Why are these forces not included in the solution? (Torque Problem)

[mathandphysics]

So please find both screenshots below taken from these youtube videos:


Why in the first problem, the tension in the string, from which a mass hangs, is not included when the teacher is placing all torques in the equation (see image)?
Why in the second problem, the normal force from the rod on the man, is not included when the teacher is placing all the torques in the equation (see image)?

Thank you.

 

[Orodruin]

You have to decide what system you are considering. If you consider only the beam then the box/man is not part of the system and you have a tensile/normal force acting as an external force on the system. If your system is the beam + box/man, then the tensile/normal forces are internal amd not part of the force/torque balace. Instead you have the gravitational force on the box/man as an external force on the system.

 

[mathandphysics]

You have to decide what system you are considering. If you consider only the beam then the box/man is not part of the system and you have a tensile/normal force acting as an external force on the system. If your system is the beam + box/man, then the tensile/normal forces are internal amd not part of the force/torque balace. Instead you have the gravitational force on the box/man as an external force on the system.

Thank you for your reply, it makes sense. While I was checking answers, I ran across this video: .

If you have time to check it, in particular from minute 2:23 till 2:51. I failed to understand what he meant by saying that it is NOT the weight of the person, but he is going to put it there anyway. I found similar videos to his, trying to point out the same idea. What do you think?

 

[kuruman]

If you have time to check it, in particular from minute 2:23 till 2:51. I failed to understand what he meant by saying that it is NOT the weight of the person, but he is going to put it there anyway. I found similar videos to his, trying to point out the same idea. What do you think?

The weight of the person is defined as the force with which the Earth attracts the person.
The force that was drawn in the free body diagram is the force that the person exerts on the board. They are NOT the same force.

You can see the difference between the two by looking at the their reaction counterparts. The Earth attracts the person with force $m_pg$ down and the person attracts the Earth with force $m_pg$ up. The person exerts a force $N_p$ down on the plank and the plank exerts a force $N_p$ up on the person. Just because $m_pg$ is equal to $N_p$ here, you cannot conclude that they are the same force. If Jimmy is 23 years old and Bobby is also 23 years old, this doesn't mean that Jimmy and Bobby are the same person.

 

[mathandphysics]

The weight of the person is defined as the force with which the Earth attracts the person.
The force that was drawn in the free body diagram is the force that the person exerts on the board. They are NOT the same force.

You can see the difference between the two by looking at the their reaction counterparts. The Earth attracts the person with force $m_pg$ down and the person attracts the Earth with force $m_pg$ up. The person exerts a force $N_p$ down on the plank and the plank exerts a force $N_p$ up on the person. Just because $m_pg$ is equal to $N_p$ here, you cannot conclude that they are the same force. If Jimmy is 23 years old and Bobby is also 23 years old, this doesn't mean that Jimmy and Bobby are the same person.

Thank you.. Given what @Orodruin said, that choosing and including forces depends on the system we are considering. How do you decide what system to take? For example in this problem and given the first reply, it seems that the author took is the (beam+box/man). How do I know which system to choose?

 

[jbriggs444]

How do I know which system to choose?

You choose a system that helps you solve the problem.

If you can find one, you want a system where you know all but one thing about that system. Then you can write down an equation and solve for that one thing.

If, having done that, you have not yet solved the problem, you at least have one less unknown. So repeat the process. Pick out a new system and solve for the next unknown.

 

[kuruman]

Thank you.. Given what @Orodruin said, that choosing and including forces depends on the system we are considering. How do you decide what system to take? For example in this problem and given the first reply, it seems that the author took is the (beam+box/man). How do I know which system to choose?

You can choose as system anything you want. It depends on what the problem is asking you to find. It also depends on whether you know the acceleration or not. Generally, if you are asked to find a specific force, your system should be an object on which the unknown force is exerted by another object that is not part of your chosen system. For example, if you are asked to find the force exerted by the man on the plank, your system should be the plank.

Sometimes, you may have to use two separate systems in tandem (see example below).

Example

Mass $m_1$ is on top of mass $m_2$. The bottom mass is pulled along a frictionless surface with constant force $F$ and the to masses accelerate as one. Given quantities are $m_1$, $m_2$ and $F$.
Find the force of static friction on the top mass.
Analysis
If you draw a force diagram choosing as system the top mass, you will realize that the only external force in the horizontal direction is the unknown force of static friction $f_s$. You can apply Newon's second law to get $$f_s=m_1~a\tag{1}$$which does not answer the question because you don't know the acceleration $a$.

So you choose a second system consisting of the two masses together. Here, the only external horizontal force is $F$ and $$F=(m_1+m_2)a. \tag{2}$$Now from equation (2) you can get an expression for the acceleration $a$ in terms of the given quantities, put that in equation (1) and you have the answer.

(This is an illustration of what @jbriggs444 is talking about and posted before I did.)

 

[jbriggs444]

Sometimes, you may have to use two separate systems in tandem (see example below).

The example given by @kuruman shows a "top down" or "goal oriented" approach to problem solving. You have a problem that asks for a particular quantity. For example:

Find the force of static friction on the top mass.

That's your goal. The top of your problem solving tree.

You choose a system that includes that quantity as one of its unknowns. The obvious system to choose is the top mass. You write down an equation for that system that includes your goal quantity. Maybe the equation will involve energy. Maybe momentum. Maybe something else.

But you realize that you do not know enough to solve the equation. You have one or more additional unknowns that, if you knew their values, would allow you to solve the equation and solve your problem.

Those unknowns become your new goals. quantities become your new goal. You work your way down through the problem solving tree until you've found values for the needed unknowns and can solve the original equation.


There is an alternate strategy that can sometimes be useful. This is a "bottom up" or "resource oriented" approach.

Instead of focusing on what the is asked for, you can look instead at the information you are given. This time you are looking to see what else you can calculate from what you already know. You work your way up the problem solving tree from the bottom until you have calculated the desired quantity at the top.


My usual strategy to problem solving is more along the lines of "meet in the middle". I'll work from the top down to figure out which quantities, if I could calculate them, would allow me to solve the problem. And I'll work from the bottom up to figure out which quantities, given the information at hand, could be calculated. If there is some set of quantities that meet both criteria, those are the ones that I actually need to calculate and use

 

[Lnewqban]

I failed to understand what he meant by saying that it is NOT the weight of the person, but he is going to put it there anyway.

The explanation is at minute 1:30.
He is referring to the weight of the person in the same way he previously referred to the weight of the board: as forces not touching each body directly.

The feet of the diver exert one of the several forces directly touching the board and inducing torques.

In this statically balanced situation, the magnitude of that vertical force exerted by the feet on the board happens to be exactly as much as the force induced remotely by gravity in the center of mass of the diver.

 

[mathandphysics]

You can see the difference between the two by looking at the their reaction counterparts. The Earth attracts the person with force mpg down and the person attracts the Earth with force mpg up. The person exerts a force Np down on the plank and the plank exerts a force Np up on the person. Just because mpg is equal to Np here, you cannot conclude that they are the same force. If Jimmy is 23 years old and Bobby is also 23 years old, this doesn't mean that Jimmy and Bobby are the same person.

Then why didn't he just put Np instead of mg, and I was checking the new answers here and they are saying that in this case and given the chosen system, the drawn weight force is considered as an external force and needs to be included. I am not sure how both of these statement complete each other?

 

[Orodruin]

Then why didn't he just put Np instead of mg, and I was checking the new answers here and they are saying that in this case and given the chosen system, the drawn weight force is considered as an external force and needs to be included. I am not sure how both of these statement complete each other?

He is skipping s bit and being sloppy. If you want to be painfully thorough you first do the force equilibrium for the load, which tells you that the weight of the load is equal in magnitude and opposite in direction to the normal force from the beam on the load. By Newton’s third load, the weight is therefore equal to the normal force from the load on the beam. This is sloppily referred to as the weight of the load acting on the beam.

The thing is that ”everybody” allegedly understands this.

 

[mathandphysics]

He is skipping s bit and being sloppy. If you want to be painfully thorough you first do the force equilibrium for the load, which tells you that the weight of the load is equal in magnitude and opposite in direction to the normal force from the beam on the load. By Newton’s third load, the weight is therefore equal to the normal force from the load on the beam. This is sloppily referred to as the weight of the load acting on the beam.

The thing is that ”everybody” allegedly understands this.

Excuse me, but here is where I am confused. If I want to be thorough as you said, then the fact that I would consider "normal force from the beam on the load" means that this force is no longer internal force as mentioned in previous replies. However given what I have understood from the above comments is that the load and the beam are considered as a system by the teacher in the above youtube videos.

 

[Orodruin]

Excuse me, but here is where I am confused. If I want to be thorough as you said, then the fact that I would consider "normal force from the beam on the load" means that this force is no longer internal force as mentioned in previous replies. However given what I have understood from the above comments is that the load and the beam are considered as a system by the teacher in the above youtube videos.

Typically, what is considered is the system of the beam only. As I said earlier, which forces are considered depends on what system you consider. If you consider beam+load, the weight is an external force on that system because it acts on the load, which is part of the system. In that case there is no ambiguity in saying that the weight acts on the system.