Variable angle modified Atwood's machine I am an AP Physics teacher and am looking to do a conceptual lab wherein I have a near frictionless car (with a force sensor to measure the tension T) on a track with a pulley, the string attached to which is also attached to a motorized constant velocity car (it has significantly more mass than the frictionless car) on the table about 1 foot below the track.  This causes the angle between the string and the horizontal to decrease over time as the car moves forward.  I want my students to investigate the relationship between $\frac {d\theta}{dt}$ and $\frac {dT}{dt}$.  I think the math may be beyond their capabilities, but I think they can get a good qualitative understanding of the relationship.  What I like about this scenario is that the answer is intuitive (as $\theta$ decreases, T increases), but the physics of why provoke deep analysis.
My questions are:

*

*when the CV car is moving (not when is speeding up), is the thrust force (F) due to static friction very small because the internal resistance forces are small?


*can I consider the CV car to provide a constant force as the angle changes, despite the increasing tension T (and therefore increasing backwards component of tension)?


*how can I relate F & T in a free-body diagram of the CV car while it is moving (as a corollary, does F increase as the $T_x$ component increases to compensate and keep the car moving at constant velocity or would the velocity drop slightly as the car progresses)?
The problem I am encountering is how are the thrust force F, angle, and tension T related.  My first thought was $F = T cos \theta$, but this doesn't seem to fit when $\theta$ = 90º.  I suspect this has to do with my FBD according to the questions I asked above.  Any help would be very much appreciated!
Edit:  I added a picture as requested.  This is not quite the same, as I am simply going to prop up the track with lab jacks and have the CV car on the table rather than the floor, but the general principle is the same.
 A: Let's choose coordinates first. Call $x_1$ the horizontal position of the upper mass ($m_1$) relative to some arbitrary zero point. Similarly, call $x_2$ the horizontal position of the CV cart, whose mass we'll denote by $m_2$. Finally, let $\theta$ be the angle the string makes with the horizontal. Newton's $2^\text{nd}$ Law for the two particles then reads
$$
\begin{align}
T = m_1\ddot{x_1} \\
F - T\cos\theta = m_2\ddot{x_2} \tag{1}\\
T\sin\theta + N = m_2g
\end{align}
$$
Here, $T$ is the tension in the string and $F$ and $N$ are the static friction and normal forces, respectively, exerted on the CV cart's wheels by the floor.
Notice that $\dot{x_1}$ equals the component of the CV cart's velocity along the radial line from the pulley to it, i.e.,
$$\dot{x_1} = \dot{x_2}\cos\theta.$$
So
$$\ddot{x_1} = -\dot{x_2}\dot{\theta}\sin\theta + \ddot{x_2}\cos\theta.$$
Supposing that the CV cart does, indeed, move at constant velocity, $\ddot{x_2} = 0$, so $\ddot{x_1} = -\dot{x_2}\dot{\theta}\sin\theta$, and Eqs. (1) become
$$
\begin{align}
T = -m_1\dot{x_2}\dot{\theta}\sin\theta \\
F = T\cos\theta \\
T\sin\theta + N = m_2g
\end{align}
$$
Now the question becomes, how large is the static friction force? All we know is its maximum possible value, $F^\text{max} = \mu N$, where $\mu$ is the coefficient of static friction between the wheels and the floor. When $F = F^\text{max}$, the CV cart's wheels are just about to lose traction, and it's velocity is just about to start changing. To learn more about this breakdown point, we can substitute $F = \mu N$ into the equations of motion. After some algebra, we get
$$-\dot{x_2}\dot{\theta}\sin\theta\left(\sin\theta + \frac{1}{\mu}\cos\theta\right) = \frac{m_2g}{m_1}. \tag{2}$$
Something you might want to do with your students is check the limiting cases. For instance, when $m_2 << m_1$, Eq. (2) reduces to
$$\tan\theta \approx -\frac{1}{\mu}.$$
This would require that $\theta$ be outside the range $(0, \pi/2)$. But the setup mandates that $\theta$ live inside this range, so in the case when the CV cart is much lighter than the upper mass, we never have to worry about reaching the friction breakdown point.
But the CV cart might start to slip for a different reason. The static friction force $F$ exerts a torque on the cart's wheels. If that torque becomes too large, the cart's engine won't be able to counteract it, and the angular velocity of the wheels will start to change.
I hope my analysis answers some of your questions.
