Why Is the Normal Force Excluded in Beam Static Equilibrium Calculations?

[cardinalboy]

Homework Statement

A 3.0 m long rigid beam with a mass of 130 kg is supported at each end. A 65 kg student stands 2.0 m from support 1. How much upward force does each support exert on the beam?

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Homework Equations

Net Forces = F1 + F2 - w(student) - w(beam) = 0

where F1, F2 are the upward forces of support 1 and support 2; w indicates weights of beam and student

The Attempt at a Solution

I was able to finally solve this question but there is something I am still not clear on. During the initial part of our physics course (dynamics combined with kinematics, for example) we used normal forces when a mass was resting on top of a surface, including it in the force diagram even when there was no acceleration in the "y" direction and so the normal force was not relevant to the problem being solved. However, in statics problems similar to the one above, the normal force supporting a mass on a beam is not included. For example, in the problem given above, the student contributes to the weight of the beam, however the beam does not provide a normal force on the student - or at least it is not included when summing the vertical forces in the problem, anyway. In these types of "beam" static equilibrium problems, why is it that only the weight of the mass is included when summing forces, and yet never the normal force on the mass? Isn't the beam providing a normal force on the mass?

 

[rl.bhat]

In all the static problems, we are dealing with the equilibrium condition. And there are two conditions. 1) Total downward force = Total upward force. 2) Total clockwise moment = Total anticlockwise moment. In the above problem, take one support as the reference point and take the moments of beam, boy and reaction of the second support from that point.. Student can stand on the beam due to the normal reaction of the beam, but that does not contribute to the moment.

 

[Doc Al]

However, in statics problems similar to the one above, the normal force supporting a mass on a beam is not included. For example, in the problem given above, the student contributes to the weight of the beam, however the beam does not provide a normal force on the student - or at least it is not included when summing the vertical forces in the problem, anyway. In these types of "beam" static equilibrium problems, why is it that only the weight of the mass is included when summing forces, and yet never the normal force on the mass?

Generally, we choose to analyze the forces acting on the beam, not on the mass. The mass exerts a normal force on the beam, which is equal to the weight of the mass.

Isn't the beam providing a normal force on the mass?

Absolutely: The beam most certainly exerts a normal force on the mass, which is equal and opposite to the normal force that the mass exerts on the beam. That normal force is equal to the weight of the mass, since the net force on the mass must be zero.

 

[cardinalboy]

Thanks very much for your help. It helps when I consider this problem from the perspective of the forces on the beam only. When doing so, it makes sense that we would be concerned only with the forces acting on the beam, and therefore the normal force exerted by the beam on the mass would not be important.

However, it still seems that I am simply ignoring some forces in order to get a solution to the problem. In the problem I gave above, if the normal force provided by the beam on the student IS included in the force and torque equations, then the (upward) normal force exerted by the beam on the student would be equal and opposite to the (downward) force due to gravity on the student (weight) and so they would cancel one another out when writing the force equation:

Net Forces = F1 + F2 + N(student) - w(student) - w(beam) = F1 + F2 - w(beam) = 0

where N(student) is the normal force exerted by the beam on the student.

The rotational equilibrium equation gives (assuming the left support is the pivot point):

Net Torque = [2 m][N(student)] - [2 m][w(student)] - [1.5 m][w(beam)] + [3 m][F2] = [3 m][F2] - [1.5 m][w(beam)] = 0

When solving for the upward forces provided by the supports using these 2 equations, the solution is F1 = F2. Clearly this cannot be correct for the right support must be providing a greater upward force than the left support due to the proximity of the student to the right support.

Where is my logic failing me here? I am not sure that I would have known to exclude the normal force on the student if I hadn't gotten assistance on this problem. I would like to clarify the error in my logic so that I don't make the same false assumptions in future problems. Thanks in advance!

 

[Doc Al]

However, it still seems that I am simply ignoring some forces in order to get a solution to the problem. In the problem I gave above, if the normal force provided by the beam on the student IS included in the force and torque equations, then the (upward) normal force exerted by the beam on the student would be equal and opposite to the (downward) force due to gravity on the student (weight) and so they would cancel one another out when writing the force equation:

Net Forces = F1 + F2 + N(student) - w(student) - w(beam) = F1 + F2 - w(beam) = 0

where N(student) is the normal force exerted by the beam on the student.

You simply need to decide what body you are analyzing. If you are analyzing the beam, then you must stick to forces acting on the beam. In the above, you added an extraneous force of N(student) which acts on the student and not on the beam. (For some reason you are concerned about the force the beam exerts on the student, but not about the forces that the beam exerts on the supports. )

Note that you are not ignoring the normal force between student and beam. The force you call "w(student)" is actually the normal force that the student exerts on the beam. (It happens to equal the student's weight, since the sum of the forces on the student must add to zero.)

Note that it's perfectly OK to consider the "beam + student" as single system. In which case we ignore the internal forces between beam and student and just deal with their weights (and the forces F1 & F2). Of course, you get the same answer.

 

[cardinalboy]

(For some reason you are concerned about the force the beam exerts on the student, but not about the forces that the beam exerts on the supports. )

Ahhh! Good point! You got me there... Thanks for pointing that out to me. It's always good to better understand my errors.

Note that it's perfectly OK to consider the "beam + student" as single system. In which case we ignore the internal forces between beam and student and just deal with their weights (and the forces F1 & F2). Of course, you get the same answer.

Yes, I like this approach better...it simplifies things for me to think about a single system as opposed to separate systems. Thank you very much for taking the time to help me out. It is much appreciated!