Net force and moment on a wooden bar fixed at one end

[brotherbobby]

Homework Statement

The diagram below shows a wooden bar of mass 800 g (0.8 kg) length 0.6 m pinned at its left end. A force of 50 N is applied at it right end in the downward direction. (a) Calculate the net force and net moment on the bar about P. (b) What is the acceleration of the center of mass (CM) of the bar O about P?

Relevant Equations

Newton's law for a system : $\Sigma \vec F_E= M \vec a_{\text{CM}}$
Torque on a system about a point : $\Sigma \vec {\tau}_E = I_{\text{CM}} \vec {\alpha}$

Here is the diagram to the right.

(a) Clearly, owing the fact that the bar is pinned at P, the net force on the bar is zero: $\boxed{\Sigma \vec F = 0}$. The pin P applies an equal and opposite force to the one applied. This force keeps the bar from translating to a different position in space. (We assume tacitly that the pin P is capable of applying a force as large as necessary to keep the bar from moving in a straight line or translating).

Net moment of the bar about P : $\boxed{\Sigma \vec M = -50\times 0.6 = -30\; \text{Nm}}$. (Negative owing to the clockwise moment).

(b) Here comes my problem. According to Newton's law, the acceleration of the CM : $\color{blue}{\boxed{\vec{a}_{\text{CM}} = 0}}$, the net force on the bar being zero. But again, the angular acceleration $\vec \alpha = \frac{-30}{I}$. The moment of inertia of the bar about the CM is $I = \frac{1}{12}ML^2 = \frac{1}{12}\times 0.8\times 0.6^2 = 0.024 \; \text{kg m}^2$. Hence the angular acceleration of the bar $\vec \alpha = \frac{-30}{0.024} = -1250 \; \text{rad/s}^2$. We note that the entire bar would have the same angular acceleration $\vec \alpha$.

The acceleration of the CM is : $\vec a_{\text{CM}} = r \vec \alpha = 0.3 \times (-1250) = -375\; \text{m/s}^2$, hence $\color{red}{\boxed{\vec a_{\text{CM}} = -375\; \text{m/s}^2}}$ !

Refer to the two boxes above in colours blue and red for the same quantity : $\vec a_{\text{CM}}$

Why are they different?

 

[Cutter Ketch]

Clearly, owing the fact that the bar is pinned at P, the net force on the bar is zero: Σ→F=0ΣF→=0\boxed{\Sigma \vec F = 0}.

Then why is the center of mass accelerating?

 

[Cutter Ketch]

Question: In the diagram did you draw in that upward 50N at p, or was that given in the problem?

 

[brotherbobby]

Then why is the center of mass accelerating?

The CM accelerates using the other formula : $\vec \alpha = \Sigma \vec \tau /I$.

This is my problem. On using $\vec a = \Sigma \vec F / M$, $\vec a$ = 0.

 

[brotherbobby]

Question: In the diagram did you draw in that upward 50N at p, or was that given in the problem?

I drew the upward 50 N at P . My reason for doing that is that the bar is fixed at P. It can rotate about P, but not translate from P.

 

[Cutter Ketch]

It can rotate about P, but not translate from P

Your equations of translational and rotational motion are both simultaneously and independently true. Regardless of the rotational motion, the center of mass is accelerating. The net force is not zero.

 

[brotherbobby]

Your equations of translational and rotational motion are both simultaneously and independently true. Regardless of the rotational motion, the center of mass is accelerating. The net force is not zero.

Hmm... I forgot the weight of the beam. This weight has a moment about the pinned point on the left P.
But even if I take the weight into account, the net force has to be 0. Or else the system would translate, wouldn't it?

 

[Cutter Ketch]

Hmm... I forgot the weight of the beam. This weight has a moment about the pinned point on the left P.
But even if I take the weight into account, the net force has to be 0. Or else the system would translate, wouldn't it?

No, you don’t need to reformulate the problem. Your problem is perfectly valid without weight. Perhaps the bar and the forces are in the horizontal plane, or perhaps this is happening in free fall or deep space. Anyway, it doesn’t matter. There is nothing wrong formulating a problem with just one force and a pin. You are not obligated to say it is in a vertical plane under gravity and include the weight of the bar.

You say the net force has to be zero or the system would translate. This is true. What I am trying to explain is that the system IS translating! The center of mass is accelerating. The net forces are not zero.

Look at it another way. If your diagram were correct, if the net force was zero and there were one 50N force down on one end of the stick and one 50N force up on the other, about which point along the stick would it rotate?

 

[Lnewqban]

Another thing to consider:
For constant angular acceleration of point O to happens, the force F must remain applied in a direction perpendicular to the bar as it rotates around the pivot.
The total acceleration of point O will be formed by two components: centripetal and tangential.

 

[haruspex]

Another thing to consider:
For constant angular acceleration of point O to happens, the force F must remain applied in a direction perpendicular to the bar as it rotates around the pivot.
The total acceleration of point O will be formed by two components: centripetal and tangential.

As I read the question, we are only asked for the instantaneous acceleration at the initial position (presumably stationary). We do not care that the angular acceleration will not be constant, nor that there will later be a centripetal acceleration.

 

[brotherbobby]

Look at it another way. If your diagram were correct, if the net force was zero and there were one 50N force down on one end of the stick and one 50N force up on the other, about which point along the stick would it rotate?

The bar should rotate about its center, O. Only I don't see how, for it is pinned at its left end at P.

There is another curiosity. If you take any point along the bar, like the X I show alongside, the net moment about it is negative (clockwise). The moment is similarly negative (but different) about any point on the bar. Hence the bar must rotate in the clockwise direction.

 

[gmax137]

Whenever I write a paragraph that starts out "clearly {blah blah blah}" and then ends up with an apparent contradiction, I put "{blah blah blah}" high on my list of things to question.

 

[Cutter Ketch]

The bar should rotate about its center, O. Only I don't see how, for it is pinned at its left end at P.

Right, IF you were correct, the bar would rotate about its center. The bar does NOT rotate about its center, ergo you are incorrect.

I’m trying to tell you something here, and you don’t want to hear it. That your approach produces results that disturb you and seem contradictory should be a big red flag that something in your understanding is broken and you should listen to what others are telling you.

I’ll say it one more time. The center of mass accelerates. The sum of the forces in the translational equation is not zero. The upward reaction force of the pin is not 50N. Get rid of that error in your thinking and you will stop getting answers that bother you and seem wrong.

 

[haruspex]

The bar should rotate about its center, O. Only I don't see how, for it is pinned at its left end at P.

It depends what you mean by "rotating about" a point.

Yes, point O is the instantaneous centre of rotation, so you can think of the rod's motion as rotation about O.

You can also represent the motion as the sum of a rotation about its mass centre and a linear translation of its mass centre. This is useful since you can relate the accelerations of those to the sums of torques and forces respectively.

A third concept is angular momentum about a point. Even a body which is moving linearly, not rotating, has angular momentum about a point that is not on the line of motion through its mass centre. Conversely, if a body is moving linearly and rotating about its centre there will be a point in space about which it (instantaneously) has no angular momentum.

 

[brotherbobby]

I’ll say it one more time. The center of mass accelerates. The sum of the forces in the translational equation is not zero. The upward reaction force of the pin is not 50N. Get rid of that error in your thinking and you will stop getting answers that bother you and seem wrong.

Yes thank you. I need to think again on this question right from the start. I will get back to you with a different set of results along the lines of what you said above. Let me put those points down using mathematics where I can.

1. The center of mass accelerates : $\vec a_{\text{CM}} \neq 0$. Thus $\Sigma \vec F_e = 0$, external forces don't add up to zero.
2. The "upward" force at the pin P is not zero : $F_P \neq 0$. Of course this follows from (1) above. I ignore the weight of the bar.

Let me see what I can make of this. The trick I suppose is to begin from $F_P \neq 0$ and move forward.

The other bit is using $\vec \alpha_{\text{CM}} = \Sigma \vec \tau / I$. This will lead to $\vec a_{\text{CM}} = r \vec \alpha_{\text{CM}} = r \times \Sigma \vec \tau / I$.

I will get back to you.

 

[Cutter Ketch]

well, shoot. The formatting falls apart when I quote it, but I wanted to quote your

$a_{cm} = r \alpha_{cm}$

And say: Yes, the crucial point!

 

[haruspex]

The formatting falls apart when I quote it

Yes, it's annoying. Subscripts and superscripts using those buttons get lost when quoting too.
It's ok when you use the Reply button. Sometimes I use that then delete everything I don't want to quote.