Normal force in man-on-ladder torque questions?

[miaou5]

Homework Statement

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!

 

[ehild]

Good question! The man stands on a step of the ladder which is (hopefully) horizontal. He exerts some normal force F_N 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 F_N-mg=0, F_N=mg. At the end, the man exerts a force, equal to its weight, on the ladder.

ehild

 

[haruspex]

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.

 

[miaou5]

I get it now! Thanks so much to both of you. Whew, thinking about these things can be a real doozy sometimes

 

[Nickie]

The normal force, or Fn, is a reaction force that acts perpendicular to the surface of contact between two objects. In the case of the man-on-ladder scenario, the normal force is acting between the man's feet and the ladder. This force is necessary for the man to maintain his balance and prevent him from falling off the ladder. However, in terms of torque, the normal force does not play a significant role.

Torque is a measure of the rotational force applied to an object. In this scenario, the only force that is creating torque is the weight of the man (mg). The normal force, being perpendicular to the surface of contact, does not contribute to the rotational force. Therefore, it is not included in the torque calculation.

In the case of loop-the-loop roller coaster problems, the normal force does play a role because the roller coaster car is in contact with the track at all times and the normal force is constantly changing to keep the car on the track. This is not the case for the man-on-ladder scenario as the man is not in contact with the ladder at all times and the normal force remains constant.

I hope this explanation helps to clarify the role of the normal force in torque calculations.