript $a_x$ and $a_y$ to all accelerations in the answer since they are different in different directions as Rasim mentions. $\endgroup$ – Steeven Dec 12 '18 at 17:15
  • $\begingroup$ @RasimMuradov If the tension is $T=500\;\mathrm N$, then because this tension pulls in two places on the same object, the total tension force on that object must naturally be $1000\;\mathrm N$. The tension is not only caused by the falling $m_2$ but also by the fixed hooks in the wall at the right side. $\endgroup$ – Steeven Dec 12 '18 at 17:18
  • $\begingroup$ You are very welcome, @RasimMuradov. Glad it helped. $\endgroup$ – Steeven Dec 12 '18 at 17:27

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