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Connected Bodies - Ropes & Tow Bar Simplified Revision Notes

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3.2.2 Connected Bodies - Ropes & Tow Bar

Newton's Third Law

Statement: This law states that every force has an equal and opposite reaction force.

In connected bodies problems (ropes or tow bars), multiple objects are linked and move together.

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Here's a summary of the key steps:

  1. Identify all forces: Draw separate free body diagrams for each object. Include forces like tension (in ropes), the weight of each object, normal reaction, and applied forces.
  2. Apply F=maF = ma for each body: For each object, write separate equations for the net force using F=maF = ma. Take care to maintain consistent acceleration for all bodies since they are connected.
  3. Consider tension: For ropes or tow bars, tension acts in opposite directions on the connected objects. For example, if object A pulls on object B with tension TT, then object B pulls on object A with the same magnitude of TT, but in the opposite direction.
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Examples:

  1. A Book on a Table:
  • The weight of the book acts downward.
  • The table exerts an equal and opposite reaction force upward.
  1. A Ball Falling Towards Earth:
  • When the ball is dropped, it is pulled towards the Earth by gravity.
  • Simultaneously, the Earth is pulled towards the ball with the same magnitude of force.

Problem:

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Given:

  • Two particles P and Q of masses 5 kg and 3 kg, respectively, are connected by a piece of light inextensible string.
  • Particle P is pulled by a horizontal force of 40 N along a rough horizontal plane.
  • Particle P experiences a 10 N frictional force, and Q experiences a 6 N frictional force. Questions:
  1. Find the acceleration of the particles.
  2. Find the tension in the string.

Solution:

Step 1: Draw a Diagram with All Forces Acting on Particles

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  • For Particle PP:
  • Horizontal force=40 Nforce = 40 \text{ N} to the right.
  • Tension TT in the string acting to the left.
  • Frictional force=10 Nforce = 10 \text{ N} to the left.
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  • For Particle QQ:
  • Tension TT in the string acting to the right.
  • Frictional force=6 Nforce = 6 \text{ N} to the left.

Step 2: Resolve forces and use FR=MaF_R=Ma for each particle

Note: Choosing the direction of motion as the positive direction usually leads to easier calculations.

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For Particle PP:

40T10=ma30T=5a(Equation 1)40 - T - 10 = ma \quad \Rightarrow \quad 30 - T = 5a \quad \text{(Equation 1)}
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For Particle QQ:

T6=3a(Equation 2)T - 6 = 3a \quad \text{(Equation 2)}

Step 3: We now have 22 Simultaneous equations that we can solve to find a TT

  1. Add Equation 11 and Equation 22:
30T+T6=5a+3a24=8a30 - T + T - 6 = 5a + 3a \quad \Rightarrow \quad 24 = 8a a=248=:success[3ms2]a = \frac{24}{8} = :success[3 \, \text{ms}^{-2}]
  1. Substitute a=3m/s2a = 3 \, \text{m/s}^2 back into Equation 22 to find T**:**
T6=3(3)T=9+6=:success[15N]T - 6 = 3(3) \quad \Rightarrow \quad T = 9 + 6 = :success[15 \, \text{N}]

Note:

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  • The string being light is important as if it were not, it would add extra weight to the particles, affecting accelerations.
  • The string being inextensible (i.e., not elastic) is important as this implies both particles accelerate at the same rate.
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