# Homework Help: Free Body Diagram lab report -- General Question

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1. Feb 8, 2017

### Justin Webb

< 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.

Last edited by a moderator: Feb 8, 2017
2. Feb 8, 2017

### Staff: Mentor

Welcome to the PF.

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. Feb 8, 2017

### Justin Webb

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. Feb 8, 2017

### Staff: Mentor

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. Feb 8, 2017

### Justin Webb

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. Feb 8, 2017

### Staff: Mentor

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. Feb 8, 2017

### Justin Webb

Ok I'm gonna 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. Feb 8, 2017

### Justin Webb

That's what I got after doing a bit of research. Sorry the first one spazzed out.

9. Feb 8, 2017

### Staff: Mentor

10. Feb 8, 2017

### Justin Webb

I fixed it

11. Feb 8, 2017

### Staff: Mentor

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. Feb 8, 2017

### Justin Webb

We haven't gone over cross products yet. We've done resolving force vectors with trig.

13. Feb 8, 2017

### Staff: Mentor

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. Feb 8, 2017

### Justin Webb

We havent 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. Feb 9, 2017

### Nidum

Would the more intuitive notation used in technical diagrams be ok for this problem ?