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 Kinematics and Dynamics Kinematics and Dynamics Physics Help Forum Sep 1st 2008, 01:54 PM #1 Senior Member   Join Date: Apr 2008 Posts: 815 [SOLVED] Free body diagram problem Time to give up for me... I've been thinking for days on this exercise and yesterday I could solve a part from it. It is asked to determine what is the acceleration of the blocks and what is the tension of the cord. Then it is asked to solve for $\displaystyle m_1=0.2 kg$, $\displaystyle m_2=0.18 kg$, $\displaystyle \alpha=30°$ and $\displaystyle \beta=60°$. I've found that the acceleration of the blocks is $\displaystyle a= \frac{m_2g \sin \beta -m_1g \sin \alpha}{m_1+m_2}$. But I'm stuck to find a general formula for the tension of the cord. I've been into the wrong direction and found that it is worth $\displaystyle m_1g \sin \alpha +m_2g \sin \beta$ but I know it's not true because what I did was wrong. I detail a bit more : I added the 2 tensions that each block create on each other, but in each case I was affirming that the system was not moving. ($\displaystyle a=0$). I'm not able to figure out the tension if the blocks are moving. __________________ Isaac If the problem is too hard just let the Universe solve it.   Sep 1st 2008, 03:37 PM   #2

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 Originally Posted by arbolis Time to give up for me... I've been thinking for days on this exercise and yesterday I could solve a part from it. It is asked to determine what is the acceleration of the blocks and what is the tension of the cord. Then it is asked to solve for $\displaystyle m_1=0.2 kg$, $\displaystyle m_2=0.18 kg$, $\displaystyle \alpha=30°$ and $\displaystyle \beta=60°$. I've found that the acceleration of the blocks is $\displaystyle a= \frac{m_2g \sin \beta -m_1g \sin \alpha}{m_1+m_2}$. But I'm stuck to find a general formula for the tension of the cord. I've been into the wrong direction and found that it is worth $\displaystyle m_1g \sin \alpha +m_2g \sin \beta$ but I know it's not true because what I did was wrong. I detail a bit more : I added the 2 tensions that each block create on each other, but in each case I was affirming that the system was not moving. ($\displaystyle a=0$). I'm not able to figure out the tension if the blocks are moving.
Good job on the acceleration. Notice that the string is (presumably) ideal, so the tension along it at any point will be the same. From the FBD on m1 (you can do a similar problem for m2) we know that
$\displaystyle -m_1g~sin( \alpha ) + T = m_1 a$
and we know that
$\displaystyle a= \frac{m_2g \sin \beta -m_1g \sin \alpha}{m_1+m_2}$

So....
$\displaystyle -m_1g~sin( \alpha ) + T = m_1 \left ( \frac{m_2g \sin \beta -m_1g \sin \alpha}{m_1+m_2} \right )$

Solving this is trivial, as is the simplification as none can really be made. So from here it's "plug'n'chug."

I'm curious how you got the acceleration without recognizing that the tensions were equal. There is a way to do this without considering tensions, but it is a much more complicated problem to visualize and I'm assuming you didn't do the problem in this way.

-Dan
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See the forum rules here.   Sep 1st 2008, 06:58 PM   #3
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Thanks a lot topsquark, you are encouraging.
I'll check out the details tomorrow, but I think I understood what you did.
At first I was trying to solve the problem with a single Cartesian graph with a unique origin. But I read in the book "Fundamentals of Physics" that I must separate the problem into 2 "systems", so I forgot about my first idea. For me it looked much more simple putting all the equations with respect to the same reference system. Especially because when I have vectors, say for example $\displaystyle P_1=3 i + 7j$ and $\displaystyle P_2=\sin \zeta 3i$ and that I want to sum up these vectors, if they are expressed into two different reference systems I can't add them that easily while if they are expressed with respect to the same referential system it's just a common addition. I'm not sure you can understand what I'm trying to say...
 I'm curious how you got the acceleration without recognizing that the tensions were equal. There is a way to do this without considering tensions, but it is a much more complicated problem to visualize and I'm assuming you didn't do the problem in this way.
Okay, checking what I did : "The acceleration is only occurring in the x-axis. So I get that the force that matters to the movement of the bodies is only acting on the x-axis. $\displaystyle \sum F_x=a_xm \Leftrightarrow a(m_1+m_2)=m_2g\sin \beta -m_1g\sin \alpha \Leftrightarrow a=\frac{m_2g\sin \beta -m_1g\sin \alpha}{m_1+m_2}$." Here note that the x-axis from the 1st body is not the same that the x-axis from the 2nd body. In other words I followed what all students do : separate the problem into 2 separated Free Body Diagram, as you did. Is it clear?
When I assume that the acceleration is only occurring along the x-axis it is because I know that the bodies won't leave out the triangle, otherwise they would be flying and the normal force would then disappear and we would fall into a physically impossibility.

EDIT: Ah topsquark, I know that "the tensions were equal". In fact I wrote that $\displaystyle T_1+T_2=T$ where $\displaystyle T$ is the total tension in the cord and $\displaystyle T_1$ is the tension created by the first body while $\displaystyle T_2$ is the tension created by the second one. T would be the sum of the 2 tensions which is also the total tension in the cord at any point in it. But I know I didn't do well on that part, so that's why I'll check out what you answered to me.

EDIT2: Oh... you are saying that in my example $\displaystyle T_1$ would be equal to $\displaystyle T_2$? If so... then I don't really understand why, but I will investigate.
But now my question is : Do my calculus of the acceleration is wrong because I neglected the tensions? It's like I assumed that they would cancel out... without knowing it. Ah yes... I can't do what I did. I must show that $\displaystyle T_1=T_2$ before doing what I did.
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Last edited by arbolis; Sep 1st 2008 at 07:13 PM.   Sep 2nd 2008, 04:01 PM   #4

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 Originally Posted by arbolis Thanks a lot topsquark, you are encouraging. I'll check out the details tomorrow, but I think I understood what you did. At first I was trying to solve the problem with a single Cartesian graph with a unique origin. But I read in the book "Fundamentals of Physics" that I must separate the problem into 2 "systems", so I forgot about my first idea. For me it looked much more simple putting all the equations with respect to the same reference system. Especially because when I have vectors, say for example $\displaystyle P_1=3 i + 7j$ and $\displaystyle P_2=\sin \zeta 3i$ and that I want to sum up these vectors, if they are expressed into two different reference systems I can't add them that easily while if they are expressed with respect to the same referential system it's just a common addition. I'm not sure you can understand what I'm trying to say... And for about Okay, checking what I did : "The acceleration is only occurring in the x-axis. So I get that the force that matters to the movement of the bodies is only acting on the x-axis. $\displaystyle \sum F_x=a_xm \Leftrightarrow a(m_1+m_2)=m_2g\sin \beta -m_1g\sin \alpha \Leftrightarrow a=\frac{m_2g\sin \beta -m_1g\sin \alpha}{m_1+m_2}$." Here note that the x-axis from the 1st body is not the same that the x-axis from the 2nd body. In other words I followed what all students do : separate the problem into 2 separated Free Body Diagram, as you did. Is it clear? When I assume that the acceleration is only occurring along the x-axis it is because I know that the bodies won't leave out the triangle, otherwise they would be flying and the normal force would then disappear and we would fall into a physically impossibility. EDIT: Ah topsquark, I know that "the tensions were equal". In fact I wrote that $\displaystyle T_1+T_2=T$ where $\displaystyle T$ is the total tension in the cord and $\displaystyle T_1$ is the tension created by the first body while $\displaystyle T_2$ is the tension created by the second one. T would be the sum of the 2 tensions which is also the total tension in the cord at any point in it. But I know I didn't do well on that part, so that's why I'll check out what you answered to me. EDIT2: Oh... you are saying that in my example $\displaystyle T_1$ would be equal to $\displaystyle T_2$? If so... then I don't really understand why, but I will investigate. But now my question is : Do my calculus of the acceleration is wrong because I neglected the tensions? It's like I assumed that they would cancel out... without knowing it. Ah yes... I can't do what I did. I must show that $\displaystyle T_1=T_2$ before doing what I did.
You actually did the problem in the more conceptually difficult way. But you did it correctly, which is more than I can say for most people. Effectively you used the fact that the pulley does nothing more than change the direction of the forces. So you "unwrapped" the pulley to put the whole problem in a straight line, which you called the x axis. This can be difficult to visualize correctly. And in this form we don't need to worry about the tensions because the tension on m1 is in the opposite direction to the tension on m2, so when you add them vectorally they cancel out of the problem. The downside is that you cannot calculate the tension using this method because of the cancellation.

The way I would encourage my students to do this problem would be using a FBD for each mass. The trick is to assume a direction for the acceleration of the masses and stick to it. That is to say if m1 is accelerating downward then m2 is going to have to be accelerating upward. The rest is applying Newton's 2nd.

Let m2 be accelerating down the slope. For the FBD for m1 I would choose +x up the slope (I almost always choose +x in the direction of the acceleration) and +y out of the slope. So we get
$\displaystyle \sum F_x = -w_1~sin( \alpha ) + T_1 = m_1 a$

Similarly for the FBD for m2 I have +x down the slope. Then
$\displaystyle \sum F_x = w_2~sin( \beta ) - T_2 = m_2 a$

Since the string is ideal we have $\displaystyle T_1 = T_2 = T$ and, of course $\displaystyle w_1 = m_1 g$ and $\displaystyle w_2 = m_2 g$. If we add these two equations the tension cancels out of the system and we can easily solve for a and get your result. Then we can plug the acceleration back into either of these to get T.

-Dan
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See the forum rules here.   Sep 5th 2008, 09:40 AM #5 Senior Member   Join Date: Apr 2008 Posts: 815 Thank you very much topsquark! I'm sorry if I had not answered before but I lost my internet connection during the last days. During all this time I solved some problems and now I'm much more used to this kind of problem than before. (I'm very glad about that, now I'm not that afraid when I face a problem!) And about your recommendation about how to solve these kind of problems, I totally agree with you. I should have written all the forces acting on each body and not to neglect the tension since I was asked to find it. __________________ Isaac If the problem is too hard just let the Universe solve it.   Sep 5th 2008, 01:51 PM   #6

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 Originally Posted by arbolis Thank you very much topsquark! I'm sorry if I had not answered before but I lost my internet connection during the last days. During all this time I solved some problems and now I'm much more used to this kind of problem than before. (I'm very glad about that, now I'm not that afraid when I face a problem!) And about your recommendation about how to solve these kind of problems, I totally agree with you. I should have written all the forces acting on each body and not to neglect the tension since I was asked to find it.
There's nothing wrong with doing it the other way. It depends on your skill level and what you want out of the problem. If, for example, all you wanted was the acceleration then your original way is much more efficient. -Dan
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