# Static Equilibrium: Normal Force

Consider the following situation where an object (for example a person) is at rest at point D such that the system is in static equilibrium:

Why is there no normal force present at point D? To my understanding the normal force occurs when an object comes in contact with another object.

What is fundamentally different from points A and B (plank resting on block) than point D (object resting on plank)?

I ask this since this is the solution my professor has given me, but still don't understand it clearly and this is a big issue i have while solving static equilibrium problems.

• Why do you think there is no normal force at D? There is, and it is marked with a red arrow. Perhaps it's the diagram that is misleading, labeling the normal force by $mg$ might obscure the fact that that force is the normal force of the person's shoes on the plank. – garyp Jan 11 '18 at 19:10
• Nice question, you should study the shear and moments or mqn diagrams. – user98038 Apr 16 '18 at 20:08

## 2 Answers

The forces marked in red are the forces which act on the plank. Forces acting on other objects are not shown because they are not relevant. This is what is called a Free Body Diagram in which only the forces acting on the object in question (the plank) are considered.

The force shown at D is the normal force which the man exerts on the plank. Its magnitude is the same as the gravitational pull on the man, $m_0 g$.

Probably you are misled by the fact that the force at D is marked $m_0 g$. This force is not the gravitational pull on the man, because that force acts on the man. But because the man himself is in equilibrium, the force which the plank exerts on the man must equal $m_0 g$. And because forces come in equal and opposite pairs the reaction force which the man exerts on the plank is also $m_0 g$.

The forces shown at A and B are the normal forces exerted by the supports at those points.

So there is no difference between forces at A, B and at D. At all three points there are objects in contact with the plank, and there are normal reaction forces which those objects exert on the plank.

In the description above, the plank is the free body whose equilibrium is being studied. An alternative way of looking at the problem is to consider the plank and the man together as the free body which is modelled as two masses $m_p$ and $m_0$ attached rigidly to a massless rod.

The force at D is then the gravitational pull on the man at D, just as the force at the midpoint is the gravitational pull on the plank. The force at D is no longer a normal reaction force because the man and the plank are now taken to be part of the same body. As with the stresses along the plank which keep it rigid, the reaction forces between the man and the plank are now considered to be internal forces. Only external forces which act on the free body are relevant to the analysis.

• So there is actaully a normal force at point D, but it does not exert on the plank? Thanks, this makes alot of sense. – Bert de Saffel Jan 11 '18 at 19:00
• That is right. There are two equal and opposite normal reaction forces at D : the force of the man on the plank (shown) and the force of the plank on the man (not shown). Forces always occur in equal and opposite pairs, and act on different objects. Only one force in this pair acts on the plank. – sammy gerbil Jan 11 '18 at 19:02
• @BertdeSaffel There is a normal force on the plank at point D. This is stated in the first sentence of the second paragraph of this answer. – garyp Jan 11 '18 at 19:09
• @kamran Normal means perpendicular to the contact surface. The contact surfaces in this example are horizontal, so the contact forces are vertical. – sammy gerbil Jan 11 '18 at 19:44
• @kamran This is a physics website, not a structural engineering website. Terms used on this site have their usual meanings in physics, unless stated otherwise. – sammy gerbil Jan 11 '18 at 19:49

This beam does not have any normal force (axial force). Neither on supports nor at point D.

For equilibrium we have Sum Force Horizontal=0

and none of the loading is along horizon, therefor we do not have normal reactions or force anywhere.

If we consider vertical loads then thats a different case and the diagram shows we have two point loads, P and O and they have created two reactions on the supports upward. the sum of those two reactions of course is equall to sum of the two vertical load. And to calculate the two reactions we can assume point A as reference and find the sum of moment of loads P and O about A and divide by distance AB to get the reaction at B.