2nd codition of equilibrium

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2nd codition of equilibrium

  1. 1. ROTATIONAL EQUILIBRIUM THE SECOND CONDITION OF EQUILIBRIUM
  2. 2. Introduction <ul><li>When a body is in equilibrium, it is either at rest or in uniform motion. According to Newton’s first law, only the application of a resultant force can change this condition. We have seen that, if all forces acting on such body intersect at a single point and their vector sum is zero, the system must be in equilibrium. When a body is acted on by forces that do not have a common line of action , it may be in translational equilibrium but not in rotational equilibrium. It means that it may not move to the right or left or up or down, but it may still rotate </li></ul>
  3. 3. Introduction <ul><li>Consider the forces exerted on a cross wrench, two equal opposing forces F are applied to the right and to the left. The first condition of equilibrium tells us that the vertical and horizontal forces are balanced. Hence, the system is said to be in equilibrium. If the same forces are applied however, the wrench has the tendency to rotate. This is true even though the vector sum is still zero. Clearly, we need a second condition for equilibrium to cover rotational motion. </li></ul>
  4. 4. DIAGRAM OF CROSS WRENCH F F AXIS OF ROTATION
  5. 5. Definition of Terms <ul><li>A line of action of a force is an imaginary line extended indefinitely along a vector in both directions. </li></ul><ul><li>Axis of rotation is an imaginary line passing through a point perpendicular to the direction of rotation. </li></ul>
  6. 6. MOMENT ARM <ul><li>The perpendicular distance from the axis of rotation to the line of action of force is called moment arm of the force. It is the moment arm that determines the effectiveness of a given force in causing rotational motion. If we exert a force at increasing distances from the center of a wheel, it becomes easier to rotate the wheel about its center. </li></ul>
  7. 7. The moment arm of a force is the perpendicular distance from the line of action of the force to the axis of rotation.
  8. 8. THE MOMENT ARM A B C F F F
  9. 9. EXAMPLES OF MOMENT ARMS and Sign Convention NEGATIVE TORQUE F r
  10. 10. EXAMPLES OF MOMENT ARMS and Sign Convention POSITIVE TORQUE F r
  11. 11. EXAMPLES OF MOMENT ARMS and Sign Convention POSITIVE TORQUE F r
  12. 12. EXAMPLES OF MOMENT ARMS and Sign Convention POSITIVE TORQUE F r
  13. 13. TORQUE <ul><li>The tendency to produce a change in rotational motion. </li></ul><ul><li>It is also called moment of force . Rotational motion is affected by both the magnitude of a force F and its moment arm r . </li></ul>
  14. 14. TORQUE <ul><li>We will define torque τ as the product of a magnitude of a force F and its moment arm r . </li></ul>Torque = force x moment arm τ = Fr
  15. 15. Remember: <ul><li>That r is measured perpendicular to the line of action of the force F . </li></ul><ul><li>That the units of torque are the units of force times distance, e.g. newton-meter [N-m] and pound-feet [lb-ft]. </li></ul><ul><li>That if the force F tends to produce counterclockwise rotation about an axis, the torque will be positive . Clockwise torques will be considered negative . </li></ul>
  16. 16. ILLUSTRATIVE EXAMPLE 1 <ul><li>A force of 250 N is exerted on a cable wrapped around a drum that has a diameter of 120 mm. What is the torque produced about the center of the drum? </li></ul>
  17. 17. PLAN: <ul><li>Make a drawing and extend the line of action of the force. Determine the moment arm of the drum. The moment arm is therefore equal to the radius of the drum. </li></ul>250 N 0.12m
  18. 18. SOLUTION: <ul><li>Notice that the line of action of the 250-N force is perpendicular to the diameter of the drum. The moment arm is therefore equal to the radius of the drum. </li></ul><ul><li>D 120 mm </li></ul><ul><li>r = — = ———— or r = 60 mm = 0.06 m </li></ul><ul><li>2 2 </li></ul>
  19. 19. SOLUTION: <ul><li>The magnitude of the torque if found by the equation: </li></ul><ul><li>T = Fr = (250N)(0.06m) </li></ul><ul><li>= 15.0 N-m </li></ul><ul><li>Finally, we determine that the sign of the torque is negative because it tends to cause clockwise motion about the center of the drum. Thus, the answer is written as: </li></ul><ul><li>T = - 15.0 N-m </li></ul>
  20. 20. ILLUSTRATIVE EXAMPLE 2 <ul><li>A mechanic exerts a 20-lb force at the end of a 10-in. wrench, as shown in the diagram. If this pull makes an angle of 60 º with the handle, what is the torque produced on the nut? </li></ul><ul><li>PLAN </li></ul><ul><li>Sketch the problem and determine the moment arm, multiply it by the magnitude of the force, and affix the appropriate sign. </li></ul>
  21. 21. SOLUTION: <ul><li>First, we draw a neat sketch, extend the line of action of the 20-lb force and draw in the moment arm. </li></ul>60 º 20 lb r
  22. 22. From the figure, we obtain: <ul><li>r </li></ul><ul><li>Sin 60 º = ——— </li></ul><ul><li>10 in. </li></ul><ul><li>R = 10 in. (sin 60 º) = 8.66 in </li></ul><ul><li>T = Fr = (20 lb)(8.66 in.) = 173 lb-in. </li></ul><ul><li>Or 14.4 lb-ft. </li></ul>
  23. 23. THE SECOND CONDITION OF EQUILIBRIUM <ul><li>THE ALGEBRAIC SUM OF ALL TORQUES ABOUT ANY AXIS MUST BE ZERO. </li></ul><ul><li>It tells us that the clockwise torques are exactly balanced by the counterclockwise torques </li></ul><ul><li>Στ = 0 </li></ul><ul><li>Στ cw + Στ ccw =0 </li></ul>
  24. 24. Second Condition of Equilibrium <ul><li>Since the rotation is not changing, we may choose whatever point we wish as an axis of rotation. As long as the moment arms are measured to the same point for each force, the resultant torque will be zero. Choosing the axis of rotation at the point of application of an unknown force is a way to simplify problems. If a particular force has a zero moment arm, it does not contribute to torque, regardless of its magnitude. </li></ul>
  25. 25. Illustrative Example 3 <ul><li>A 300-N girl and a 400-N boy stand on a platform supported by posts A and B as shown below. The platform itself weighs 200N. What are the forces exerted by the supports on the platform? </li></ul>
  26. 26. Sketch of problem 300 N 200 N 400 N 6.00 m 4.00 m 4.00 m 12 m A B
  27. 27. Plan <ul><li>Draw a free-body diagram to show the forces acting on the platform. Determine the CG of the platform and apply the 1 st and 2 nd conditions of equilibrium. </li></ul>2m 6m 4m 4m 300 N 200 N 400 N A B
  28. 28. Solution <ul><li>Σ Fx = 0: A + B – 300N =200 N – 400 N = 0 </li></ul><ul><li>A + B = 900 N </li></ul><ul><li>Στ = 0: </li></ul><ul><li>-A(12m) + 300N(10m)+ 200N(4m)-400N(4m) =0 </li></ul><ul><li>(-12m)A + 3000Nm + 800Nm - 1600Nm =0 </li></ul><ul><li>2200Nm = (12m)A </li></ul><ul><li>A = 183 N </li></ul><ul><li>A+B =900N </li></ul><ul><li>B = 900N-183N </li></ul><ul><li>B=717N </li></ul>
  29. 29. THANK YOU FOR LISTENING!.

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