Repeat Problem 14–23 with ß = 120°. Problem 14–23: Figure P14.23 shows a slider-crank mechanism....

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Repeat Problem 14–23 with ß = 120°. Problem 14–23: Figure P14.23 shows a slider-crank mechanism. Link 2 rotates clockwise at a constant 200 rad/s. The weight of link 2 is negligible, link 3 is 3 lb,...

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Repeat Problem 14–23 with ß = 120°.

Problem 14–23:

Figure P14.23 shows a slider-crank mechanism. Link 2 rotates clockwise at a constant 200 rad/s. The weight of link 2 is negligible, link 3 is 3 lb, and link 4 is 2 lb. The radius of gyration of link 3 relative to the center of gravity is 3 in. For , determine the following:

1. The linear acceleration of link 4 and the center of gravity of link 3,

2. The angular acceleration of link 3,

3. The inertial force and torque of the coupler link,

4. The pin forces at B and C, and

5. The torque to drive the mechanism in this position.

Answered Same Day Dec 25, 2021

David · Accepted Answer

Solution:
Given data: 
Angular speed of crank (link 2), ω = 200 rad/s 
Angle of crank, β = 120o 
Let, 
  r = Radius of crank , = 1.5’’ 
  l = Length of connecting rod, = 4’’ 
   n = Obliquity ratio = l/r = 4’’/1.5’’ = 2.67 
mR = Mass of the reciprocating parts, piston, = 2 lb 
     WR = Weight of the reciprocating parts = mR.g 
  m = Mass of the connecting rod, = 3 lb 
  β = Angle made by crank with I.D.C., = 120o 
1. The linear acceleration of the piston (link 4) is given by the formula below: 
The acceleration of the reciprocating parts (piston) aP,

Repeat Problem 14–23 with ß = 120°. Problem 14–23: Figure P14.23 shows a slider-crank mechanism. Link 2 rotates clockwise at a constant 200 rad/s. The weight of link 2 is negligible, link 3 is 3 lb,...

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