U3AOS1 Topic 5: Connected Bodies

Definition


Systems in which two or more items are connected in a way that makes their motions reliant on one another are referred to as connected bodies, sometimes known as connected particles. This idea is fundamental to mechanics, especially when applying Newton's principles to analyze forces and motion.

Several bodies that link up in a way that allows them to move together and affect one another's motion are referred to as connected bodies in physics. These connections might be elastic (like threads or ropes), rigid (like rods or bars), or a mix of the two. According to the laws of mechanics, the link insures that the motion of one body influences the motion of the others.


Examples


Most commonly, pulleys, tow bars, ropes, or strings are used to connect connected bodies. For example, in a basic system with two masses linked by a string over a pulley, the movement of one mass has an immediate effect on the other. Connected motion can be defined by this mutual dependence.



Mechanics of Connected Bodies

The mechanics of connected bodies often involve analyzing the forces acting on each body, such as tension in the string and gravitational forces. The key principles include:

  • Tension: The force exerted by a rope or string on the connected masses.


  • Newton's Second Law: This law (F = ma) is applied to each body in the system to derive equations that describe the motion and forces in the system.




Pulleys


The mechanics of connected bodies often involve analyzing the forces acting on each body, such as tension in the string and gravitational forces. The key principles include:


Equations of Motion


Free-body diagrams are essential tools for visualizing the forces acting on each body. By applying Newton's second law to these diagrams, we can derive the equations of motion. For example, in an Atwood machine, two masses are connected by a string over a pulley. 


  • For vertical motion over a pulley:

When the bodies move vertically,


                            m1g−T=m1a

                            T – m2g = m2a


Here, T is the tension in the string, g is the acceleration due to gravity, and a is the acceleration of the system.

By adding both equations, we get,

                             m1g – m2g = m1a + m2a

                             (m1 – m2)g = (m1 + m2) a

                              a = (m1 – m2)g / (m1 + m2)

Putting the value ‘a’ in equation > T – m2g = m2a

                               T = (2m1m2)g / (m1 + m2)




  • For horizontal motion over a pulley: 

When the body moves vertically and the other moves horizontally:

                         m1g – T = m1a

                         T = m2a

By putting the value of T in above equation:

                          m1g – m2a = m1a

                          m1g = (m1+m2)a


                          a = m1g / (m1 + m2)

And thus,

                          T = m1m2g / (m1+m2)



Application


Connected bodies are widely used in various fields, including engineering and physics. They are essential in designing mechanical systems such as elevators, cranes, and conveyor belts. Understanding the motion of connected bodies helps in optimizing these systems for efficiency and safety.

Understanding connected bodies is crucial in various fields, including mechanical engineering, robotics, and biomechanics.

  1. Mechanical Engineering: Engineers use the principles of connected bodies to design machines and structures that can withstand forces and perform specific functions. This includes everything from simple tools to complex machinery like engines and cranes.


  1. Robotics: In robotics, the analysis of connected bodies helps in designing robotic arms and other mechanisms that require precise control of motion and force distribution.


  1. Biomechanics: The human body itself can be analyzed as a system of connected bodies, where bones act as rigid links and muscles as flexible connectors. Understanding this system is essential for fields like prosthetics and physical therapy.


    Creating a free-body diagram


    Creating a free-body diagram involves several steps. Follow these steps to ensure accuracy:


    1. Identify the Object of Interest.

    Sketch the object being analyzed in isolation.

    1. Determine All Forces Acting on the Object:

    Consider all forces, including gravity, normal force, friction, tension, and applied forces.

    1. Draw the Object in Isolation:

    Represent the object with a simple shape, like a box or a dot.


    1. Represent Forces with Arrows:

    Draw arrows emanating from the object's center to show the direction and relative magnitude of each force. Label each arrow with the appropriate force symbol (e.g., Fg​ for gravitational force).

    1. Use Equations to Represent Forces:

    For each force, write the corresponding equation. For example:

    • Gravitational Force: Fg​=mg


    • Normal Force: FN​=mgcos(θ)


    • Frictional Force: Ff​=μFN​


    • Tension: T=mgsin⁡(θ)(if applicable)


    1. Summarize with Newton’s Second Law:

    Apply Newton's second law to set up equations of motion.


     For a body in equilibrium: ∑Fx=0 and ∑Fy=0


     For a body in motion:   F=ma


Analysis of Forces

In problems involving connected bodies, forces can act vertically, horizontally, or along inclined planes. Each scenario requires a specific approach to analyzing the forces and resulting accelerations. For inclined planes, the gravitational component along the plane and the normal force must be considered.


Mathematical Analysis

To analyze connected bodies, one must apply the principles of Newtonian mechanics, which involve forces, torques, and the resulting accelerations. Key equations and concepts include:

  1. Newton's Laws of Motion: These laws describe the relationship between the motion of an object and the forces acting on it. For connected bodies, Newton's Second Law (F=maF = maF=ma) is particularly important, as it allows for the calculation of acceleration when multiple forces are acting on a system.


  1. Equilibrium Conditions: For a system of connected bodies in equilibrium, the sum of forces and the sum of torques must be zero. This condition is used to solve for unknown forces or tensions in the connections.


  1. Kinematic Constraints: These constraints describe the geometric relationships between the motions of connected bodies. For example, in a pulley system, the movement of one part of the rope determines the movement of other parts.


Conclusion

The study of connected bodies is fundamental in mechanics. It involves understanding the interactions between linked objects and applying Newton's laws to predict their motion. By mastering these principles, one can solve complex problems involving multiple interconnected components in both academic and real-world applications.



Example 1
Two blocks, AA and BB, of masses mA=2 kgm_A = 2 \, \text{kg} and mB=3 kgm_B = 3 \, \text{kg} respectively, are connected by a light string over a frictionless pulley. If block AA is on a frictionless table and block BB is hanging off the edge, find the acceleration of the system and the tension in the string.

= Let TT be the tension in the string and aa be the acceleration of the system.

For block AA: T=mAaT = m_A a

For block BB: mBg−T=mBam_B g - T = m_B a

Combining the equations: T=mAaT = m_A a mBg−T=mBam_B g - T = m_B a

Substitute TT from the first equation into the second: mBg−mAa=mBam_B g - m_A a = m_B a a(mA+mB)=mBga (m_A + m_B) = m_B g

a=mBgmA+mB=3×9.82+3=29.45=5.88 m/s2a = \frac{m_B g}{m_A + m_B} = \frac{3 \times 9.8}{2 + 3} = \frac{29.4}{5} = 5.88 \, \text{m/s}^2

Now, find the tension: T=mAa=2×5.88=11.76 NT = m_A a = 2 \times 5.88 = 11.76 \, \text{N}


Example 2
A 5 kg block on a table is connected to a 3 kg block hanging off the side of the table by a light string over a frictionless pulley. The coefficient of kinetic friction between the 5 kg block and the table is 0.2. Find the acceleration of the system and the tension in the string.

m1​=5kg,       m2​=3kg,         μk​=0.2

Let TT be the tension and aa be the acceleration.

For block on the table: T−μkm1g=m1aT - \mu_k m_1 g = m_1 a

T−0.2×5×9.8=5aT - 0.2 \times 5 \times 9.8 = 5 a

T−9.8=5aT - 9.8 = 5 a


Combine the equations: T−9.8=5aT - 9.8 = 5 a 29.4−T=3a29.4 - T = 3 a

Add these two equations: 29.4−9.8=8a29.4 - 9.8 = 8 a 19.6=8a19.6 = 8 a

a=19.68=2.45 m/s2a = \frac{19.6}{8} = 2.45 \, \text{m/s}^2

Now, find the tension: T=5a+9.8=5×2.45+9.8=12.25+9.8=22.05 NT = 5 a + 9.8 = 5 \times 2.45 + 9.8 = 12.25 + 9.8 = 22.05 \, \text{N}


Example 3
Two blocks, AA and BB, with masses 4 kg4 \, \text{kg} and 6 kg6 \, \text{kg}, are connected by a string over a frictionless pulley. Block AA is on an inclined plane with an angle of 30∘30^\circ and a coefficient of friction of 0.1. Block BB hangs vertically. Find the acceleration and the tension.

mA​=4kg,         mB​=6kg,          θ=30∘,        μ=0.1

For block AA on the incline: T−mAgsin⁡θ−μmAgcos⁡θ=mAaT - m_A g \sin \theta - \mu m_A g \cos \theta = m_A a

For block BB: mBg−T=mBam_B g - T = m_B a

Combine the equations: T−4×9.8×sin⁡30∘−0.1×4×9.8×cos⁡30∘=4aT - 4 \times 9.8 \times \sin 30^\circ - 0.1 \times 4 \times 9.8 \times \cos 30^\circ = 4 a T−19.6−3.39=4aT - 19.6 - 3.39 = 4 a T−22.99=4aT - 22.99 = 4 a

T−22.99=4aT - 22.99 = 4 a

6 58.8−T=6a58.8 - T = 6 a

Add the equations: 58.8−22.99=10a58.8 - 22.99 = 10 a 35.81=10a35.81 = 10 a a=3.581 m/s2a = 3.581 \, \text{m/s}^2

Find the tension: T=4a+22.99=4×3.581+22.99=14.324+22.99=37.314 NT = 4 a + 22.99 = 4 \times 3.581 + 22.99 = 14.324 + 22.99 = 37.314 \, \text{N}



Example 4
A 7 kg block on a frictionless table is connected by a string to a 5 kg block hanging off the table. What is the acceleration of the blocks and the tension in the string?

m1​=7kg,      m2​=5kg

For the block on the table: T=m1aT = m_1 a

For the hanging block: m2g−T=m2am_2 g - T = m_2 a

Combine the equations: T=7aT = 7 a 5×9.8−T=5a5 \times 9.8 - T = 5 a 49−T=5a49 - T = 5 a

Substitute TT: 49−7a=5a49 - 7 a = 5 a

49=12a49 = 12 a

a=4912=4.083 m/s2a = \frac{49}{12} = 4.083 \, \text{m/s}^2

Find the tension: T=7a=7×4.083=28.581 NT = 7 a = 7 \times 4.083 = 28.581 \, \text{N}


Example 5
Two blocks, each of mass 3 kg3 \, \text{kg}, are connected by a string over a pulley. One block is on a frictionless incline of 45∘45^\circ, and the other hangs vertically. Find the acceleration and tension.

m1​=       m2​=3kg,       θ=45∘

For block on the incline: T−m1gsin⁡θ=m1aT - m_1 g \sin \theta = m_1 a T−3×9.8×sin⁡45∘=3aT - 3 \times 9.8 \times \sin 45^\circ = 3 a

T−3×9.8×22=3aT - 3 \times 9.8 \times \frac{\sqrt{2}}{2} = 3 a

T−20.79=3aT - 20.79 = 3 a

For the hanging block: m2g−T=m2am_2 g - T = m_2 a

3×9.8−T=3a3 \times 9.8 - T = 3 a 29.4−T=3a29.4 - T = 3 a


Add the equations: 29.4−20.79=6a29.4 - 20.79 = 6 a

8.61=6a8.61 = 6 a a=8.616=1.435 m/s2a = \frac{8.61}{6} = 1.435 \, \text{m/s}^2

Find the tension: T=3a+20.79=3×1.435+20.79=4.305+20.79=25.095 NT = 3 a + 20.79 = 3 \times 1.435 + 20.79 = 4.305 + 20.79 = 25.095 \, \text{N}

Example 6
A 4 kg block is on a table connected by a string to a 2 kg block hanging off the side. If the table has a friction coefficient of 0.3, find the acceleration of the system and the tension in the string.

m1​=4kg,     m2​=2kg,      μ=0.3

For the block on the table: T−μm1g=m1aT - \mu m_1 g = m_1 a T−0.3×4×9.8=4aT - 0.3 \times 4 \times 9.8 = 4 a T−11.76=4aT - 11.76 = 4 a

For the hanging block: m2g−T=m2am_2 g - T = m_2 a 2×9.8−T=2a2 \times 9.8 - T = 2 a 19.6−T=2a19.6 - T = 2 a

Combine the equations: T−11.76=4aT - 11.76 = 4 a


Combine the equations: T−11.76=4aT - 11.76 = 4 a 19.6−T=2a19.6 - T = 2 a

Add the equations: 19.6−11.76=6a19.6 - 11.76 = 6 a

Example 7
Two blocks, AA and BB, of masses mA=2 kgm_A = 2 \, \text{kg} and mB=3 kgm_B = 3 \, \text{kg} respectively, are connected by a light string over a frictionless pulley. If block AA is on a frictionless table and block BB is hanging off the edge, find the acceleration of the system and the tension in the string.

= Let TT be the tension in the string and aa be the acceleration of the system.

For block AA: T=mAaT = m_A a

For block BB: mBg−T=mBam_B g - T = m_B a

Combining the equations: T=mAaT = m_A a mBg−T=mBam_B g - T = m_B a

Substitute TT from the first equation into the second: mBg−mAa=mBam_B g - m_A a = m_B a a(mA+mB)=mBga (m_A + m_B) = m_B g

a=mBgmA+mB=3×9.82+3=29.45=5.88 m/s2a = \frac{m_B g}{m_A + m_B} = \frac{3 \times 9.8}{2 + 3} = \frac{29.4}{5} = 5.88 \, \text{m/s}^2

Now, find the tension: T=mAa=2×5.88=11.76 NT = m_A a = 2 \times 5.88 = 11.76 \, \text{N}


Example 8
A 5 kg block on a table is connected to a 3 kg block hanging off the side of the table by a light string over a frictionless pulley. The coefficient of kinetic friction between the 5 kg block and the table is 0.2. Find the acceleration of the system and the tension in the string.

m1​=5kg,       m2​=3kg,         μk​=0.2

Let TT be the tension and aa be the acceleration.

For block on the table: T−μkm1g=m1aT - \mu_k m_1 g = m_1 a

T−0.2×5×9.8=5aT - 0.2 \times 5 \times 9.8 = 5 a

T−9.8=5aT - 9.8 = 5 a


Combine the equations: T−9.8=5aT - 9.8 = 5 a 29.4−T=3a29.4 - T = 3 a

Add these two equations: 29.4−9.8=8a29.4 - 9.8 = 8 a 19.6=8a19.6 = 8 a

a=19.68=2.45 m/s2a = \frac{19.6}{8} = 2.45 \, \text{m/s}^2

Now, find the tension: T=5a+9.8=5×2.45+9.8=12.25+9.8=22.05 NT = 5 a + 9.8 = 5 \times 2.45 + 9.8 = 12.25 + 9.8 = 22.05 \, \text{N}


Example 9
Two blocks, AA and BB, with masses 4 kg4 \, \text{kg} and 6 kg6 \, \text{kg}, are connected by a string over a frictionless pulley. Block AA is on an inclined plane with an angle of 30∘30^\circ and a coefficient of friction of 0.1. Block BB hangs vertically. Find the acceleration and the tension.

mA​=4kg,         mB​=6kg,          θ=30∘,        μ=0.1

For block AA on the incline: T−mAgsin⁡θ−μmAgcos⁡θ=mAaT - m_A g \sin \theta - \mu m_A g \cos \theta = m_A a

For block BB: mBg−T=mBam_B g - T = m_B a

Combine the equations: T−4×9.8×sin⁡30∘−0.1×4×9.8×cos⁡30∘=4aT - 4 \times 9.8 \times \sin 30^\circ - 0.1 \times 4 \times 9.8 \times \cos 30^\circ = 4 a T−19.6−3.39=4aT - 19.6 - 3.39 = 4 a T−22.99=4aT - 22.99 = 4 a

T−22.99=4aT - 22.99 = 4 a

6 58.8−T=6a58.8 - T = 6 a

Add the equations: 58.8−22.99=10a58.8 - 22.99 = 10 a 35.81=10a35.81 = 10 a a=3.581 m/s2a = 3.581 \, \text{m/s}^2

Find the tension: T=4a+22.99=4×3.581+22.99=14.324+22.99=37.314 NT = 4 a + 22.99 = 4 \times 3.581 + 22.99 = 14.324 + 22.99 = 37.314 \, \text{N}



Example 10
A 7 kg block on a frictionless table is connected by a string to a 5 kg block hanging off the table. What is the acceleration of the blocks and the tension in the string?

m1​=7kg,      m2​=5kg

For the block on the table: T=m1aT = m_1 a

For the hanging block: m2g−T=m2am_2 g - T = m_2 a

Combine the equations: T=7aT = 7 a 5×9.8−T=5a5 \times 9.8 - T = 5 a 49−T=5a49 - T = 5 a

Substitute TT: 49−7a=5a49 - 7 a = 5 a

49=12a49 = 12 a

a=4912=4.083 m/s2a = \frac{49}{12} = 4.083 \, \text{m/s}^2

Find the tension: T=7a=7×4.083=28.581 NT = 7 a = 7 \times 4.083 = 28.581 \, \text{N}


Example 11
Two blocks, each of mass 3 kg3 \, \text{kg}, are connected by a string over a pulley. One block is on a frictionless incline of 45∘45^\circ, and the other hangs vertically. Find the acceleration and tension.

m1​=       m2​=3kg,       θ=45∘

For block on the incline: T−m1gsin⁡θ=m1aT - m_1 g \sin \theta = m_1 a T−3×9.8×sin⁡45∘=3aT - 3 \times 9.8 \times \sin 45^\circ = 3 a

T−3×9.8×22=3aT - 3 \times 9.8 \times \frac{\sqrt{2}}{2} = 3 a

T−20.79=3aT - 20.79 = 3 a

For the hanging block: m2g−T=m2am_2 g - T = m_2 a

3×9.8−T=3a3 \times 9.8 - T = 3 a 29.4−T=3a29.4 - T = 3 a


Add the equations: 29.4−20.79=6a29.4 - 20.79 = 6 a

8.61=6a8.61 = 6 a a=8.616=1.435 m/s2a = \frac{8.61}{6} = 1.435 \, \text{m/s}^2

Find the tension: T=3a+20.79=3×1.435+20.79=4.305+20.79=25.095 NT = 3 a + 20.79 = 3 \times 1.435 + 20.79 = 4.305 + 20.79 = 25.095 \, \text{N}

Example 12
A 4 kg block is on a table connected by a string to a 2 kg block hanging off the side. If the table has a friction coefficient of 0.3, find the acceleration of the system and the tension in the string.

m1​=4kg,     m2​=2kg,      μ=0.3

For the block on the table: T−μm1g=m1aT - \mu m_1 g = m_1 a T−0.3×4×9.8=4aT - 0.3 \times 4 \times 9.8 = 4 a T−11.76=4aT - 11.76 = 4 a

For the hanging block: m2g−T=m2am_2 g - T = m_2 a 2×9.8−T=2a2 \times 9.8 - T = 2 a 19.6−T=2a19.6 - T = 2 a

Combine the equations: T−11.76=4aT - 11.76 = 4 a


Combine the equations: T−11.76=4aT - 11.76 = 4 a 19.6−T=2a19.6 - T = 2 a

Add the equations: 19.6−11.76=6a19.6 - 11.76 = 6 a

Exercise &&1&& (&&1&& Question)

 Two masses, 6 kg and 2 kg, connected by a string over a pulley, with 6 kg on a 30° incline. Find acceleration and tension.

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Exercise &&2&& (&&1&& Question)

Two blocks, A (5 kg) on a frictionless table, and B (10 kg), hanging, connected by a string over a pulley. Find acceleration
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Exercise &&3&& (&&1&& Question)

Two bodies of mass 3 kg and 4 kg are suspended at the ends of massless string passing over a frictionless pulley. The acceleration of the system is (g=9.8m/s2)

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Exercise &&4&& (&&1&& Question)

A 2 kg block is lying on a smooth table which is connected by a body of mass 1 kg by a string which passes through a pulley. The 1 kg mass is hanging vertically. The acceleration of block and tension in the string will be

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Exercise &&5&& (&&1&& Question)

A block of mass 4 kg is suspended through two light spring balances A and B is series. Then A and B will read respectively.

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Exercise &&6&& (&&1&& Question)

Two masses m and m (m>m) are connected by massless flexible and inextensible string passed over massless and frictionless pulley. The acceleration of center of mass is

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Exercise &&7&& (&&1&& Question)

A light string passing over a smooth light pulley connects two blocks of masses m1 and m2 (vertically). If the acceleration of the system is g/8, then the ratio of the masses is

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Exercise &&8&& (&&1&& Question)

A block of mass m1 rests on a horizontal table. A string tied to the block is passed on a frictionless pulley fixed at the end of the table and to the other end of string is hung another block of mass m2. The acceleration of the system is 

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Exercise &&9&& (&&1&& Question)

Three solids of masses m1,m2 and m3 are connected with weightless string in succession and are placed on a frictionless table. If the mass m3 is dragged with a force T , the tension in the string between m2 and m3 is

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Exercise &&10&& (&&1&& Question)

Two masses 2 kg and 3 kg are attached to the end of the string passed over a pulley fixed at the top. The tension and acceleration are

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