Clarifying the Leaning Sticks Problem Solution

[PhysHobbiest]

Homework Statement

The problem along with its solution have been screenshotted and attached

Relevant Equations

T = r x F

I'm getting tan³(θ) = (1/μ) as a solution. Morin's solution seems wrong to me because it seems to me that he's assuming that the entire weight of each stick is providing torque when in reality the component of the gravitational force perpendicular to each stick should be all that is providing torque. Am I missing something or is Morin's solution incorrect?

 

[BvU]

Hello and !

he's assuming that the entire weight of each stick is providing torque

What makes you think that ?

is Morin's solution incorrect?

Don't think so.

 

[Doc Al]

Morin's solution seems wrong to me because it seems to me that he's assuming that the entire weight of each stick is providing torque when in reality the component of the gravitational force perpendicular to each stick should be all that is providing torque.

Ask yourself what the component of gravity perpendicular to the stick is.

 

[PhysHobbiest]

Hello and !

What makes you think that ?
Don't think so.

What makes me think that he's using the entire weight of each stick is his equation for normal force. Which is
N = (M_lG) (This is the gravitational force supposedly perpendicular to the stick) x (1/2)(sin(θ)) (This being the center of gravity)

 

[etotheipi]

What makes me think that he's using the entire weight of each stick is his equation for normal force. Which is
N = (M_lG) (This is the gravitational force supposedly perpendicular to the stick) x (1/2)(sin(θ)) (This being the center of gravity)

The torque of a force has magnitude $rF\sin{\theta}$. You interpret this as the component of the force perpendicular to the rod times the distance along the rod to where the force acts, or alternatively (but equivalently) as the force times the perpendicular distance between the line of action of the force and the point where the rod meets the ground.

Just on its own, what would you say is the torque of the weight of the left stick about the point where the stick meets the ground?

 

[haruspex]

See section 1 of https://www.physicsforums.com/insights/frequently-made-errors-mechanics-moments/

 

[PhysHobbiest]

I was blind but now I see. I was equating the torque on the stick to the normal force (i.e. $rF\sin{\theta} = N$)
when I really should have been balancing torques (i.e. $rF\sin{\theta} = Nsin{\theta}$). Silly mistake but it was driving me crazy. Thank you all for helping me correct my mistake!

 

[etotheipi]

I was blind but now I see. I was equating the torque on the stick to the normal force (i.e. $rF\sin{\theta} = N$)
when I really should have been balancing torques (i.e. $rF\sin{\theta} = Nsin{\theta}$). Silly mistake but it was driving me crazy. Thank you all for helping me correct my mistake!

That's not quite right. The normal force is normal to the rod and $N\sin{\theta_{N}} = N$. You haven't multiplied it by a lever arm. You are right in that you need to balance torques, but first you must make sure you actually have torques.

Call the total length of the rod $d$, and take moments about the point where the rod meets the ground. What is the torque of the weight? What is the torque of the normal force?

 

[PhysHobbiest]

That's not quite right. The normal force is normal to the rod and $N\sin{\theta_{N}} = N$.

Sorry, I should have been clearer here. sin(θ) here is the length of the rod (I should have written $lsin{\theta}$ although Morin himself omits the $l$), not for the component of the normal force perpendicular to stick, so really what I was saying is that I should have equated the torque to $rN\sin{\theta_{N}} = rNsin{(\pi/2)}$ with $r = sin{\theta}$.

Before I was equating a Force to a Torque which was incorrect.

 

[etotheipi]

Sorry, I should have been clearer here. sin(θ) here is the length of the rod (I should have written $lsin{\theta}$ although Morin himself omits the $l$), not for the component of the normal force perpendicular to stick, so really what I was saying is that I should have equated the torque to $rN\sin{\theta_{N}} = rNsin{(\pi/2)}$ with $r = sin{\theta}$.

Ah okay, glad you got it. Just be careful since you've also been using the letter $r$ for two different distances ($\frac{l\sin{\theta}}{2}$ and $l\sin{\theta}$).