Teaching Forces - Having issues with directions & vectors

[orangeblue]

Hi everyone. I'm the only physics teacher at my school, so I have nobody to bounce ideas off of. I'm having a problem with students getting confused with direction when calculating net force.

• I teach an introductory physics course. It's the first time the students have ever seen physics and they struggle a LOT at first. We start with Kinematics. To try and simplify things, I ALWAYS call up the positive direction, down negative, right positive, and left negative. The idea is to get them used to assigning a sign to direction. It usually works out fine - students mostly get it.
• In Dynamics, I start to have problems. Here's what I do:
  • I define Fnet as the SUM of the forces
  • For solving math problems, I've found that using Fnet = (bigger force) - (smaller force) works best for most scenarios. However I run into problems trying to explain why the subtraction sign is there when Fnet is a SUM.
  • I tried to tell the students we're making the direction of the bigger force the positive direction, but they just got super confused and I had a hard time explaining it.

Does anyone have a better way of teaching directions/vectors in dynamics? I feel like I'm losing my mind...

 

[kuruman]

I think you should stick to your adopted convention for signs and not worry about the magnitude of Fnet, which is what you're doing. Define Fnet always as the vector sum of all the forces. If, when you do the math, it turns out that Fnet is negative, then it points to the left (or down) and so does the acceleration. It's much cleaner that way.

 

[orangeblue]

So I agree that it would be cleaner that way; however, I tend to run into problems when teaching Atwood machines (two masses, connected by a string over a pulley).

The force of gravity on each mass is DOWN, but when we do Fnet we have to make one Fg positive and the other negative for it to work.

How would I explain that to the students? I feel it causes issues when I've been harping so much on calling down negative. I guess I just need to say pulleys change the direction of the force, essentially?

 

[kuruman]

So I agree that it would be cleaner that way; however, I tend to run into problems when teaching Atwood machines (two masses, connected by a string over a pulley).

The force of gravity on each mass is DOWN, but when we do Fnet we have to make one Fg positive and the other negative for it to work.

How would I explain that to the students? I feel it causes issues when I've been harping so much on calling down negative. I guess I just need to say pulleys change the direction of the force, essentially?

Yes, an ideal pulley changes the direction of the tension but does not affect its magnitude. Also, note that Fnet is undetermined unless you specify the system on which it acts. To be specific, in the Atwood machine case there are 3 Fnet's possible depending on what you choose as your system
1. Smaller mass: Fnet is positive (up) because the tension (up) is greater than the weight (down). The equation that follows from this is$$F_{net~1}=T-m_1g=m_1a_1$$
2. Larger mass: Fnet is negative (down) because the tension (up) is less than the weight (down). The equation that follows from this is$$F_{net~2}=T-m_2g=m_2a_2$$
Note that the net forces and accelerations are algebraic quantities, they can be positive or negative. However, the magnitudes of the two accelerations are equal, but their directions are opposite which is another way of saying that the string is inextensible. Because the smaller mass is accelerating up and the larger mass down, we write $a_1=+a$ and $a_2=-a$ where $a$ now is a magnitude with the explicit sign showing direction. Hence, the two standard Atwood machine equations found in a standard textbook that provide the standard expression for the acceleration $a$.

3. Both masses taken together as the system: The mass of the system is $(m_1+m_2)$. Here $F_{net~both}$ is the net force acting on the center of mass of the two masses. This choice of system can be treated very neatly by "straightening out" the string to vertical (the ideal pulley allows this). The smaller mass is now on top and has force $m_1g$ acting on it conventionally "up"; the larger mass is on the bottom and has force $-m_2g$ acting on it, conventionally "down". Then $$F_{net~both}=(-m_2g)+m_1g=(m_1+m_2)a_{cm}~\rightarrow~a_{cm}=-\frac{m_2-m_1}{m_2+m_1}g.$$