# Rotational Equilibrium and Normal Forces

• chemica1mage
In summary, the conversation discusses the concept of torque and rotational equilibrium in relation to a person standing on a plank. The person's weight and the normal force exerted by the plank are considered for both translational and rotational equilibrium. The clarification is made that the normal force of the board on the person is not considered as a torque-producing force, as the person can be treated as a point object. The conversation also touches on the different forces at play in two different scenarios - a person standing on the floor versus a person on a see-saw. Ultimately, it is clarified that the support force on the fulcrum is what balances out the person's weight in the see-saw scenario.
chemica1mage
Hi,

The question I have is not for a numerical answer but for clarification.

Some of the questions involving torque/rotational equilibrium describe a person standing on a plank. I know that the gravitational force of the person on the plank needs to be considered for translational and rotational equilibrium. My question is, why don't I consider the normal force exerted by the board on the person as a torque-producing force? (Your typical dynamics situation where F(normal) = F(gravity).) Or is it because the "normal force" is distributed between the two supports.

Any help on this question is greatly appreciated!

welcome to pf!

hi chemica1mage! welcome to pf!
chemica1mage said:
… why don't I consider the normal force exerted by the board on the person as a torque-producing force? (Your typical dynamics situation where F(normal) = F(gravity).) Or is it because the "normal force" is distributed between the two supports.

by the board on the person? but the person is effectively a point object … where does torque come into it?

Maybe that picture wasn't the best to demonstrate my problem. Here's another.

Why is it that for the diagram on the right (person standing on the floor), you can consider the normal force of the person, but with the diagram on the left (person on a see-saw), you don't consider the "normal force" of board pushing up on the person? Or is it because that support force gets lumped into the support on the fulcrum?
I hope that it makes a little more sense, what I'm trying to ask. Thanks in advance!

chemica1mage said:
Maybe that picture wasn't the best to demonstrate my problem. Here's another.
View attachment 55411
The left hand diagram shows all the forces acting on the board.

The right hand diagram shows all the forces acting on the person.

Thanks! That really clears things up for me. Seems so obvious, now that you say that.

## 1. What is rotational equilibrium?

Rotational equilibrium is a state in which an object is not rotating, or is rotating at a constant rate, without any external torques acting on it. This means that the sum of all the torques acting on the object is equal to zero.

## 2. How do you determine the direction of the normal force in rotational equilibrium?

In rotational equilibrium, the normal force is always perpendicular to the surface that the object is resting on. Its direction can be determined by using the right-hand rule, which states that if you point your fingers in the direction of the object's motion, and curl them towards the surface, your thumb will point in the direction of the normal force.

## 3. What factors affect the magnitude of the normal force in rotational equilibrium?

The magnitude of the normal force in rotational equilibrium depends on the weight of the object, the angle of the surface, and any other external forces acting on the object. The normal force will increase if the weight of the object increases, or if the angle of the surface becomes steeper.

## 4. Can the normal force ever be greater than the weight of an object in rotational equilibrium?

Yes, the normal force can be greater than the weight of an object in rotational equilibrium. This can happen if the object is on an inclined plane with a steep angle, or if there are other external forces acting on the object.

## 5. How do you calculate the normal force in rotational equilibrium?

The normal force in rotational equilibrium can be calculated using the formula N = mg cosθ, where N is the normal force, m is the mass of the object, g is the acceleration due to gravity, and θ is the angle of the surface. This formula takes into account the weight of the object and the angle of the surface to determine the magnitude of the normal force.

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