Identifying Action-Reaction Forces on Free Body Diagrams

Free body diagram is the diagrammatic representation of all the forces acting on a body. They are applied in mechanical engineering very commonly. This wiki deals with the following topics:

1. Basic principles of free body diagrams
2. Free body diagrams related to inclined surfaces
3. Static and kinetic friction
4. Coefficient of friction
5. Connected objects

Prerequisites:

1. You should know Newton's three laws of motion.
2. You should have a strong understanding of trigonometry, at least the basics.
3. You should know what \(x\) and \(y\) components of vectors are.
4. You should be able to form and solve simultaneous equations.

They are diagrams that represent the forces acting on an object; usually, the forces discussed are the vertical and horizontal forces. The reaction forces of the object are not included, but the force exerted by the horizontal plane to the object is crucial; this force is perpendicular to the horizontal force. It is known as the normal, denoted \(n.\)

• \(n\) in the diagram above represents the normal force, which is not the reaction force from the same body to itself, because action and reaction forces never happen at the same object. \(n\) is the force provided by the ground to the object, preventing the object from sinking into the ground.

• \(F\) in the diagram is usually the force applied to the object which might otherwise make the object accelerate/decelerate or stay in equilibrium.

• \(f\) is the friction force, which opposes the motion; there are two types, static and kinetic, but these will be discussed later on.

• \(mg\) is the weight force of the object, which acts downwards towards the object.

Horizontal and Vertical Forces:

Forces do not have to act necessarily like this. They could act on angles for that matter, but just to explain it to you, we will put it this way: the sum of horizontal forces for an object in equilibrium, whether at a constant velocity or at rest, will be zero. That means that if an object is going at the same speed, it will have a net force of zero because it is not accelerating, by the formula \(F=ma,\) in which \(a=0\) means \(F=0.\) If an object does undergo acceleration, it will have a force acting on it. For an object that is going horizontal not in the air but on the ground, its vertical forces will be equal to each other.

These equations will illustrate everything to you:

Free body diagrams related to inclined surfaces:

An inclined surface is something that looks like this (what, you wanted a complex definition?):

If you are good at pattern spotting, then you might have realized that the sine of the angle multiplied by the weight is the force causing the movement of the object. The cosine angle multiplied by the weight is the force normal (imagine the arrow cosine having a line going up as well as down). The following examples will help you:

A driver forgot to put the handbrake on his sports car. Now it is accelerating down an icy slope with negligible friction. If the weight of the car is 980 N, and the hill has an inclined angle of \(30^{\circ}\), find the force at which it is going down.

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Strategy: Since we know weight is a vector and the inclined surface is at \(30^{\circ}\), we can use what we were taught in defining the inclined surface, to solve this problem. That means taking vector components of the vector and finding the component acting along the slope (since it is parallel to the force acting on it); let's call it \(x\). Then

\[\begin{align} x\text{-component } &= mg\sin \theta \\ &=980 \text{ N } \times \sin 30^{\circ} \\ &=980 \times \dfrac{1}{2} \\ &=490 \text{ (N)}. \ _\square \end{align}\]

Well, I hope that gives you an idea, but most objects you will be dealing with will never be this easy (but if you understand this part, that is a major big step, so pat yourself in the back). The hard part is that you will have to consider other forces acting on, which are usually air resistance and friction, which take us go to our next topic.

Friction is simply the resistance to movement. It is a contact force.

Kinetic Friction: Usually, when you are first introduced to friction, it is the force that tries to stop the object from moving (don't forget the good things about friction). Well, that is called kinetic friction, the opposing force of an object while it is in motion.

Static Friction: Have you ever noticed it is harder to start pushing something than to keep it in motion? Well, if you have not, try pushing something a little heavy for you. You should notice that trying to push it to get a movement out of it was harder than keeping it moving. That is known as static friction and it is the force that keeps the object "inert" (I hope you know this definition). The static friction is always more than the kinetic friction. The maximum value of static friction is called limiting friction. Until limiting friction is reached, static friction is equal to the applied force.

There are rules relating to the movement of friction:

1. The movement is always parallel to the surface.
2. It is always against the applied force.
3. It is independent of the applied force.
4. It is independent of the area of contact with the surface.

Coefficient of Friction: The following is a table that compares coefficient of kinetic friction and coefficient of static friction between two objects:

This proves the fact that static friction is always greater than kinetic friction. Don't worry about this table because no one is asking you to memorize it. They usually give it to you in the question; sometimes you have to calculate the coefficient of friction.

A body of mass 2 kg is kept on a table with coefficient of static friction 0.2 and coefficient of kinetic friction 0.1. If a force of 20 N is applied, find the frictional force acting on the body. \(\big(\)Assume that \(g = 10\text{ m/s}^2\big).\)

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Firstly, we have to know these formulae:

• (Limiting Friction) = (Mass) \(\times\) (Coefficient of Static Friction) \(\times\) (Acceleration due to Gravity)
• (Kinetic Friction) = (Mass) \(\times\) (Coefficient of Kinetic Friction) \(\times\) (Acceleration due to Gravity).

In this case,

\[\text{Limiting Friction } = 2 \times 0.2 \times 10 = 4 \text{ N}.\]

This means that the body doesn't move till a minimum force of \(4\text{ N}\) is applied. So, in this case, since a force greater than \(4\text{ N}\) is applied, the body is in motion and the friction which now comes into play is the kinetic friction.

Now,

\[\text{Kinetic Friction} = 2\times 0.1\times 10 = 2\text{ N}.\]

Therefore, the frictional force acting on the body is 2 N. \(_\square\)

A car moves with a uniform velocity of \(54\text{ km/h}\) on a rough, horizontal road with coefficient of kinetic friction \(0.75.\) When it sees a child standing on the road \(20\text{ m}\) ahead, he applies brakes to bring the car to rest in some distance. How far is the child from the car when the car is at rest? \(\big(\)Assume that \(g= 10\text{ m/s}^2.\big)\)

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