Free Body Diagram lab report -- General Question

In summary: I fixed...In summary, the student was trying to find a way to represent torque in an FBD. They learned that the vector torque is represented by the vector from the pivot to the point of force application and the applied force. They also learned about the Right Hand Rule and how to find the vector torque.
  • #1
Justin Webb
8
0
< Mentor Note -- thread moved to HH from the technical forums to use a scpecific schoolwork problem to illustrate the general concept of torque in a FBD >

I'm writing up a lab report for Physics class and I have to draw FBDs for torque being applied to a ruler resting on a pivot point. I've drawn FBDs for force vectors before but not quite sure how to represent torque in a FBD. Thanks in advance.
 
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  • #2
Justin Webb said:
I'm writing up a lab report for Physics class and I have to draw FBDs for torque being applied to a ruler resting on a pivot point. I've drawn FBDs for force vectors before but not quite sure how to represent torque in a FBD. Thanks in advance.
Welcome to the PF. :smile:

Have you learned the vector equation for the vector torque in terms of the vector from the pivot to the point of force application and the applied force? Can you show us that vector equation? Are you familiar with the Right Hand Rule?
 
  • #3
berkeman said:
Welcome to the PF. :smile:

Have you learned the vector equation for the vector torque in terms of the vector from the pivot to the point of force application and the applied force? Can you show us that vector equation? Are you familiar with the Right Hand Rule?
Not specifically but if I'm understanding you correctly, do you mean the force multiplied by the moment arm? If so, it was 73.5g(33cm). Which I converted into Newton meters and I got .2379Nm.
 
  • #4
Justin Webb said:
Not specifically but if I'm understanding you correctly, do you mean the force multiplied by the moment arm? If so, it was 73.5g(33cm). Which I converted into Newton meters and I got .2379Nm.
So just as the forces in a FBD are vectors (so direction is important, and they may have components along the coordinate axes, so too the torque(s) shown on a FBD are vectors. It's important to get the directions right when drawing the torque vector(s).

Can you show a sketch of your FBD with the force vectors and what you think are the torque vectors? Since we will start talking now specifically about your schoolwork lab report, I'll move this thread to the Homework Help forums to continue...
 
  • #5
CHAPTER - 8.jpg

The idea was to achieve rotational equilibrium. Really I'm just not sure if I've included everything, or if I'm representing the experiment properly.
 
  • #6
So that's a ruler balanced part-way along its length on a pivot point, and those are the moment arms and applied masses, right? And I'm guessing the masses were determined experimentally to balance the ruler on that pivot, right?

So to show the two torques acting on the ruler, you would draw the position vectors from the pivot point out to the two ends of the ruler, and show the forces at those two ends based on the mass values that were hung there. The vector torques will be based on those two quantities (position vector & force vector), and will have directions determined by the Right Hand Rule.

Can you take a cut at showing those vectors, and look up the Right Hand Rule to figure out how to get the vector torque? Also, it's best to draw a coordinate system to keep things straight. Center it on the pivot point for simplicity, and label x, y, and z unit vectors along those 3 axes...
 
  • #7
berkeman said:
So that's a ruler balanced part-way along its length on a pivot point, and those are the moment arms and applied masses, right? And I'm guessing the masses were determined experimentally to balance the ruler on that pivot, right?

So to show the two torques acting on the ruler, you would draw the position vectors from the pivot point out to the two ends of the ruler, and show the forces at those two ends based on the mass values that were hung there. The vector torques will be based on those two quantities (position vector & force vector), and will have directions determined by the Right Hand Rule.

Can you take a cut at showing those vectors, and look up the Right Hand Rule to figure out how to get the vector torque? Also, it's best to draw a coordinate system to keep things straight. Center it on the pivot point for simplicity, and label x, y, and z unit vectors along those 3 axes...
Ok I'm going to have to do some research. The right hand rule and position vectors is new for me. I'll post what I come up with later tonight.
 
  • #8
CHAPTER - 8 (1).jpg
That's what I got after doing a bit of research. Sorry the first one spazzed out.
 
  • #9
Justin Webb said:
https://docs.google.com/a/student.ccm.edu/drawings/d/sMAC9IQu8shROBt-xAIOQsw/image?w=459&h=265&rev=1&ac=1
Your upload or web link did not work...
 
  • #10
berkeman said:
Your upload or web link did not work...
I fixed it
 
  • #11
Justin Webb said:
I fixed it
Nice, thanks!

So, what do you know about vectors and the vector cross product so far? Have you covered that in your classes yet?

If not, what tools have your instructors given you so far for calculating the vector torque's magnitude and direction from the magnitude and direction of the position vectors and the force vectors?
 
  • #12
berkeman said:
Nice, thanks!

So, what do you know about vectors and the vector cross product so far? Have you covered that in your classes yet?

If not, what tools have your instructors given you so far for calculating the vector torque's magnitude and direction from the magnitude and direction of the position vectors and the force vectors?
We haven't gone over cross products yet. We've done resolving force vectors with trig.
 
  • #13
Justin Webb said:
We haven't gone over cross products yet. We've done resolving force vectors with trig.
Resolving force vectors with trig is a step in the right direction.

In this problem, because the force vectors are perpendicular (90 degrees) to the position vectors, that simplifies things a bit.

But you still need to use the right hand rule to tell you what the sign of the vector cross product (direction) is.

Do you have access to a TA or office hours to ask how you are supposed to represent the vector torques in this lab? Or are you supposed to somehow figure that out without any classroom sessions about this? It seems strange that you are given this important calculation to learn with the lab, without any introduction to the vector cross product representation of the vector torque in your classroom or TA sessions...

I could provide you lots of links to illustrate how to do this (or you can find them with simple searches), but it sure seems like those topics should have been covered in the class as preparation for your lab experiment...
 
  • #14
berkeman said:
Resolving force vectors with trig is a step in the right direction.

In this problem, because the force vectors are perpendicular (90 degrees) to the position vectors, that simplifies things a bit.

But you still need to use the right hand rule to tell you what the sign of the vector cross product (direction) is.

Do you have access to a TA or office hours to ask how you are supposed to represent the vector torques in this lab? Or are you supposed to somehow figure that out without any classroom sessions about this? It seems strange that you are given this important calculation to learn with the lab, without any introduction to the vector cross product representation of the vector torque in your classroom or TA sessions...

I could provide you lots of links to illustrate how to do this (or you can find them with simple searches), but it sure seems like those topics should have been covered in the class as preparation for your lab experiment...
We haven't gone over cross products. But if it's pertinent I would like to learn more about it even if just for the sake of being prepared for future labs or higher level classes. I have class tomorrow night and I'll ask my professor about the right hand rule and cross products. Links would definitely be helpful. Thanks for all your help and patience tonight.
 
  • #15
Would the more intuitive notation used in technical diagrams be ok for this problem ?
Torque notation.jpg
 

1. What is a free body diagram?

A free body diagram is a visual representation of all the forces acting on an object. It is used to analyze and understand the motion of an object by considering all the forces that are acting on it.

2. Why is a free body diagram important?

A free body diagram helps to identify and understand the forces that are acting on an object, which is essential for solving problems involving motion and equilibrium. It also helps to visualize the direction and magnitude of each force, making it easier to analyze the motion of an object.

3. How do I draw a free body diagram?

To draw a free body diagram, you first need to identify the object and all the forces acting on it. Then, draw a simple diagram of the object, and label all the forces with arrows to represent their direction and magnitude. Make sure that all forces are labeled with their appropriate units.

4. What are the common mistakes to avoid when drawing a free body diagram?

Some common mistakes to avoid when drawing a free body diagram include not labeling the forces with their appropriate units, not including all the forces acting on the object, and not accurately representing the direction and magnitude of each force. It is also important to make sure that the diagram is neat and easy to read.

5. Can I use a free body diagram for any object?

Yes, a free body diagram can be used for any object, regardless of its shape or size. It is a useful tool for analyzing the forces acting on an object, whether it is stationary, moving at a constant velocity, or accelerating. It is important to accurately identify and label all the forces in order to draw an effective free body diagram.

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