A questionable statement in some FBD notes?

[etotheipi]

My sibling has been given a set of mechanics notes by her teacher; when I was looking through I saw something very suspicious...

"The direction of acceleration is not yet known so you draw the acceleration vector in a fixed direction on the diagram. You calculate its components in your coordinate system and compare these to the net force per unit mass in those directions."

Huh? I quite literally have no idea what this means. The acceleration is a physical observable, so it already has a fixed (if unknown) direction and you are not just free to fix it in some other arbitrary direction? I wondered if anyone here can make sense of this statement?

What I suspect they are saying is that the acceleration has two degrees of freedom (in 2D), so you can draw it with a magnitude and variable direction $\theta$ and work in terms of that. Needless to say that's not what they've written...

 

[sophiecentaur]

The choice of the word "fixed" is bad. If you replace it by 'arbitrary' then it starts to be more reasonable. It's what we do in many cases because the components will give the correct value for its magnitude and direction. But you have to choose to draw a line somewhere or you can't write any equations.

 

[etotheipi]

The choice of the word "fixed" is bad. If you replace it by 'arbitrary' then it starts to be more reasonable. It's what we do in many cases because the components will give the correct value for its magnitude and direction. But you have to choose to draw a line somewhere or you can't write any equations.

I think you are correct. I just found the whole thing slightly baffling since I've never had any reason to draw an acceleration vector on the FBD in the first place; it seems better to just draw forces and deduce kinematic quantities from algebra.

I guess it's not wrong, but drawing arrows in random directions seems to detract from the geometrical intuition of vectors, so I very much dislike the teacher's approach .

 

[robphy]

A related way to think about what I think @sophiecentaur is saying
is to
choose a coordinate system [choice of axes], an arbitrary choice that is fixed once and for all for this problem.

We do something similar when working on a circuit problem (since we don't usually don't know the directions of positive current flow in the circuit).

 

[etotheipi]

A related way to think about what I think @sophiecentaur is saying
is to
choose a coordinate system [choice of axes], an arbitrary choice that is fixed once and for all for this problem.

We do something similar when working on a circuit problem (since we don't usually don't know the directions of positive current flow in the circuit).

It is slightly distinct in this case however. For labelling a current, the arrow represents the direction of the surface vector within the wire i.e. $I = \vec{J} \cdot d\vec{A}$. So in that sense the numerical current $I$ is the component of the current density in the direction of $d\vec{A}$. We can let $d\vec{A}$ point in either direction, so the positive/negative versions are consistent with the arrow direction.

For a mechanics problem, we are free to choose any frame of reference and with it a coordinate system, with a set of (usually orthonormal) basis vectors. Now we want to establish the acceleration vector in this coordinate system. Its direction is currently unknown to us, as is perhaps also its magnitude. In two dimensions, this vector has two degrees of freedom.

So we may either identify it with a magnitude $a$ and an arbitrary angle $\theta$ to a specified reference (i.e. the horizontal), or in terms of two independent components $a_x$ and $a_y$. If $\theta$ is measured to $\vec{e}_x$, then of course these two representations are related via. $a_x = a\cos{\theta}$ and $a_y = a\sin{\theta}$.

The sloppiness on the teachers' part was in using the words fixed direction. Since the acceleration is an observable, uniquely defined in any frame of reference, and has a definite magnitude and direction. Drawing it in the FBD with a fixed angle, if it's direction is not previously known, is a mistake.

 

[kuruman]

Most 2-d FBDs are drawn to determine two things knowing everything but these two. These are components of forces or acceleration that go into the 2-d $F_{net} = ma$ equations. This is what I recommend. As best as one can, one draws all the forces acting on the system and surrounds them with a box to separate them from what is not the system. Then one draws $ma$ outside the box because it is not a force. If the acceleration is known, e.g. inclined plane, uniform circular motion, one knows its direction and can draw it in the right direction. If the acceleration is not known, then at worst there can be one component of a force that is unknown because one can have only two unknowns since there are two equations in $\vec F_{net} = m\vec a$. I believe that the direction of any unknown arrow in a FBD can be guessed at pretty accurately.

The circuit example applies when the acceleration is constrained in one dimension but its direction is not obvious unless numbers are substituted. With FBDs this is usually the case when kinetic friction is involved. For example you have mass $m_1$ on an incline connected with a string over a pulley to hanging mass $m_2$. The direction of the common acceleration depends on the sign of $(m_1 \sin\theta - m_2)$. Nevertheless it is either up or down the incline for $m_1$ and, respectively down or up for $m_2$. My point is that it is more appropriate to replace "arbitrary" with "educated guess" when you draw FBD arrows.

 

[vanhees71]

How can you draw a vector, that you don't know? Perhaps it makes sense if the wider context is given. Usually in a free-body diagram you draw forces, as the name suggests. Then you vectorially add all these forces (sometimes called "Newton's parallelogram rule", but that's nothing else than vector addition, expressed in a language where vectors have not yet been discovered by the matheamticians and physicists in Newton's time; to my knowledge vectors were added to the toolkit quite late, around 1890-1900 by Gibbs and Heaviside). The net force, together with the particle's mass then gives the resulting acceleration from $\vec{F}=m \vec{a}$.

 

[etotheipi]

How can you draw a vector, that you don't know?

Yes, I don't think it's a great idea, but usually the intended meaning is clear from the context.

If both the direction and magnitude of the acceleration of say, a particle, are unknown, then any acceleration vector you might 'draw on the FBD' must still have those two degrees of freedom available. That is to say that both the magnitude and angle to a specified reference must be indicated with variable parameters. I think this was what the teacher was trying to say.

It would of course be non-sensical to arbitrarily "fix the acceleration vector to the rightward horizontal", and then try and assert $\vec{F} = m\vec{a}$, even if $\vec{F}$ pointed upwards, or somthing. This would just be incorrect, but seemed to be what the teacher's wording was implying .

The most unambiguous way IMO is just to draw forces on a FBD.

 

[jbriggs444]

If one has to draw an unknown vector, one approach is to draw it component-wise as two (or three) orthogonal unknown vectors, each with a known direction.

The choice of either direction might be off by 180 degrees, but that just means that the associated component will be negative. Same as for the circuit situation.

 

[sophiecentaur]

drawing arrows in random directions seems to detract from the geometrical intuition of vectors

That's a bit like saying you would hesitate to use 'x' to denote a scalar quantity when you construct an equation for solving a physical question.
If you feel it interferes with intuition then perhaps intuition has to take second place until you have solved the magnitude and direction of the actual vector. At that stage, intuition might help you with the credibility test of your answer - ie you could spot a ludicrously wrong answer.

 

[etotheipi]

That's a bit like saying you would hesitate to use 'x' to denote a scalar quantity when you construct an equation for solving a physical question.
If you feel it interferes with intuition then perhaps intuition has to take second place until you have solved the magnitude and direction of the actual vector. At that stage, intuition might help you with the credibility test of your answer - ie you could spot a ludicrously wrong answer.

If I had to draw on an unknown vector, yes I would be fine drawing in a set of vector components $v^i\vec{e}_i$ or an arrow with a magnitude and (N-1) arbitrary angles, since either approach provides the required N available degrees of freedom. The key part is that they must all be variable parameters.

 

[sophiecentaur]

The key part is that they must all be variable parameters.

There is no more information in two component vectors in arbitrary directions than in a single vector with two variable parameters (2D). The four parameters which describe those two component vectors are actually interdependent and reduce to two. I think we are only talking in terms of personal preference here. There is no basis for any particular choice of reference axes - only convenience, sometimes - the vector is the vector.

 

[etotheipi]

There is no more information in two component vectors in arbitrary directions than in a single vector with two variable parameters (2D). The four parameters which describe those two component vectors are actually interdependent and reduce to two. I think we are only talking in terms of personal preference here. There is no basis for any particular choice of reference axes - only convenience, sometimes - the vector is the vector.

I am in agreement, sorry I wasn't too clear! We either have N components $v^i \vec{e}_i$, or N-1 angles and 1 magnitude. If the basis is orthonormal, then the one-to-one mapping is $v^i = \vec{v} \cdot \vec{e}_i = v\cos{\theta}$. So a 2D vector has 2 degrees of freedom only, however you look at it .

 

[kuruman]

Can you or your sibling provide a specific example that illustrates how the method as suggested by her teacher is to be applied? Surely, if the teacher suggests a method, there ought to be a sample application.

 

[etotheipi]

Can you or your sibling provide a specific example that illustrates how the method as suggested by her teacher is to be applied? Surely, if the teacher suggests a method, there ought to be a sample application.

I can't remember specific examples, but I can give an indication of what the worked examples were like. If $\vec{F}$ is the net force on a particle, then they let $\vec{a} = a\cos{\theta} \hat{x} + a\sin{\theta} \hat{y}$ for some unknown $\theta$. If the resultant force was in the positive $\hat{x}$ direction, they might then write $F_x = ma\cos{\theta}$, $F_y = 0 \implies \theta = 0$, so $F_x = ma$.

It is quite convoluted, although I suppose it is a perfectly correct method. The part I was contesting was that the teacher (probably sloppily) said you draw $\vec{a}$ at some fixed angle $\theta$, whilst of course as @sophiecentaur pointed out it is much better described as "at some arbitrary/unknown angle".

 

[Ibix]

Perhaps the teacher means to think of your unknown accelerations as always in the same arbitrary direction. That does tend to make the book-keeping easier for constraints due to connections between masses.

For example, with two masses on a string running over a pulley I recall some fellow students deducing the direction of motion by inspection and proceeding from there. That is, they implicitly thought of the accelerations as being in opposite directions. They then had trouble with more complex problems because they couldn't deduce the directions of the accelerations. One recipe to avoid this is to insist that all accelerations are always upwards ("in a fixed direction") even for cases where you can see they're not and to explicitly write the constraints.

Agree that, even if I'm on track, "the same arbitrary direction", or something of that nature, would be better than "fixed".