Lesson 5 Video 2

Post Reply
Posts: 52
Joined: Mon Oct 02, 2017 9:15 pm

Lesson 5 Video 2

Post by ArielM » Wed Apr 18, 2018 1:10 pm

Last night during Office Hours we went over a physics question that I did not understand. I also found the question in Lesson 5 Video 2, Slide 4. After reviewing the video, I still did not understand.

The question states:

What is the mass of the brass weight hanging from the end of the string if the system is at equilibrium?
A. 2.3 kg
B. 3.0 kg
C. 4.6 kg
D. 6.0 kg

I believe I understand this concept, that equilibrium means that the torque of the beam+m2 is supposed to be equal to the torque of the string at the hinge.

What I do not understand was how they solved for the problem. So torque is acting on three things: the beam, the mass 2, and the string. They said the torque of the beam was equal to 2 (where did they get this value from? 2 is the mass of the beam. I thought we were solving for torque, which is t=r L sinθ, what does mass have to do with this equation?) * 10 (I suppose this is 10m/s^2 of to acceleration due to gravity) * 0.25 (the distance of the beam from the hinge to the string attachment point). Then they set the torque of the beam equal to the torque of mass 2, which is equal to 1 (again, the mass of m2, but again... why is mass included? What does mass have to do with torque?) * 10 (acceleration due to gravity) * 0.5 (the length of the entire beam). What is confusing is that it seems like to find the torque of the beam and the torque of m2, I use the mass, the acceleration due to gravity and length of it's position on the beam. So essentially to find these two torques, beam and m2, I use the equation t= mgL? Why am I using the mass to find the torque? This is where I am confused.

Okay, so going further, both the torque of the beam and the torque of the mass 2 equal 5Nm. So the torque of the string must equal 10Nm to offset the torque(s) of the beam + m2. So we have the torque of the string, t= 10Nm, which is set equal to rFsinθ. I see that you need the force to solve for the mass. In a situation similar to randomly switching to mass, we now switch to force. How do you intuitively know when to switch from m to f to l in the equation? This is not the time for a "you get it or you don't" scenario; I'm tired of getting problems like these wrong. I literally need to know how do you know when to modify the equation and when you know that you need to use Force instead of Length at this time in the process of solving for this equation?

-Ariel Morrow
Posts: 520
Joined: Mon May 23, 2016 1:47 pm

Re: Lesson 5 Video 2

Post by NS_Tutor_Andrew » Thu Apr 19, 2018 11:29 am

Hi Ariel,

The OH recording isn't up yet so I can't address anything that came up specifically in that discussion, but in general this is a tough passage/question and you're in good company in terms of finding it difficult. I'm attaching an explanation that I wrote up using nicer equation typesetting than is possible in a plain-text forum (although this explanation solves for the WEIGHT of m1, it's easy to convert from weight to mass by dividing by 10 m/s2) and an annotated diagram of the problem. Hopefully this will help. In terms of why you're using mass to find torque, that's because torque = F * d * sin(theta), and F in this case is the force of gravity, which equals mg.

Let me know if you have further questions about this problem!
derivation of m1.png
(165.61 KiB) Not downloaded yet
sketch of problem setup.png
(214.52 KiB) Not downloaded yet
Andrew D.
Content Manager, Next Step Test Prep.
Post Reply