Normal force in man-on-ladder torque questions?

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Homework Help Overview

The discussion revolves around the concept of torque in a scenario involving a man standing on a ladder. The original poster questions the role of the normal force (Fn) exerted by the man on the ladder in relation to torque, particularly why it is not included in the torque calculations when the weight (mg) is considered.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to understand the omission of the normal force in torque calculations, contrasting it with scenarios like loop-the-loop problems where normal force is significant. Participants explore the balance of forces acting on the man and the implications for torque.

Discussion Status

Some participants provide insights into the relationship between the normal force and gravitational force, suggesting that the normal force equals the weight of the man when he is steady. The discussion appears to be productive, with participants offering different perspectives on the problem.

Contextual Notes

The original poster expresses confusion regarding the treatment of normal force in this context, indicating a potential gap in understanding how forces interact in torque scenarios. There is an implicit assumption that the ladder is horizontal and that static friction is sufficient to prevent sliding.

miaou5
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Homework Statement



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This is just a more generalized question, and not really about a specific value. The question was regarding torque, and had a man standing on a ladder like so. The only force (exerted by the man on the ladder) that provided torque was indicated to be mg, but I don’t really understand—why isn’t Fn included? Obviously there is a downwards Fn exerted by the man on the ladder, so why doesn’t the diagram account for that? (I’m a bit confused since in loop-the-loop roller coaster problems, we always have to account for downwards Fn.)

Thank you in advance!
 
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Good question! The man stands on a step of the ladder which is (hopefully) horizontal. He exerts some normal force FN on that step. The step acts with the same force, but in opposite direction on the man. Gravity also acts on the man. The man is steady, so the resultant force on him is zero. That means FN-mg=0, FN=mg. At the end, the man exerts a force, equal to its weight, on the ladder.

ehild
 
Here's another way to think about it. It needn't be a ladder; it could just be a plank, provided there is enough static friction that the man does not slide. In this case there would be a normal force, and a frictional force up the plank. But since these have to be balanced exactly by the vertical weight of the man, their resultant is vertical.
 
I get it now! Thanks so much to both of you. Whew, thinking about these things can be a real doozy sometimes :-p
 

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