Why can't force normal produce a torque?

[hellocello]

Homework Statement

A student is trying to push a desk of mass m = 110kg and horizontal width w = 0.7m along the floor. The coefficient of friction between the desk and the floor is ms = 0.56. The center of mass of the desk is at the geometric center of the rectangle shown. If the force that is applied is just barely less than what is necessary to move the desk, what is the maximum height h so that the desk won't tip over?

Homework Equations

torque= Fl
Net torque= 0 in equilibrium

The Attempt at a Solution

net torque= 0
F_applied*h - F_friction*l +F_gravity*l -F_normal*l =0

The friction force creates no torque if the desk is rotating around an axis on the ground because l will be zero. Then the gravity and normal forces cancel out. But my textbook doesn't include the normal force in the equation- why?

 

[PhanthomJay]

hellocello, welcome to Physics Forums

The Attempt at a Solution

net torque= 0
F_applied*h - F_friction*l +F_gravity*l -F_normal*l =0

The friction force creates no torque if the desk is rotating around an axis on the ground because l will be zero. Then the gravity and normal forces cancel out. But my textbook doesn't include the normal force in the equation- why?

A few comments here should be noted.

(1.) First be careful about the use of the letter variable 'l'. The 'l' is the perpendicular distance from the line of action of a particular force to the point on the axis about which you are summing torques. The l's are not necessarily the same, so don't use the same designation 'l' for the specific perpendicular distance to a specific force.

(2.) You seem to be summing torques about the left bottom corner of the desk. This is fine, you can choose any point for bodies in equilibrium where sum of torques about any point = 0. If this is the chosen point, then the 'l' for the friction force is 0, the 'l' for weight is .35 m, the 'l' for the applied force is 'h', and the l for the normal force is (??...see (3) below). I think you understand this OK, but i just want to emphasize my comment in (1) above, to avoid confusion.

(3) In your equation, you noted that the torque from the friction force is zero, which is correct, and that the torques from the weight and normal forces cancel, which is incorrect. If this was correct, you get F_applied*h = 0, which makes no sense. Your error is in the location of the Normal force. It is not located at the center of the bottom of the desk when it is about to tip over. When it's about to tip over, it has mostly lifted off the surface. Where is it located as this instant? Once you establish this location, you then also need to find the magnitude of applied force, based on info given in the problem statement.

 

[Raziel2701]

Here's my thought: If you push the desk and it starts to tip over at the legs of the desk, on the side of the desk opposite to you, then it tips because there's a force of friction resisting your applied force and so the legs act as a pivot point about which to rotate the desk.

This frictional force arises simply because you pushed the desk, it's a reactive force that depends on the coefficient of friction and also the normal force.

Now the normal force, it's always in the direction perpendicular to the surface, it's a Newton's third law pair, you push against something, and it pushes right back. So I don't see how the normal force could ever do work or torque on an object.

Also, if we look at torque as r x F, then it's clear that these two vectors are parallel, so this cross product is zero, and thus no torque.

 

[hellocello]

Okay, that makes sense, thank you!

 

[rwisz]

The normal force is perpendicular to the surface of contact between the desk and the floor, and therefore, does not produce any torque. Torque is a measure of the force's ability to cause rotational motion, and since the normal force acts in a direction perpendicular to the motion of the desk, it does not contribute to any rotational motion. In this scenario, the normal force is simply balancing out the force of gravity to keep the desk in equilibrium. Therefore, it is not included in the torque equation.