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Mayank is doing an experiment in the lab with a bob of mass 10 Kg and a string of 6 m. He tied a string with the laboratory roof and another end with the bob. He then tied another string at 3 m below the roof with the previous one as shown in the figure. Find the tension T1 in the string? g =10 m/s2

Mayank is doing an experiment in the lab with a bob of mass 10 Kg and a string of 6 m. He tied a string with the laboratory roof and another end with the bob. He then tied another string at 3 m below the roof with the previous one as shown in the figure. Find the tension T1 in the string? g =10 m/s2 Correct Answer -50√2 N

CONCEPT:

  • Equilibrium: It is a state of a body in which overall change in a body is zero.
  • Translational Equilibrium: A body is said to be in translational equilibrium if the net sum of all the forces (F) acting on a body is zero.

Σ F = 0

Fnet = 0

F1 + F2 + F3+ ............... + Fn = 0

  • Rotational Equilibrium: A body is said to be in rotational equilibrium if the net sum of all the torque (τ) acting on a body is zero.

Σ τ = 0

τnet = 0 

τ1 +τ2 + τ3 + .................. + τn = 0

CALCULATION:

Given: m = 10 Kg; g = 10 m/s2

  • First draw the FBD of the given figure

  • For the system to be in equilibrium T1 must be equal to the horizontal component but the direction must be opposite.
  • Here the direction of T1 is reversed for balancing the horizontal component.

T2 = 50√2 N

T1 = -50√2 N​ 

  • So, the value of T1 is -50√2 N
  • Negative sign only show the direction of T1

​Hence option 1 is correct

Additional Information

  • Solving the numerical problem on the equilibrium of forces always draws the FBD (Fee Body Diagram). 
  • A simplified version of the body on which all external forces are shown is called FBD  (Fee Body Diagram) of the given body.

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