Pulley Problems: How to Solve Pulley Problems: CASE – 1: CASE – 2 (For CBSE, ICSE, IAS, NET, NRA 2022)

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Title: Pulley Problems

How to Solve Pulley Problems?

For solving any pulley problem, the first step is to understand the given conditions and write down the constraint equations accordingly.

CASE – 1

Case 1
  • Let, M1 & M2 be the mass attached to the pulley A.
  • Now, consider that the mass M1 is moving down with acceleration a1 and mass M2 is moving up with acceleration a2.

From the Free Body Diagram (FBD) :

Free Body Diagram of Case 1
  • The equation of force from FBD:
  • Equation of force from FBD:

  • From the above force equation, we have three unknowns but there are only 2 equations (Equation (1) & Equation (2)
  • So we need a third equation relating the two unknowns. The relation is known as constraint equation because the motion of M1 and M2 is interconnected.
  • The following assumptions must be considered before writing the equation:
  • The string is taut and inextensible at each and every point of time.
  • The string is massless and hence the tension is uniform throughout.
  • Pulley is massless.
  • The string is inextensible hence the total change in length of the string should be zero.
  • Suppose mass goes down by distance and mass moves up by distance.
  • Change in length (Taking elongation as positive and compression as negative)
  • On equating it to zero we get, = 0 (x = displacement)
  • On Differentiating equation (3) twice we get,
  • = 0 … (3) (a = acceleration)
  • Using Equation (1) , (2) & (3) we can calculate the values of T,
  • This is an example of a Fixed Pulley System.

CASE – 2

Case 2
  • Consider the following pulley system:
  • First, we have to relate the tension between string 1 & string 2,

Consider F. B. D of pulley B:

Free Body Diagram of Case 2

(Since the pulley is massless)

  • Now, Consider the accelerations as shown in the below figure:
  • For String 2:
  • Movable pulley B and Block M2 will have same acceleration otherwise the string will stretch or compress.

… (5)

  • Increase in length to the left of fixed pulley A =
  • Decrease in length to the right of fixed pulley A =
  • Decrease in length to the right of movable pulley B
  • And, .
  • Therefore, the total change in length = (Since, )
  • On differentiating the above equation twice we get,

… (6)

  • Similarly, if we write force equation on both the blocks:

Using equations (5) , (6) , (7) and (8) ,

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