Solving Forces & Torques: The Weight Problem

  • Thread starter Thread starter BlackMamba
  • Start date Start date
  • Tags Tags
    Forces Torques
Click For Summary

Homework Help Overview

The discussion revolves around a problem involving forces and torques, specifically calculating the torque produced by the weight of a jet engine suspended beneath an airplane wing. The weight is given as 9300 N, and participants are trying to understand the correct approach to find the torque with respect to a specified rotational axis.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the formula for torque and question the use of trigonometric functions in their calculations. There is confusion regarding the application of cosine versus sine in determining the lever arm and the correct geometry involved.

Discussion Status

Several participants are actively engaging with the problem, offering suggestions and questioning each other's reasoning. There is a recognition of potential geometric misunderstandings, and some guidance has been provided regarding the definition of torque and the importance of the lever arm.

Contextual Notes

Participants express frustration over their inability to arrive at the correct answer, indicating a possible lack of clarity in the problem setup or their understanding of the geometric relationships involved. The original poster has referenced a drawing that is not visible to others, which may contribute to the confusion.

BlackMamba
Messages
187
Reaction score
0
I can't understand why I can't get this problem right. It's irritating me, so any help would be greatly appreciated.

The Problem: The drawing shows a jet engine suspended beneath the wing of an airplane. The weight W of the engine is 9300 N and acts as shown in the drawing. The rotational axis in the drawing is perpendicular to the plane of the paper. With respect to this axis, find the magnitude of the torque due to the following force: the weight.

Drawing

I know how to find torque using Torque = Fl

Initially my equation consisted of:

Torque = 9300(2.50)cos 32 - But my answer was wrong. So I then thought perhaps I should being using gravity in there since W = mg. In using that my second equation consisted of:

Torque = 91140(2.50)cos 32 - But again my answer was wrong.

What am I missing?? This should not be that difficult that's why I'm getting frustrated that I can't figure out something simple, that I'm not just doing right.
 
Physics news on Phys.org
Try dividing by cos32 instead of multiplying. Not promising anything, just saying give it a try.

I seem to remember doing problems like this in the past and that was the answer.
 
You are having problem with geometry, look at the triangle again!
 
BlackMamba said:
I know how to find torque using Torque = Fl

Initially my equation consisted of:

Torque = 9300(2.50)cos 32
Why do you multiply by cos 32 ?
So I then thought perhaps I should being using gravity in there since W = mg.
The weight is given; no need to calculate it.
 
I have no idea why I multiply by cos 32. I was just following an example in the book that was similar. Is it supposed to be sin?
 
BlackMamba said:
I have no idea why I multiply by cos 32. I was just following an example in the book that was similar. Is it supposed to be sin?
You tell me. :smile: What's the definition of torque? What distance is involved?
 
Torque is the magnitude of the force times the lever arm. The lever arm being the distance that is invloved.

So...I still don't know what I'm doing wrong.
 
BlackMamba said:
Torque is the magnitude of the force times the lever arm. The lever arm being the distance that is invloved.

So...I still don't know what I'm doing wrong.
The lever arm is the perpendicular distance from the axis of rotation to the line of action of the force.

As Cyclovenom noted, you are having a geometry/trig problem. (Examine the triangle involved, find the side that represents the lever arm, and determine which trig function you need.)
 
Ok, I'll give it another go. Thank you for your help. It's much appreciated. :)
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
2K
Replies
7
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 35 ·
2
Replies
35
Views
3K
  • · Replies 22 ·
Replies
22
Views
3K
Replies
1
Views
2K
Replies
4
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
Replies
6
Views
2K
  • · Replies 14 ·
Replies
14
Views
3K