Tips & Tricks To Solve The Mechanics Problems Using Free Body Diagram (FBD) Part: 1

Free-body diagrams are diagrams used to show the relative magnitude and direction of all forces acting upon an object in a given situation. A free-body diagram is a special example of the vector diagrams that were discussed in an earlier unit. The drawing of a free-body diagram is an important step in the solving of mechanics problems since it helps to visualize all the forces acting on a single object. The net external force acting on the object must be obtained in order to apply Newton's Second Law to the motion of the object.

In physics a free body diagram (force diagram or FBD) is a graphical illustration used to visualize the applied forces, movements, and resulting reactions on a body in a steady state condition (no acceleration of the system). They depict a body or connected bodies with all of the applied forces and moments, as well as reactions, that act on that/those bodies. The body may consist of multiple internal members, for example, a truss, or be a compact body such as a beam. A series of free bodies and other diagrams may be necessary to solve complex problems.

Free body diagrams are used to visualize the forces and moments applied to a body and calculate the resulting reactions, in many types of mechanics problems. Most free body diagrams are used both to determine the loading of individual structural components as well as calculating internal forces within the structure in almost all engineering disciplines from Bio-mechanics to Structural. In the educational environment, learning to draw a free body diagram is an important step in understanding certain topics in physics, such as statics, dynamics and other forms of classical mechanics. One professor at Cornell University earned the nickname "Free-Body Perkins" because of his passion for teaching the importance of using free body diagrams to solve mechanics problems. He went so far as to stop random students in the hall, saying, “You! Come in my office! Draw a free body diagram.

A block rests on a horizontal surface:

(1) A block rests on a friction-less horizontal surface and the block pulled horizontally with a force $F$:

From free body diagram we get, $R = mg$ and from Newton's law $F = ma$
Thus, $a = \frac{F}{m}$


(2) A block rests on a horizontal surface where co-efficient of friction between the block and the surface is $\mu$ and the block pulled horizontally with a force $F$:

From free body diagram we get, $R = mg$ and from Newton's law $F - f = ma$
or, Acceleration $a = \frac{{F - f}}{m} = \frac{{F - \mu R}}{m} = \frac{{F - \mu mg}}{m}$


(3) A block rests on a horizontal surface and a pull is acting at an angle $\theta$ to the horizontal in upward direction:

From free body diagram we get, $R + F\sin \theta = mg$
or, $R = mg - F\sin \theta$
And the net effective force on the block along horizontal direction is $F\cos \theta$, Thus from Newton's law we get, $ma = F\cos \theta$
or, Acceleration $a = \frac{{F\cos \theta }}{m}$


(4)A Block rest on a horizontal plane and a pushing force $F$ acts downward direction at an angle $\theta$ to the horizontal:

From free body diagram we get, $R = mg + F\sin \theta$
And the net effective force of the block along horizontal direction is $F\cos \theta$
Or, acceleration $a = \frac{{F\cos \theta }}{m}$


Motion of a Block on Inclined Plane:

(1) A Block placed on a frictionless inclined plane with angle of inclination $\theta$. Here the inclined plane is at rest and the body slides down the inclined plane.

From free body diagram we get, $R = mg\cos \theta$
And net effective force on the block is $F = mg\sin \theta$
From Newton's second law of motion we get, $F = ma$
Therefore, $ma = mg\sin \theta$
Or, $a = g\sin \theta$


(2) A Block placed on a inclined plane with angle of inclination $\theta$ and whose co-efficient of friction between the surface of the block and inclined plane is $\mu$. Here the inclined plane is at rest and the body slides down the inclined plane.

From free body diagram we get, $R = mg\cos \theta$
And, friction force $f = \mu R = \mu mg\cos \theta$ acts upward opposes the motion.
Therefore net effective force on the block is $F = \left( {mg\sin \theta - \mu mg\cos \theta } \right)$
From newton's second law of motion, $F = ma$
So, $ma = \left( {mg\sin \theta - \mu mg\cos \theta } \right)$
Or, $a = g\left( {\sin \theta - \mu \cos \theta } \right)$


(3) Retardation of a block slides up through a rough a inclined plane of angle of inclination $\theta$ and co-efficient of friction between block and the inclined surface is $\mu$:

From free body diagram we get, $R = mg\cos \theta$
Here, friction force $f = \mu R = \mu mg\cos \theta$ acts downward opposes the upward motion of the block.
And, net effective force on the block is $F = mg\sin \theta + \mu mg\cos \theta$.
From Newton's second law of motion we get, $F = ma$
Therefore, $ma = mg\sin \theta + \mu mg\cos \theta$
or, $a = g\left( {\sin \theta + \mu cos\theta } \right)$


(4) A Block placed on a frictionless inclined plane with angle of inclination $\theta$. Here the inclined plane is in an acceleration $b$ and the body slides down the inclined plane.

Here, the body lie in ARF (Accelerated reference frame), an inertial force $mb$ acts on it's in the opposite direction .
From free body diagram, we get $R = mg\cos \theta + mb\sin \theta$

The net effective force on the block is
$F = mg\sin \theta - mb\cos \theta$
From Newton's second law of motion we get, $F = ma$
Therefore, $ma = mg\sin \theta - mb\cos \theta$
Or, $a = g\sin \theta - b\cos \theta$

Special Notes:
The condition for the body to be rest relative to the inclined plane if, $a = 0$
Thus, $g\sin \theta - b\cos \theta = 0$
Or, $b\cos \theta = g\sin \theta$
Or, $b = g\frac{{\sin \theta }}{{\cos \theta }} = g\tan \theta$

Motion of a Block in Contact:

(1) Two body in contact and a force apply on the left side of the block:

Here $F$ be the applied force on ${m_1}$ and $f$ be the contact force between ${m_1}$ and ${m_2}$, then from free body diagram we get,

For mass ${m_1}$:

$F - f = {m_1}a$ ... ... ... (i)

For mass ${m_2}$:

$f = {m_2}a$ ... ... ... (ii)

From equation (i) and (ii) we get,
$F - {m_2}a = {m_1}a$
Or, $\left( {{m_1} + {m_2}} \right)a = F$
Or, $a = \frac{F}{{\left( {{m_1} + {m_2}} \right)}}$

And the contact force $f = {m_2}a = \frac{{{m_2}F}}{{\left( {{m_1} + {m_2}} \right)}}$


(2) Two body in contact and force apply on the right side of the block:

Here $F$ be the applied force on ${m_2}$ and $f$ be the contact force between ${m_1}$ and ${m_2}$, then from free body diagram we get,

For mass ${m_2}$:

$F - f = {m_2}a$ ... ... ... (i)

For mass ${m_1}$:

$f = {m_1}a$ ... ... ... (ii)

From equation (i) and (ii) we get,
$F - {m_1}a = {m_2}a$
Or, $\left( {{m_1} + {m_2}} \right)a = F$
Or, $a = \frac{F}{{\left( {{m_1} + {m_2}} \right)}}$

And the contact force $f = {m_1}a = \frac{{{m_1}F}}{{\left( {{m_1} + {m_2}} \right)}}$


(3) Three body are in contact and a force apply to the left side of the block:

Here $F$ is the applied force on ${m_1}$ and ${T_1}$ be the contact force between the bodies of masses ${m_1}$ and ${m_2}$, and ${T_2}$ be the contact force between the bodies of masses ${m_2}$ and ${m_3}$.
Then from free body diagram we get,

For mass ${m_1}$:

$F - {T_1} = {m_1}a$ ... ... ... (i)

For mass ${m_2}$:

${T_1} - {T_2} = {m_2}a$ ... ... ... (ii)

For mass ${m_3}$:

${T_2} = {m_3}a$ ... ... ... (iii)

From equation (i), (ii) and (iii) we get,
${T_1} - {m_3}a = {m_2}a$
Or, ${T_1} = \left( {{m_2} + {m_3}} \right)a$

So, $F - \left( {{m_2} + {m_3}} \right)a = {m_1}a$
Or, $F = \left( {{m_1} + {m_2} + {m_3}} \right)a$
Or, $a = \frac{F}{{\left( {{m_1} + {m_2} + {m_3}} \right)}}$

And, contact force between the masses ${m_2}$ and ${m_3}$ is ${T_2} = {m_3}a = \frac{{{m_3}F}}{{\left( {{m_1} + {m_2} + {m_3}} \right)}}$

And, the contact force between ${m_1}$ and ${m_2}$ is ${T_1} = \left( {{m_2} + {m_3}} \right)a = \frac{{\left( {{m_2} + {m_3}} \right)F}}{{\left( {{m_1} + {m_2} + {m_3}} \right)}}$


(4) Three body are in contact and a force apply to the right side of the block:

Here $F$ is the applied force on ${m_3}$ and ${T_1}$ be the contact force between the bodies of masses ${m_2}$ and ${m_3}$, and ${T_2}$ be the contact force between the bodies of masses ${m_1}$ and ${m_2}$.
Then from free body diagram we get,

For mass ${m_1}$:

${T_2} = {m_1}a$ ... ... ... (i)

For mass ${m_2}$:

${T_1} - {T_2} = {m_2}a$ ... ... ... (ii)

For mass ${m_3}$:

$F - {T_1} = {m_3}a$ ... ... ... (iii)

From equation (i), (ii) and (iii) we get,
${T_1} - {m_1}a = {m_2}a$
Or, ${T_1} = \left( {{m_1} + {m_2}} \right)a$

Therefore, $F - \left( {{m_1} + {m_2}} \right)a = {m_3}a$
Or, $F = \left( {{m_1} + {m_2} + {m_3}} \right)a$
Or, $a = \frac{F}{{\left( {{m_1} + {m_2} + {m_3}} \right)}}$

And the contact force between the masses ${m_1}$ and ${m_2}$ is ${T_2} = {m_1}a = \frac{{{m_1}F}}{{\left( {{m_1} + {m_2} + {m_3}} \right)}}$

And the contact force between ${m_2}$ and ${m_3}$ is ${T_1} = \left( {{m_1} + {m_2}} \right)a = \frac{{\left( {{m_1} + {m_2}} \right)F}}{{\left( {{m_1} + {m_2} + {m_3}} \right)}}$


(5) Four body are in contact and a force apply on the right side block:

Let A, B, C and D four body of masses ${m_1}$, ${m_1}$, ${m_1}$ and ${m_1}$ are in contact. Now the force $F$ is applied on mass ${m_4}$.
Here, the acceleration of the system is $a = \frac{F}{{\left( {{m_1} + {m_2} + {m_3} + {m_4}} \right)}}$

And the contact force between ${m_1}$ and ${m_2}$ is $\frac{{{m_1}F}}{{\left( {{m_1} + {m_2} + {m_3} + {m_4}} \right)}}$

The contact force between ${m_2}$ and ${m_3}$ is $\frac{{\left( {{m_1} + {m_2}} \right)F}}{{\left( {{m_1} + {m_2} + {m_3} + {m_4}} \right)}}$
The contact force between ${m_3}$ and ${m_4}$ is $\frac{{\left( {{m_1} + {m_2} + {m_3}} \right)F}}{{\left( {{m_1} + {m_2} + {m_3} + {m_4}} \right)}}$

Motion of a Block connected in Massless String:


(1) Two body system and a pulling force acts on the right-side:

Let A and B the two body of masses ${m_1}$ and ${m_2}$ are connected by a massless string and a pulling force $F$ is applied on the body A.
If the tension of the string between the two body is $T$, then
from the free body diagram we get,

For mass ${m_1}$:

$F - T = {m_1}a$ ... ... ... (i)

For mass ${m_2}$:

$T = {m_2}a$ ... ... ... (ii)

From equation (i) and (ii) we get,
$F - {m_2}a = {m_1}a$
Or, $\left( {{m_1} + {m_2}} \right)a = F$
or, $a = \frac{F}{{\left( {{m_1} + {m_2}} \right)}}$

And the tension of the string between masses ${m_1}$ and ${m_2}$ is $T = {m_2}a = \frac{{{m_2}F}}{{\left( {{m_1} + {m_2}} \right)}}$


(2) Two Body System and a pulling force acts on the Left-side:

Let A and B the two body of masses ${m_1}$ and ${m_2}$ are connected by a massless string and a pulling force $F$ is applied on the body ${m_1}$. If the tension of the string between the two body is $T$ then,
From free body diagram,

For the mass ${m_1}$:

$F - T = {m_1}a$ ... ... ... (i)

For the mass ${m_2}$:

$T = {m_2}a$ ... ... ... (ii)

From equation (i) and (ii) we get,
$F - {m_2}a = {m_1}a$
Or, $F = \left( {{m_1} + {m_2}} \right)a$
Or, $a = \frac{F}{{\left( {{m_1} + {m_2}} \right)}}$

And, the tension of the string between the masses ${m_1}$ and ${m_2}$ is $T = {m_2}a = \frac{{{m_2}F}}{{\left( {{m_1} + {m_2}} \right)}}$


(3) Three Body System and a pulling force acts on the Right-side:

Let A, B and C be the three body of masses ${m_1}$, ${m_2}$ and${m_3}$ are connected by a massless string and a pulling force $F$ is applied on a body C.
If the tension of the string between ${m_2}$ and ${m_3}$ is ${T_2}$ and the tension of the string between the body ${m_1}$ and ${m_2}$ is ${T_1}$ then,
From free body diagram we get,

For mass ${m_1}$:

${T_1} = {m_1}a$ ... ... ... (i)

For mass ${m_2}$:

${T_2} - {T_1} = {m_2}a$ ... ... ... (ii)

For mass ${m_3}$:

$F - {T_2} = {m_3}a$ ... ... ... (iii)

From equation (i), (ii) and (iii) we get,
${T_2} - {m_1}a = {m_2}a$
Or, ${T_2} = \left( {{m_1} + {m_2}} \right)a$

And, $F - \left( {{m_1} + {m_2}} \right)a = {m_3}a$
Or, $F = \left( {{m_1} + {m_2} + {m_3}} \right)a$
Or, $a = \frac{F}{{\left( {{m_1} + {m_2} + {m_3}} \right)}}$

And the tension of the string between the mass ${m_1}$ and ${m_2}$ is ${T_1} = {m_1}a = \frac{{{m_1}F}}{{\left( {{m_1} + {m_2} + {m_3}} \right)}}$

And the tension of the string between the mass ${m_2}$ and ${m_3}$ is ${T_2} = \left( {{m_1} + {m_2}} \right)a = \frac{{\left( {{m_1} + {m_2}} \right)F}}{{\left( {{m_1} + {m_2} + {m_3}} \right)}}$


(4) Three Body System and a pulling force acts on the Left-Side:

Let A, B and C be the three bodies of masses ${m_1}$, ${m_2}$ and ${m_3}$ are connected by a massless string and a pulling force $F$ is applied on the body A.
If the tension of the string between ${m_2}$ and ${m_3}$ is ${T_2}$ and the tension of the string between the bodies ${m_1}$ and ${m_2}$ is ${T_1}$, then
From free body diagram we get,

For mass ${m_1}$:

$F - {T_1} = {m_1}a$ ... ... ... (i)

For mass ${m_2}$:

${T_1} - {T_2} = {m_2}a$ ... ... ... (ii)

For mass ${m_3}$:

${T_2} = {m_3}a$ ... ... ... (iii)

From equation (i), (ii) and (iii) we get,
$a = \frac{F}{{\left( {{m_1} + {m_2} + {m_3}} \right)}}$

And the tension between masses ${m_1}$ and ${m_2}$ is ${T_1} = \frac{{\left( {{m_2} + {m_3}} \right)F}}{{\left( {{m_1} + {m_2} + {m_3}} \right)}}$

And the tension between masses ${m_2}$ and ${m_3}$ is ${T_2} = \frac{{{m_3}F}}{{\left( {{m_1} + {m_2} + {m_3}} \right)}}$

Learn Free Body Diagram Part: 1

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Learn Free Body Diagram Part: 3

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