Moment Arm: Determine Torque Vector & Qty

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Homework Help Overview

The discussion revolves around determining the torque vector and its quantity based on a provided sketch. The subject area includes concepts of torque, force application, and rotational mechanics.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • The original poster attempts to calculate torque using a specific force and distance, but questions the application point of the force. Participants discuss the significance of the force's line of action and its impact on torque calculations.

Discussion Status

Participants are exploring the implications of force application on torque direction and magnitude. Some guidance has been offered regarding the irrelevance of the path taken by the force application in determining torque direction.

Contextual Notes

There is an emphasis on understanding the relationship between the force's point of application and the resulting torque about the axle. The original poster's calculations and assumptions are being scrutinized for clarity.

Pinon1977
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Homework Statement


I'm trying to determine the torque vector and quantity of said vector per the sketch attached.

Homework Equations



T=fd[/B]

The Attempt at a Solution



T = 100 lbs x 2 f2f
T= 200 ftlbs

But my real question is where is that Force being applied at is it the attachment point to the axle or is it relative to its position in space and the vector would be straight down?
 

Attachments

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Pinon1977 said:
where is that Force being applied
The weight acts at the mass centre of the mass, so you would draw the force vector straight down from there. This is important for determining the torque it has about the axle, but for the purpose of figuring out the linear force at the axle the line of action does not matter, only the direction.
 
Ok, noted. So the 200ft lbs in this instance would be twisting the stationary axle in a CCW direction? Even though its attached to the opposite side of the acting force?
 
Pinon1977 said:
Ok, noted. So the 200ft lbs in this instance would be twisting the stationary axle in a CCW direction? Even though its attached to the opposite side of the acting force?
Yes. The path the rigid member takes from axle to point of application of force is irrelevant. You could make the member an entire disc centred on the axle.
 

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