Free Body Diagram Definition
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
1. How do you make a free body diagram and what does it represent?
Ans: FBDs are excellent tools for describing the relative magnitude and direction of all forces operating on an item in a particular condition. The meticulous drafting of a free-body diagram is the initial step in evaluating and explaining most physical processes. We represent the item of interest by sketching all the forces operating on it and resolving all force vectors into x– and y–components when generating a free body diagram. Each object in the problem should have its own free body diagram. In various sorts of mechanics issues, free body diagrams are used to illustrate the forces and moments applied to a body and to calculate the subsequent responses.
2. What is the significance of a free-body diagram?
Ans: The free body diagram's objective is to deconstruct a given problem by using only the essential information. Students can use this figure to build up computations to identify unknown variables, such as force directions, moments, or force magnitudes. Students can use a Free-Body Diagram to vividly envision a certain problem in its entirety or to closely investigate a specific component of a more complicated topic. So, in general, FBD is a highly valuable tool for visualizing and solving engineering challenges. It should be noted that a sequence of free body diagrams may be necessary to solve a difficult problem.
