Magnitude of torque due to gravity

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Homework Help Overview

The problem involves calculating the magnitude of torque on a pole due to gravity, with specific parameters including the mass and length of the pole, as well as the angle at which it is held.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the angle at which the pole is held and the point of application of the force. Questions arise regarding the correct interpretation of the torque formula and the distribution of mass along the pole.

Discussion Status

Participants are exploring different interpretations of the torque calculation, with some offering insights into the relationship between the angle and the force applied. There is a focus on clarifying assumptions about the setup and the parameters involved.

Contextual Notes

There is a lack of visual representation of the problem, which may affect the clarity of the discussion. Participants are also navigating the implications of uniformly distributed mass in their reasoning.

dorian_stokes
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Homework Statement


A person carries a long pole of mass 9 kg and length 4.4 m. Find the magnitude of the torque on the pole due to gravity.



Homework Equations

sin(60) and cos(60), Torque=Force_gravity*r*sin(theta)



The Attempt at a Solution

 
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Hi dorian_stokes! :smile:
dorian_stokes said:
A person carries a long pole of mass 9 kg and length 4.4 m. Find the magnitude of the torque on the pole due to gravity.

erm :redface: …

where is he holding the pole? and at what angle to the horizontal? :confused:​
 
he's holding it at the end and the pole is at a 60 degree angle. Sorry for not explaining it, I don't have a picture.
 
(just got up :zzz: …)

ok, then …
dorian_stokes said:
A person carries a long pole of mass 9 kg and length 4.4 m. Find the magnitude of the torque on the pole due to gravity.

… torque = force times perpendicular-distance-to-the-line-of-application-of-the-force :wink:

(and cos = adj/hyp, sin = opp/hyp)

… what do you get? :smile:​
 
wouldn't it follow the formula torque=-mgrsin(x)? because we're given the angle that the pole makes with the ground, which makes the force that is perpendicular to the center of mass of the pole, opposite to the angle that we're given... and because the distribution of the pole's mass is uniformly distributed, our r is actually r/2. right?
 

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