Calculating Net Torque for an Inclined Wheel: A Problem in Applied Physics

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Homework Help Overview

The discussion revolves around calculating the net torque on an inclined wheel about its axle, given specific distances and forces. The problem involves understanding the effects of angles on torque calculations.

Discussion Character

  • Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculation of torque using the formula τ=f⋅d, with attempts to sum individual torque contributions. There is a focus on the need to consider angles in the calculations, particularly for components not aligned with the horizontal or vertical axes.

Discussion Status

Some participants are seeking clarification on the calculations presented, noting missing details in the original poster's explanation. There is an emphasis on ensuring all contributions to torque are accounted for, and a suggestion to reconsider the direction of the forces involved.

Contextual Notes

Participants are working under the assumption that the positive direction for torque is counterclockwise, and there is a mention of a specific angle (150°) that may affect the calculations. The original poster's computations have been questioned for completeness.

Erenjaeger
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Homework Statement


Find the net torque on the wheel in the figure below about the axle through O, taking a = 13.0 cm and b = 31.0 cm. (Assume that the positive direction is counterclockwise.)
pic: https://fbcdn-sphotos-c-a.akamaihd...._=1470130918_26fa32f17821823200ba75439cef0a52
[/B]

Homework Equations

τ=f⋅d[/B]

The Attempt at a Solution


just sum all for torque, I thought you could just go
τ1=9.0N ⋅ 0.31m = 2.79 n⋅m
τ2=10.0N⋅0.31 = 3.1 N⋅ m
but the torque component with the angle shown, I calculated it the same and then summed all the values of torque together, and it said it was wrong, so I am assuming that you have to add something like a cosine or sine into that component because it isn't in a horizontal or vertical direction but 150° below the horizontal (positive x axis) any tips on how to solve this,?
 
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Erenjaeger said:
I calculated it the same
This does not actually show us what you did. Please specify the values you put in and the result you got.
 
Orodruin said:
This does not actually show us what you did. Please specify the values you put in and the result you got.
i don't understand how it doesn't show what i did? i completely explained what i did. the two net torques i gave how i calculated them and got 2.79 N.m and 3.1 N.m and said i did the same for the third component so obviously following that i went 12.0N x 0.13m which gave me 1.56 N.m, then i said i summed all my values for torque together, thus 2.79+3.1+1.56 which equals 7.45N.m but that value was incorrect. Does that explain it anymore or was there something else you need to know??
 
Erenjaeger said:
i completely explained what i did.
No, you left out this initially:
Erenjaeger said:
following that i went 12.0N x 0.13m

Erenjaeger said:
2.79+3.1+1.56
Think about directions (more precisely, "sense").
 
Erenjaeger said:
i don't understand how it doesn't show what i did? i completely explained what i did.
No you did not, you left out a crucial part, your computation of the last contribution and the final result. Haruspex has already given you a good hint. Try to interpret it and ask again if you still cannot work it out.
 

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