Calculating Torque & Lever Arm for a Lunch Tray

Click For Summary

Homework Help Overview

The discussion revolves around calculating the torque and lever arm for a lunch tray being held with various items on it. The problem involves understanding the forces acting on the tray, including the mass of the tray, a plate of food, and a cup of coffee, as well as the lever arms associated with these forces.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to calculate the force exerted by the thumb using torque and force tables, but expresses uncertainty about their calculations. Some participants suggest using static equations and checking the axis of rotation for the torque calculations.

Discussion Status

The discussion is ongoing, with participants providing guidance on checking the axis of rotation and the directions of the moments of force. There is no explicit consensus on the correct approach yet, but suggestions for reconsidering certain aspects of the problem have been made.

Contextual Notes

The original poster mentions difficulties with uploading a diagram, which may limit the clarity of the problem setup. There is also an implication of homework constraints, as the poster is seeking help without revealing complete solutions.

drewfus
Messages
2
Reaction score
0
A lunch tray is being held in one hand. The mass of the tray itself is 0.201 kg, and its center of gravity is located at its geometrical center. On the tray is a 1.33 kg plate of food and a 0.285 kg cup of coffee. Assume lever arm of the thumb is 0.0620 m, lever arm of the finger is 0.113 m, lever arm of the plate 0.231 m, lever arm of the coffee cup is 0.361 m and lever arm of the tray is 0.383 m. Obtain the force T exerted by the thumb. This force acts perpendicular to the tray, which is being held parallel to the ground.

The picture that I have (but could not upload for some reason) had the thumb on top of the tray have a force T exerted in the -y direction when drawn with a force diagram and the finger had a +y force.

I plugged into a force table and then calculated a torque table, and I substituted with the two equations and got an answer of -133 Newtons, but I checked it and it was wrong.

Any ideas on what I did wrong?
 
Physics news on Phys.org
torque is T=F * D

just take the moment of all forces at one point and then use statics equations to find the rections
 
From which axis of rotation the arms are measured. Check axis of rotation you have taken and the directions of the moments of force
 
thanks guys i appreciate it.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
Replies
4
Views
8K
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
Replies
1
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
Replies
3
Views
10K
  • · Replies 13 ·
Replies
13
Views
13K