Magnitude of torque due to gravity

AI Thread Summary
To calculate the torque on a 9 kg pole of length 4.4 m held at a 60-degree angle, the torque formula is applied: Torque = Force_gravity * r * sin(theta). The force due to gravity is the weight of the pole, which is 9 kg multiplied by the acceleration due to gravity (approximately 9.81 m/s²). The effective distance (r) from the pivot point to the center of mass is half the length of the pole, or 2.2 m. The discussion emphasizes the need to consider the angle and the uniform mass distribution of the pole in the torque calculation. Understanding these parameters is crucial for accurately determining the torque due to gravity.
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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