Finding the net torque about the axle?

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Homework Help Overview

The problem involves calculating the net torque about the axle of a rotating disk, given specific forces and distances. The disk has a diameter of 29.5 cm and rotates counterclockwise, with multiple forces applied at different points.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculation of torque using the formula involving force and distance from the pivot point. There are attempts to verify the calculations for each force and the direction of torque. Questions arise regarding the accuracy of trigonometric functions used in the calculations.

Discussion Status

Participants are actively engaging in verifying calculations and addressing potential errors in the original poster's approach. There is acknowledgment of a possible miscalculation related to the use of radians in trigonometric functions.

Contextual Notes

The original poster expresses uncertainty about their calculations and seeks clarification on the correct approach. There is a focus on ensuring the proper application of mathematical principles without reaching a definitive conclusion.

Kyp
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Homework Statement


The 29.5cm diameter disk rotates counterclockwise on an axle through its center. F1=20.8N, F2=58.6N, F3=58.6N, F4=20.8N, and d=5.27cm. What is the net torque about the axle? Let the counterclockwise direction be positive.

Homework Equations


Torque = force (perpendicular) x distance (from pivot point)

The Attempt at a Solution


F1 = 0 N*m
F2 = 58.6 N x 0.1475m = -8.6435 N*m (clockwise)
F3 = 0.0527m x 58.6 sin45 N = 2.627777283 N*m (counterclockwise)
F4 = 0.0527m x 20.8 N = 1.09616 N*m (counterclockwise)
Therefore torque is -4.9196 N*m.
However that is apparently not the answer... What am i doing wrong? :/
 

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check.. F3 = 0.0527m x 58.6 sin45 N = 2.627777283 N*m
 
Kyp said:

Homework Statement


The 29.5cm diameter disk rotates counterclockwise on an axle through its center. F1=20.8N, F2=58.6N, F3=58.6N, F4=20.8N, and d=5.27cm. What is the net torque about the axle? Let the counterclockwise direction be positive.

Homework Equations


Torque = force (perpendicular) x distance (from pivot point)

The Attempt at a Solution


F1 = 0 N*m
F2 = 58.6 N x 0.1475m = -8.6435 N*m (clockwise)
F3 = 0.0527m x 58.6 sin45 N = 2.627777283 N*m (counterclockwise)
F4 = 0.0527m x 20.8 N = 1.09616 N*m (counterclockwise)
Therefore torque is -4.9196 N*m.
However that is apparently not the answer... What am i doing wrong? :/
Everything is correct, you miscalculated something. sin(45)?
 
I had my calculator in radians... Thank you guys :/
 

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