Tension Exercise on frictionless inclined plane

AI Thread Summary
The discussion revolves around analyzing a system of two masses connected by a string over a frictionless pulley. The participants focus on drawing free body diagrams and applying Newton's Second Law to derive equations for both masses. A key point raised is the need to distinguish between the accelerations of the two masses, as they are linked by the string but can be analyzed using different coordinate systems. The correct tension in the string and the acceleration of the masses are derived, leading to the calculation of the final velocity of the blocks after mass m2 has fallen a height H. The final consensus confirms the approach and calculations are correct.
Kernul
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Homework Statement


A mass ##m_1## is attached to a second mass ##m_2 > m_1## by an Acme (massless, unstretchable) string. ##m_1## sits on a frictionless inclined plane at an angle ##\theta## with the horizontal; ##m_2## is hanging over the high end of the plane, suspended by the taut string from an Acme (frictionless, massless) pulley. At time ##t = 0## both masses are released from rest.
a) Draw the force/free body diagram for this problem.
b) Find the acceleration of the two masses.
c) Find the tension T in the string.
d) How fast are the two blocks moving when mass m2 has fallen a height H (assuming that m1
hasn’t yet hit the pulley)?
Immagine.png


Homework Equations


Tension
Newton's Second Law

The Attempt at a Solution


First thing I drew the free body diagram this way:
photo_2016-12-19_19-15-27.jpg

Now I write the forces acting on both masses with Newton's Second Law, knowing that the accelerations of both masses are the same, so ##a = a_1 = a_2##.
First mass:
$$\begin{cases}
F_{1 x} = m_1 a_x = T - m_1 g sin \theta \\
F_{1 y} = m_1 a_y = N - m_1 g cos \theta = 0
\end{cases}$$
Second mass:
$$\begin{cases}
F_{2 x} = m_2 a_x = m_2 g sin \theta - T sin \theta \\
F_{2 y} = m_2 a_y = m_2 g cos \theta - T cos \theta
\end{cases}$$
Now what I don't understand is how I find ##a_y##. Because the only way for ##m_1 a_y = 0## is that ##a_y = 0## but this would go in contrast with the fact that in the second mass ##a_y## is something, since the mass moves along the y-axis. Or it is ##0## because the mass moves at a constant speed? In that case I would have these:
First mass:
$$\begin{cases}
a_x = \frac{T}{m_1} - g sin \theta \\
N = m_1 g cos \theta
\end{cases}$$
Second mass:
$$\begin{cases}
a_x = g sin \theta - \frac{T}{m_2} sin \theta \\
m_2 g cos \theta = T cos \theta
\end{cases}$$
With ##m_2 g cos \theta = T cos \theta## becoming ##T = m_2 g##.
Is this way correct?
 
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Kernul said:

Homework Statement


A mass ##m_1## is attached to a second mass ##m_2 > m_1## by an Acme (massless, unstretchable) string. ##m_1## sits on a frictionless inclined plane at an angle ##\theta## with the horizontal; ##m_2## is hanging over the high end of the plane, suspended by the taut string from an Acme (frictionless, massless) pulley. At time ##t = 0## both masses are released from rest.
a) Draw the force/free body diagram for this problem.
b) Find the acceleration of the two masses.
c) Find the tension T in the string.
d) How fast are the two blocks moving when mass m2 has fallen a height H (assuming that m1
hasn’t yet hit the pulley)?
View attachment 110562

Homework Equations


Tension
Newton's Second Law

The Attempt at a Solution


First thing I drew the free body diagram this way:
View attachment 110563
Now I write the forces acting on both masses with Newton's Second Law, knowing that the accelerations of both masses are the same, so ##a = a_1 = a_2##.
First mass:
$$\begin{cases}
F_{1 x} = m_1 a_x = T - m_1 g sin \theta \\
F_{1 y} = m_1 a_y = N - m_1 g cos \theta = 0
\end{cases}$$
Second mass:
$$\begin{cases}
F_{2 x} = m_2 a_x = m_2 g sin \theta - T sin \theta \\
F_{2 y} = m_2 a_y = m_2 g cos \theta - T cos \theta
\end{cases}$$
Now what I don't understand is how I find ##a_y##. Because the only way for ##m_1 a_y = 0## is that ##a_y = 0## but this would go in contrast with the fact that in the second mass ##a_y## is something, since the mass moves along the y-axis. Or it is ##0## because the mass moves at a constant speed? In that case I would have these:
First mass:
$$\begin{cases}
a_x = \frac{T}{m_1} - g sin \theta \\
N = m_1 g cos \theta
\end{cases}$$
Second mass:
$$\begin{cases}
a_x = g sin \theta - \frac{T}{m_2} sin \theta \\
m_2 g cos \theta = T cos \theta
\end{cases}$$
With ##m_2 g cos \theta = T cos \theta## becoming ##T = m_2 g##.
Is this way correct?
I can't really see the details of your diagram too well, but why are you assuming that block 2 has an acceleration in the x-direction?
 
Kernul said:

Homework Statement


A mass ##m_1## is attached to a second mass ##m_2 > m_1## by an Acme (massless, unstretchable) string. ##m_1## sits on a frictionless inclined plane at an angle ##\theta## with the horizontal; ##m_2## is hanging over the high end of the plane, suspended by the taut string from an Acme (frictionless, massless) pulley. At time ##t = 0## both masses are released from rest.
a) Draw the force/free body diagram for this problem.
b) Find the acceleration of the two masses.
c) Find the tension T in the string.
d) How fast are the two blocks moving when mass m2 has fallen a height H (assuming that m1
hasn’t yet hit the pulley)?
View attachment 110562

Homework Equations


Tension
Newton's Second Law

The Attempt at a Solution


First thing I drew the free body diagram this way:
View attachment 110563
Now I write the forces acting on both masses with Newton's Second Law, knowing that the accelerations of both masses are the same, so ##a = a_1 = a_2##.
First mass:
$$\begin{cases}
F_{1 x} = m_1 a_x = T - m_1 g sin \theta \\
F_{1 y} = m_1 a_y = N - m_1 g cos \theta = 0
\end{cases}$$
Second mass:
$$\begin{cases}
F_{2 x} = m_2 a_x = m_2 g sin \theta - T sin \theta \\
F_{2 y} = m_2 a_y = m_2 g cos \theta - T cos \theta
\end{cases}$$
Now what I don't understand is how I find ##a_y##. Because the only way for ##m_1 a_y = 0## is that ##a_y = 0## but this would go in contrast with the fact that in the second mass ##a_y## is something, since the mass moves along the y-axis. Or it is ##0## because the mass moves at a constant speed? In that case I would have these:
First mass:
$$\begin{cases}
a_x = \frac{T}{m_1} - g sin \theta \\
N = m_1 g cos \theta
\end{cases}$$
Second mass:
$$\begin{cases}
a_x = g sin \theta - \frac{T}{m_2} sin \theta \\
m_2 g cos \theta = T cos \theta
\end{cases}$$
With ##m_2 g cos \theta = T cos \theta## becoming ##T = m_2 g##.
Is this way correct?
you do not have to use the same coordinate system for the two masses. You may use a vertical y-axis for mass 2 in which case there are only forces along y for mass 2 and the equation is simple. The key point is to use indices to distinguish the accelerations of the two masses, you should use ##a_{x1}, a_{y1}, a_{x2},a_{y2}##. You should notice that then the values of ##a_{x1}## and ##a_{y2}## are not independent, they are related by a simple equation.
 
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Elvis 123456789 said:
I can't really see the details of your diagram too well, but why are you assuming that block 2 has an acceleration in the x-direction?
It's because I used the same diagram for both masses, so I end up having the second mass moving on the x-axis too.

nrqed said:
you do not have to use the same coordinate system for the two masses. You may use a vertical y-axis for mass 2 in which case there are only forces along y for mass 2 and the equation is simple. The key point is to use indices to distinguish the accelerations of the two masses, you should use ##a_{x1}, a_{y1}, a_{x2},a_{y2}##. You should notice that then the values of ##a_{x1}## and ##a_{y2}## are not independent, they are related by a simple equation.
But I've been taught that since the two masses are linked by a string, we should use one single coordinate system for both.
Anyway, following your way, the first one remains the same and the second one becomes like this:
$$\begin{cases}
a_{x 2} = 0 \\
a_{y 2} = g - \frac{T}{m_2}
\end{cases}$$
Being not independent because of the string, ##a_{x 1}## and ##a_{y 2}## are equal and so we have:
$$\frac{T}{m_1} - g sin \theta = g - \frac{T}{m_2}$$
With some passages I end up with:
$$T = \frac{m_1 m_2}{m_1 + m_2} g (1 + sin \theta)$$
Substituting this into one of the two I get the acceleration and substituting it into this equation ##v_f = \sqrt{2 a H}## I get the last point of the exercise.
Am I right?
 
Kernul said:
It's because I used the same diagram for both masses, so I end up having the second mass moving on the x-axis too.But I've been taught that since the two masses are linked by a string, we should use one single coordinate system for both.
Anyway, following your way, the first one remains the same and the second one becomes like this:
$$\begin{cases}
a_{x 2} = 0 \\
a_{y 2} = g - \frac{T}{m_2}
\end{cases}$$
Being not independent because of the string, ##a_{x 1}## and ##a_{y 2}## are equal and so we have:
$$\frac{T}{m_1} - g sin \theta = g - \frac{T}{m_2}$$
With some passages I end up with:
$$T = \frac{m_1 m_2}{m_1 + m_2} g (1 + sin \theta)$$
Substituting this into one of the two I get the acceleration and substituting it into this equation ##v_f = \sqrt{2 a H}## I get the last point of the exercise.
Am I right?
Looks good!
 
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