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DYNAMICS - EXAMPLE


Double Block Motion
  Example

 

Two different blocks are connected through a pull system and released. What is tension in the cable? The coefficient of static and kinetic friction between block A and the ground is 0.4 and 0.25, respectively. Assume the pulley is massless and frictionless.

 

   
    Solution


Check if Blocks are in
Static Equilibrium

 

Two actions can take place with this system. First, the coefficient of static friction is sufficient to keep the block from moving (No Motion Condition). This situation will occur when

     f ≥ T

The tension in the cable is simply half the weight of block B, or 10 lb. The friction force, f, will be μN, or 0.4(10 lb) = 4 lb. Thus, f is not larger than the tension, and the blocks will move. This is not surprising since this is a dynamics course.

     

Block Motion
 

Since the blocks will move, each will have an acceleration that directly affects the cable tension. Free-body diagrams with acceleration vectors for each block helps organize all actions affecting the blocks. Summing forces and accelerations give
     Block A:  T - f = aAmA
     Block B:  20 - 2T = aBmB

The friction is simply N = 0.25(10) = 2.5 lb using the kinetic coefficient of friction. But the acceleration terms need to be related to reduce the unknowns to just two.

The pulley effectively reduces the motion of B to half the motion of A, or in equation form,
      xB = 0.5xA

This can be integrated twice to give,
     aB = 0.5aA

Substituting known information into basic equations give,
     Block A:  T - 2.5 = aA (10/g)
     Block B:  20 - 2T = 0.5aA (20/g)

Combining,
     20 - 2T = 0.5 [(T - 2.5) (g/10)] (20/g)
     20 - 2T = T - 2.5
     22.5 = 3T

Solving gives,

T = 7.5 lb

It is interesting to note, if there was no friction, the equation becomes,

     20 - 2T = 0.5 [(T - 0) (g/10)] (20/g)
     20 = 3T

or Tno friction = 6.667 lb

     
   
 
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