Physical interpretation of free body force diagram

[sassora]

Homework Statement

Free body force diagram question, where I have selected the best answer but the diagram doesn't ring completely true (see diagram uploaded). I think mathematically the diagram is sufficient but it doesn't properly represent the situation. Do you agree?

Homework Equations

The Attempt at a Solution

The normal reaction force must be somewhat counteracting the weight of the climber. So to show the weight perpendicular to the normal reaction force seems very odd, as if they are not linked. Without any weight there would be no normal reaction force.

I think this issue would be resolved if the normal reaction force was pointing slightly upward, so the climber would press onto the wall. Does that sound right?

 

[arpon]

The person is pushing the surface (in the direction perpendicular to the surface). That is why the surface is exerting equal normal reaction force on the person in the opposite direction.

 

[kuruman]

I think this issue would be resolved if the normal reaction force was pointing slightly upward, so the climber would press onto the wall. Does that sound right?

It does not sound right. The meaning of "normal" force is the same as "perpendicular" force to the surface that is exerting it. If you want to quibble with details, because of the rock's curvature, it looks like the left foot of the climber experiences a normal force that is up and to the left while the right foot experiences a force that is down and to the left. Leaving that aside, the climber does not need to press into the wall. Because the rope is at an angle, if the wall were not there, he would swing like a pendulum except that the wall stops him. If we assume a perfectly vertical surface, the normal force will have to be horizontal to cancel the horizontal component of the tension in the rope (assuming that the climber is at equilibrium). The vertical weight is canceled by the vertical component of the tension plus static friction. Tension and surface forces adjust themselves to provide the observed acceleration. Free body diagrams help you visualize all forces needed to apply Newton's second law. Gravity is easy to draw, but contact forces are not always obvious.

 

[CWatters]

I disagree with some of the above. The free body diagram makes assumptions that do not agree with the photo for the following reasons.

The reaction force acts at his feet which in the photo is below his centre of mass. If the reaction force at his feet were horizontal then there would be a net clockwise torque about his centre of mass. That cannot be the case if he is in static equilibrium.

To counter this there must be a friction force acting vertically at his feet. The vector sum of the normal force and friction forces must point from his feet through his centre of mass.

 

[CWatters]

The normal reaction force must be somewhat counteracting the weight of the climber.

See my post above. It's not the normal force at his feet that carries some of the weight. It's the friction force.

To get the free body diagram in the answer they must have assumed the climber or at least his legs are horizontal (and level with his centre of mass) so that there is no need for a friction force at his feet.

What were the other answers?

 

[sassora]

Yes, I was quibbling with the details to have a better understanding of what I am assuming.

Do you mean that the normal reaction force is due to the horizontal component of tension? I can see that these would cancel out in equilibrium but the wall cannot have a normal reaction if no force is applied directly on it. Isn't that the point of the normal reaction force?

The tension force and the weight force (force of gravity) actually exist, but the components are just abstract to help deal with the problem.

 

[kuruman]

The tension force and the weight force (force of gravity) actually exist, but the components are just abstract to help deal with the problem.

The contact force generated by the rock face also exists. If the rock face were not there, the climber would be hanging vertically in equilibrium or swinging like a pendulum if not in equilibrium. I agree that the components are abstract in the sense that they can be different depending on the axes you choose to describe them. However, they help sort things out when you add forces and torques to determine the net force and torque.

 

[CWatters]

Do you mean that the normal reaction force is due to the horizontal component of tension? I can see that these would cancel out in equilibrium but the wall cannot have a normal reaction if no force is applied directly on it. Isn't that the point of the normal reaction force?

In short. Yes.

The total Reaction Force at the feet (red on my diagram) can be broken down into horizontal and vertical components (green). The horizontal component we call the Normal Force and in this case it will indeed be equal to the horizontal component of the tension in the rope. The vertical component is the friction force.

 

[sassora]

I've taken the question from here: http://qualifications.pearson.com/c...2013/Exam-materials/6PH01_01_que_20150519.pdf
(it's question 1). The only plausible answer is C (the one attached in my first post), I'd get the mark but I'm trying to learn this inside out. Thanks for your responses by the way.

@CWatters Your diagram (at least the red parts) are what I thought the free body diagram would look like. It seems to be a level more complex than the question considers but it seems to make more sense as a whole.

@kuruman If we assume that the person is laying down holding the rope. I can see that the normal reaction force is horizontal. It seems like an oversimplification since once the person has stop swinging on the pendulum, the person is practically just suspended vertically on a rope and no contact force would be needed to maintain this (and hence no reaction force).

 

[kuruman]

I think we all agree. This is what I said,

If we assume a perfectly vertical surface, the normal force will have to be horizontal to cancel the horizontal component of the tension in the rope (assuming that the climber is at equilibrium).

For a vertical rock, the contact force at the wall has two components, (1) the normal force that is perpendicular to the face, therefore in the horizontal direction; (2) Static friction that is parallel to the face, therefore vertical. As @CWatters indicated, both components are necessary to maintain equilibrium. Nevertheless, this does not invalidate the fact that the normal component of the contact force is perpendicular to the rock face. The posted free body diagram (choice C) is incomplete in that it shows only the component of the contact force that is perpendicular to the face and omits the component that is parallel. However, what is drawn is drawn correctly. I believe that the point of this question is not a detailed analysis of the kind discussed here. Judging from the other choices, it looks like the point of the question is to determine whether the respondent (a) understands that there ought to be a normal contact force and (b) that this normal contact force is perpendicular to and directed away from the rock face.

 

[sassora]

I agree. I've met the requirements of the question, we have helpfully analysed the situation a little further. I think @CWatters point was important about the assumption that they must have assumed the climber or at least his legs are horizontal (and level with his centre of mass) so that there is no need for a friction force at his feet. Which isn't what the picture shows.

I like Physics but it stuff like this causes misunderstanding and difficultly i.e. about what level of detail people happen to be considering when they discuss a concept.

Thanks this was helpful, I will probably be posting the use of definitions / concepts as I work through them!

 

[CWatters]

I like Physics but it stuff like this causes misunderstanding and difficultly i.e. about what level of detail people happen to be considering when they discuss a concept.

Indeed. That's one of my pet hates.

Back in the 1970's when I was at school I studied Physics and Applied Maths. In physics we were taught that friction force was independent of area and the coefficient of friction had the appropriate units. Then later in Applied Maths we were presented with a question that had different units for the coefficient of friction and when we questioned this we were told it was because the friction force was proportional to area, which isn't usually true. We were told that if the units weren't mentioned we were to remember which exam we were sitting and answer accordingly.

 

[kuruman]

Back in the 1970's when I was at school I studied Physics and Applied Maths. In physics we were taught that friction force was independent of area and the coefficient of friction had the appropriate units. Then later in Applied Maths we were presented with a question that had different units for the coefficient of friction and when we questioned this we were told it was because the friction force was proportional to area, which isn't usually true. We were told that if the units weren't mentioned we were to remember which exam we were sitting and answer accordingly.

Back in the same era I was similarly taught. It took me a while to distinguish modeling from reality. I think I understand modeling now, but I am not sure about reality.

 

[Blanci1]

The problem could be a bit oversimplified. For example maybe his centre of mass is not at the pivot point and the guys arms up the rope give some fine adjustments. In another Pearson exam an assumption of 90 degrees of legs to rope was made also pretty random and not very convincing. Exam problems should be more clearly defined . This is more like a project . Too many nuances for an exam!