Step-by-Step Guide: Drawing Free Body Diagrams in Physics
The free body diagram (FBD) is one of the most effective tools for addressing a static problem. A free body diagram is a visual, dematerialized, symbolic depiction of the body (structure, element, or fragment of an element) that has had all connecting "parts" removed. A FBD is a handy way to simulate the structure, structural element, or segment under consideration. It is a method of conceptualising the structure and its composite pieces in order to begin an examination. The structure's physical components are all eliminated. This is not done at random, but rather according to a certain procedure.
A body, or a section of one, is represented by a single line. Each connection is either represented entirely by a junction with discrete attributes, or it is substituted by a collection of forces and moments representing the activity at that connection. Internal pressures observed at a node (connection or joint) can be substituted by representational exterior forces when that "part" links to the other member in the FBD. Every load is represented as a force system.
Free Body Diagram Definition
No system, natural or man-made, consists of a single body or is complete in and of itself. A single body or portion of the system, on the other hand, can be separated from the remainder by roughly accounting for its influence on the rest of the system. A free body diagram is a graphical representation of a single body or a subsystem of bodies that is isolated from its surroundings and depicts all of the forces acting on it.
It also depicts the body of interest as well as the forces operating on it. You can carefully specify the body (or object under consideration) to which you are applying mechanical equations and the forces that must be addressed using a free body diagram. So creating a free body diagram makes it easy for us to comprehend the forces, torques, and moments, and it also shows you how to use the right principles to solve your problem.
Free Body Diagram Examples
There are several examples for a free body diagram and some of which are given below as,
Consider a book lying on a table having smooth horizontal surface so the free body diagram of the book alone consists of its weight (w=mg), acting through the centre of gravity and normal reaction (N) exerted on the book by the surface.
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Consider an object lying on a frictional horizontal surface then its free body diagram consists of weight of an object (w=mg), frictional force (f) and normal reaction (N) of an object.
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Consider two blocks of masses m1, m2 attached by a string and pulled by a force (F). The following system is showing as,
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If T denotes the tension in the string, then free body diagrams of the blocks will be given as,
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Here, R1 is the normal or contact force acting between the block of mass m1 and surface, m1g is the weight of the block, T is the tension in the string and F is the force applied on a body.
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Here R2 is the normal or contact force acting between the block of mass m2 and surface, m2g is the weight of the block, T is the tension in the string and F is the force applied on a body.
How to Draw a Free Body Diagram ?
Following points need to be considered while drawing a free body diagram and these are given below as,
Begin by determining the contact forces. By highlighting the item, we can see what it is touching. Draw a dot when something contacts the outline; there must be at least one contact force where there is a dot. Draw force vectors at the contact locations to show how the force pushes or pulls on the item (including correct direction).
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Draw a dot to represent the item once we've discovered the contact forces. We are only interested in the forces operating on our item, not the forces the object exerts on other things.
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Make a coordinate system with the positive directions labelled. Align the axes with the inclination if the object is on an incline.
Draw an arrow heading away from the dot to represent the contact forces. Ascertain that the arrow lengths are proportionate to one another. All forces should be labelled.
Long-range forces should be drawn and labelled. Unless there is an electric charge or magnetism present, this will typically be weight.
Your acceleration vector should be drawn and labelled to the side of the dot, not touching it. Write a=0 if there is no acceleration.
Free Body Diagram Problems and Solutions
Problem 1. The pulley in the mechanism depicted in the figure is smooth. String is massless and inextensible. Find acceleration of the system a, tensions T1 and T2 .
(Take g=10 m/s2)
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Solution: In this problem the net pulling force will be equal to the weight of 4 kg and 6 kg block on one side minus the weight of 2 kg block on the other side. Therefore we can write,
Net pulling force = weight of 4 kg block + weight of 6 kg block - weight of 2 kg block
After putting the values of weight of different blocks we get,
Net pulling force = 6g +4g-2g=8 g=8 × 10=80 N
Now, according to Newton’s second law we can write the expression of acceleration as,
a=(Net pulling force)/(Total mass)=80/12=20/3 m/s2
In order to calculate T1 and T2 we have to draw free body diagrams for it. For T1, the FBD of 2 kg block is drawn as,
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We may deduce Newton's second law of motion by writing,
T1-20=2a
After putting the value of a we can obtain the value of T1 as,
T1=20+2× (20/3)=100/3 N
For T2, the FBD of 6 kg block is drawn as,
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We may deduce Newton's second law of motion by writing,
60-T2=6a
After putting the value of a we can obtain the value of T2 as,
T2=60-6a=60-6(20/3)=60/3 N
Hence the values of a , T1 and T2 are 20/3 m/s2, 100/3 N and 60/3 N.
Problem 2. All surfaces in the system depicted in the figure are smooth. String is massless and inextensible. Determine the system's acceleration (a) and the tension (T) in the string.. (Take g=10 m/s2)
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Solution: Here, weight of 2 kg is perpendicular to motion (or a). Hence, it will not contribute to net pulling force. Only weight of 4 kg block will be included.
Net pulling force = 4g=4 ×10=40 m/s2
We may deduce Newton's second law of motion by writing acceleration a as,
a=(Net pulling force)/(Total mass)=40/(4+2)=40/6=20/3 m/s2
In order to calculate tension (T), we have to draw the free body diagram of 4 kg block which is given below as,
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According to FBD, the equation of motions are,
40-T=4a
Now upon putting the value of a, we can obtain value of T as,
T=40-4a=40-4(20/3)=40/3 N
Hence, the values of acceleration (a) and tension (T) are 20/3 m/s2 and 40/3 N.
FAQs on Free Body Diagram Explained for Students
1. What is a Free-Body Diagram (FBD) in Physics?
A Free-Body Diagram, or FBD, is a simplified, schematic representation used in physics and engineering to visualise all the external forces acting on a single object. The object is shown isolated from its surroundings, or "free" of its environment. This allows for a clear analysis of forces to determine the object's state of motion, such as its acceleration or whether it is in equilibrium.
2. What are the essential steps to draw an accurate Free-Body Diagram?
To draw an accurate FBD, follow these steps:
Isolate the Body: First, identify the specific object you want to analyse and draw it in isolation, free from all other objects and surfaces.
Represent as a Point: For simplicity, you can often represent the object as a single point or a simple shape. This point represents the object's centre of mass.
Identify and Draw Forces: Identify all the external forces acting on the object (e.g., gravity, normal force, tension, friction). Draw each force as a vector arrow originating from the point. The arrow's direction must show the direction of the force, and its length should ideally be proportional to its magnitude.
Label the Forces: Clearly label each force vector with a symbol (e.g., 'Fg' for gravity, 'N' for normal force, 'T' for tension).
Establish a Coordinate System: Draw an x-y coordinate axis to help resolve the forces into their components for calculations.
3. What is the importance of drawing a Free-Body Diagram in solving mechanics problems?
The primary importance of a Free-Body Diagram is that it translates a complex physical situation into a simpler, organised problem. It helps students to:
Systematically identify all relevant forces acting on an object and avoid missing any or including incorrect ones (like internal forces).
Clearly visualise the direction and nature of each force, which is crucial for setting up equations.
Apply Newton's Laws of Motion correctly by setting the sum of forces in each direction equal to mass times acceleration (ΣF = ma).
Analyse complex systems by drawing separate FBDs for each component part.
4. What common forces should be included in a Free-Body Diagram for a typical Class 11 problem?
In a typical Class 11 Physics problem, the most common forces to represent on a Free-Body Diagram are:
Weight (W or Fg): The force of gravity acting vertically downwards towards the centre of the Earth.
Normal Force (N or FN): The perpendicular contact force exerted by a surface on an object to prevent it from passing through. It always acts perpendicular to the surface.
Tension (T): The pulling force transmitted through a string, rope, cable, or chain. It always acts along the direction of the string, away from the object.
Frictional Force (f): The force that opposes the relative motion or tendency of motion between surfaces in contact. It acts parallel to the surface. It can be static (fs) or kinetic (fk).
Applied Force (Fapp): Any external push or pull applied to the object.
5. How does a Free-Body Diagram help in applying Newton's Second Law of Motion?
A Free-Body Diagram is the critical first step for applying Newton's Second Law (ΣF = ma). By isolating the body and representing all external forces as vectors, the FBD allows you to translate the visual diagram into mathematical equations. You can sum the force components along the chosen x and y axes. For the x-axis, you write ΣFx = max, and for the y-axis, ΣFy = may. Without an FBD, correctly identifying and summing all the vector forces for these equations would be extremely difficult and prone to error.
6. What are some common mistakes to avoid when creating a Free-Body Diagram?
Students often make a few common mistakes:
Including Internal Forces: An FBD must only show external forces acting on the body. Forces that parts of the body exert on each other are internal and should not be included.
Incorrect Force Direction: The direction of forces like the normal force (always perpendicular to the surface) and friction (always opposing motion or intended motion) must be drawn correctly.
Drawing "ma" as a Force: The term 'ma' from Newton's second law is the result of the net force, not a force itself. It should never be drawn on the FBD.
Forgetting a Force: It is easy to forget forces like weight or the normal force. A systematic check of all possible interactions (contact and non-contact) can prevent this.
7. Why is it crucial to isolate the object from its surroundings in a Free-Body Diagram?
The act of isolating the object is the core principle that gives the diagram its power and its name—"free-body." It is crucial because physics principles like Newton's Laws of Motion apply to a single body or a defined system. By "freeing" the body from its surroundings (like tables, ropes, or other objects), you can focus exclusively on the forces that the surroundings exert on that specific body. This prevents confusion, such as incorrectly including forces that the body exerts on its environment, which are part of Newton's third law action-reaction pairs but do not determine the body's own acceleration.
8. How does the Free-Body Diagram for an object on an inclined plane differ from one on a flat surface?
The key difference lies in the orientation of the forces relative to the coordinate system.
On a Flat Surface: Typically, the weight (Fg) acts vertically down, and the normal force (N) acts vertically up, directly opposing the weight if no other vertical forces are present. Friction acts horizontally. A standard horizontal/vertical x-y axis works perfectly.
On an Inclined Plane: The weight (Fg) still acts vertically down. However, the normal force (N) acts perpendicular to the inclined surface, not vertically. To simplify calculations, the coordinate system is usually tilted, with the x-axis parallel to the incline and the y-axis perpendicular to it. This means the weight vector must be resolved into two components: one parallel to the incline (Fg sinθ) and one perpendicular to it (Fg cosθ).
