Hello all

About the question, please look at the diagram, fig 1 on the left.

http://img.photobucket.com/albums/v124/justamatur/darmon/askin.gif

I just had this question about that problem on the left (back in high school I only dealt with the one on the right).
So, suppose we have a cart moving to the right, with a mass M being suspended with a solid stick which can rotate, and with such imbalance shown will obviously bring M down. If we are to give the cart force so that M will stay on its position (at the same angle), how's the formula for the acceleration we have to give to the system? Suppose frictions are negligible.

No, I'm not in aquiz or something (heck, we'll never get any question about Dynamics here in School of Computer Engineering), but I'm just curious... Thanks.

I'm not 100% sure about this, cause I never do the sinuses and cosines right, but I would say a = g / sinα

From this, it looks like the problem may be more complex than we think. It's not the exact same situation, but looks similar enough.

The following comes from here:
http://aemes.mae.ufl.edu/~uhk/DYNAMICS.html

THE FALLING ROD PROBLEM: In class today we discussed one of the more interesting problems encoutered in dynamics, namely, the behaviour of a uniform rod initially standing vertically on a smooth floor. I summarize the governing kinetic and kinematic conditions governing the rod HERE (http://aemes.mae.ufl.edu/~uhk/ROD3.JPG). Note that the rod's angle relative to the vertical as a function of time is determined by a solution of a highly non-linear second order differential equation for theta as a function of time which can only be solved numerically. Alternatively, you can plot the square of the angular velocity versus angle directly as done HERE (http://aemes.mae.ufl.edu/~uhk/ROD4.jpg). This last result is also possible to obtain directly by use of energy methods. A numerical integration(using Runge-Kutta) for q versus time shows that it takes about 0.88 sec for the rod to hit the floor when L=1meter and the rod is started from rest at
q(0)=0.01 rad. This time increases with increasing rod length L and decreasing acceleration of gravity g.

You need to make sure that the net torque on the system around the bottom right contact point with the ground is zero, so that the thing doesn't tip over. It should be simply a matter of calulating the torques due to the two masses and the force F.

content deleted as it aimed only to inflame other posters.

At the very least thanks to superluminal for reminding me that I miss an important factor in my diagram: the length of the rod.

CURIOUCITY,
As I was expected, we never will get answer from our “experts” of anti-SRT. So, it is time to give a right solution of your problem.
1. First of all, let me start with notice made by James R:
You need to make sure that the net torque on the system around the bottom right contact point with the ground is zero, so that the thing doesn't tip over. It should be simply a matter of calulating the torques due to the two masses and the force F.

This is very important notice, because your system is much more complex than is seems to be at the quick look on it. There can be situations as shown in Fig1 and Fig.2 on this picture (http://www.sciforums.com/attachment.php?attachmentid=3767&stc=1). James R’s notice gives a recipe how to avoid the consideration of such situation. I can only add that there are possible even oscillatory motions under action of a constant force F. I guess that there even can be some kind of so called “parametric resonance” in such system: all depends upon masses, rod’s length, position of poinf O and size of body m….
2. Keeping in mind noticed 1., let us consider situation when there is no torque trying to tern our system around any of points, so that system moves straightforward as on your picture on the left. Therefore, we see the motion shown in Fig.3 of my picture, when both bodies are moving with the same acceleration a.
3. To solve your problem in this case, just as any similar problem, all what we need is to write Newton’s 2 law for each body. For that we should find all forces acting on each of bodies. (As you recognized, I mean rod rigid, but massless)
4. On the body M with mass M there act two forces: force of a gravitation, P = Mg, and reaction of rod, N=? (we do not know yet its value).As you see on picture (http://www.sciforums.com/attachment.php?attachmentid=3767&stc=1)

Ma = N*cos θ = Mg*cotan θ ………………………..(1)

It immediately gives

a = g * cotan θ …………………………………(2)
and
N = Mg/ sin θ ……………………………………….…(3)


On the body m, having mass m, acts the reaction of rob (with same value N, but directed opposite), driven force F, a gravitational force P' = mg and reaction of the table (last two forces are not shown on picture!). As you see, force N can be represented as sum of two forces: f = N * cos θ and P = Mg. Three forces -this force P, a gravitational force P' = mg and reaction of the table are compensating each other, so that they do not influent on motion with acceleration a. So, the Newton’s second law for body m will have a form:

ma = F – f = F – Ma ……………………………………(4)

What immediately gives

a = F/(M+m)……………………………………………(5)

and therefore

cotan θ = F/[g*(M+m)] ……………………………….(6)

This is full solution of your problem.
Good luck, my friend.

Thanks. I'll spend some time studying your idea and let you know whichever part I'm confused on later.

continuation of personal argument deleted.

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