Tension Exercise on frictionless inclined plane

[Kernul]

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)?

Homework Equations

Tension
Newton's Second Law

The Attempt at a Solution

First thing I drew the free body diagram this way:

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?

 

[Elvis 123456789]

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?

 

[nrqed]

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.

 

[Kernul]

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.

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?

 

[nrqed]

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!