Welcome Guest Search | Active Topics |

Tag as favorite
AAMC Question Pack Question 95
emily_5557
#1 Posted : Friday, May 21, 2021 4:58:18 PM
Rank: Member

Groups: Registered
Joined: 6/5/2020
Posts: 12

Thanks: 0 times
Was thanked: 0 time(s) in 0 post(s)
Hello! I was super confused about Question 95 from the Physics AAMC question pack and wanted to get some help understanding the answer.

Question
A 0.5-kg uniform meter stick is suspended by a single string at the 30-cm mark. A 0.2-kg mass hangs at the 80 cm mark. What mass hung at the 10-cm mark will produce equilibrium?

I'm still having trouble understanding the process and explanation. I understand that the stick is in rotational equilibrium so the torque clockwise = torque counter clockwise and that the torque expression can be simplified to torque = FL as all the forces are perpendicular. I am not sure how they considered the torque and force on the string being held at the 30 cm mark. First I thought that since this was the point of rotation for the stick, that it would not be considered in the equilibrium expression. Then when the answer considered it in the equilibrium expression, they used the distance to the centre of mass of the stick, while the other 2 masses distances were to the 30 cm point of rotation. As well the answer considered the string torque as contributing to the same direction of rotation as the mass being held at 80 cm, I thought that it would be in the same rotational direction as the mass being held at 10 cm.

Any help is appreciated thank you! :)
INSTR_Katrina_128
#2 Posted : Friday, May 21, 2021 10:52:58 PM
Rank: Advanced Member

Groups: Registered
Joined: 5/18/2021
Posts: 73

Thanks: 0 times
Was thanked: 0 time(s) in 0 post(s)
Hi Emily!

Here's an image I've made that will hopefully make this question easier for you: https://i.ibb.co/250PCFC/Capture.png. The key with this question is that everything has to be calculated RELATIVE TO THE FULCRUM. The fulcrum is where we are hanging the meter stick, which is at the 30 cm mark. So, I will calculate all of my distances RELATIVE to this value.

For example, even though the center of mass is at the 50-cm mark, it is 20 cm from the fulcrum!

Setting up a torque balance around that point, we get

torque on the left = torque on the right

(F)(20) = (5)(20) + (2)(50)

20F = 100 + 100

F = 200/20 = 10

Solving for the mass, we get

F = mg
m = F/g = 10/10 = 1 kg

The correct answer is D.

Let me know if that helps! 😄
emily_5557
#3 Posted : Monday, May 24, 2021 11:29:15 PM
Rank: Member

Groups: Registered
Joined: 6/5/2020
Posts: 12

Thanks: 0 times
Was thanked: 0 time(s) in 0 post(s)
Hi Katrina! Thanks so much, I can see that I was mainly tripped up on how to include the ruler in the torque equation and didn't realize that the fulcrum and the centre of mass were different but now I understand :)
Users browsing this topic
Guest (2)
Tag as favorite
You cannot post new topics in this forum.
You cannot reply to topics in this forum.
You cannot delete your posts in this forum.
You cannot edit your posts in this forum.
You cannot create polls in this forum.
You cannot vote in polls in this forum.

Clean Slate theme by Jaben Cargman (Tiny Gecko)
Powered by YAF | YAF © 2003-2009, Yet Another Forum.NET
This page was generated in 0.080 seconds.