Solution to the Second Rotational Equilibrium Quiz
Again, expecting something to be really hard and wasting the whole period on it, and burning away time that was supposed to be spent on the practice finals, is a terrible choice. And it is a choice. It's a choice that was made when the student decided that NOT ALL definitions don't matter. And that happened over a matter of more than a week. Knowing definitions are what make this a 5-minute quiz solution. If you disagree, it means you didn't learn all definitions. I put then in writing as a checklist.
It was a three force problem with a two-torque balance. So that's as simple as these problems can be. But people who refuse to change their prioritization of definitions will never see that. (In this case, it's definition of where the force is drawn, definition of origin, and definition of leverage as it relates to lever-arm. And definition of which torques are out of the page and which are in. And torque equilibrium. This is all old news by now, which is why I'm saying 5-minute solution.)
A perfectly done FBD would make all of the following simple and quick to see. But not everyone does that; not everyone has trained themselves to know where forces go, and not everyone has trained themselves to identify leverages by definition. So all the people who don't do those things, they will take infinite time solving this quiz and still have nonsense, because they live in a world of making simple things impossible. Those who don't live in the world of impossibility, because they prioritize the definitions I tell them to, will do the following in just a couple of minutes, because they will have made a simple FBD in about 1 minute.
I gave a known mallet force vector located a given number of centimeters from the pivot axis. The values of these quantities were (0.94 kg)g and 6 cm, (0.88 kg)g and 8 cm, (1.69 kg)g and 5 cm, or (1.175 kg)g and 4 cm.
I told people to solve for an unknown ruler mass, and I told people that the ruler was 100 cm long.
So a person who has followed directions over the last 2 weeks has placed the unknown mg vector at the 50 cm mark on the ruler. Then the two-torque balance expression (using the givens (0.94 kg)g and 6 cm as an example) became:
(Torque-CCW) - (Torque-CW) = 0
Torque-CCW = Torque-CW
"[(0.94 kg)g](6 cm) = (mg)(50 cm)" to cancel the g and solve for m would be a pretty good solution, but it is wrong.
A quality F.B. diagram shows that the leverage that goes with mg is not 50 cm. It is 50 cm times the cosine of 20 degrees. Working out that the leverage would be 50 cm times the cosine of 20 degrees is something that would be done on scratch paper, and it's a component job that I've shown countless times in class by now. (It uses SOHCAHTOA.) If you don't do the manipulation on scratch paper to get components like 50 cm times the cosine of 20 degrees, but I've done it in front of you countless times by now, well, when are you going to start doing it? (And those who don't know that such a component is important, well then you don't know the definition of leverage, and that's what I'm talking about. I've explained it many times. When are you going to change this if you haven't yet?
Correct last step: [(0.94 kg)g](6 cm) = (mg)[(50 cm)cos(55)] Cancel g, solve for m.
Answer: m = 120 g
This was a 5 minute quiz.