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Force 2

Newton’s third law of motion 1. Force 2. Next Slide. Newton’s third law of motion. Newton’s third law of motion. For every action force there is an equal and opposite reaction force. Action and reactions are:. 1. equal in magnitude,. 2. opposite in direction.

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Force 2

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  1. Newton’s third law of motion 1 Force 2 Next Slide Newton’s third law of motion • Newton’s third law of motion For every action force there is an equal and opposite reaction force • Action and reactions are: 1. equal in magnitude, 2. opposite in direction 3. acting on different bodies

  2. Newton’s third law of motion 2 Force 2 Next Slide Newton’s third law of motion • Tension • Nature of tension Diagram • Reaction force by a surface • Nature of reaction force by a surface Diagram

  3. Force diagrams 1 Force 2 Next Slide Free body diagram (Labeled force diagram) • Shows all forces acting on an object • Different kinds of force: 1. Pulling or pushing force 2. Friction (f) 3. Weight (W) and W = mg 4. Tension (T) 5. Reaction force (R)

  4. Force diagrams 2 Force 2 Next Slide Free body diagram (Labeled force diagram) • Example 1 Diagram • Example 2 Diagram • Example 3 Diagram • Net force (resultant force, unbalanced force) comes from the addition of all forces in your free body diagram. Then Newton’s second law may be used to find the acceleration.

  5. Force diagrams 3 Force 2 Next Slide Free body diagram (Labeled force diagram) • Horizontal surface example 1 Calculation • Horizontal surface example 2 Calculation • Suspending body example Calculation • Inclined plane example Calculation • Two-body example Calculation

  6. Pressure 1 Force 2 Next Slide Pressure • Definition Force acting at right angles to a unit area • Unit : pascal (Pa) • Examples Calculation

  7. Turning effect of a force 1 Force 2 Next Slide Turning effect of a force • Force can be used to turn an object which can freely rotate about a fixed point • Moment (torque) of a force is a measure the turning effect of the force about a fixed point. • Unit : newton-metre (Nm)

  8. Turning effect of a force 2 Force 2 Next Slide Turning effect of a force • Directions of turning effect • Clockwise moment or Anti-clockwise moment • Example Calculation • Definition of Equilibrium (no rotation) : Clockwise moment = Anti-clockwise moment • Example Calculation • Daily examples Photo

  9. Turning effect of a force 3 Force 2 Next Slide Turning effect of a force • Conditions for equilibrium 1. Clockwise moment = Anti-clockwise moment (about any point) 2. The sum of the forces in one direction must be equal to the sum of the forces in the opposite direction • Examples Calculation • Daily Examples Photo

  10. END of Force 2

  11. tension tension Back to Newton’s third law of motion 2 Force 2 Click Back to • A rectangular block is suspended by a string. • Tension is always along the string. • Action and reaction pairs in Newton’s third law

  12. reaction force acting on the floor Back to Newton’s third law of motion 2 Force 2 Click Back to • A rectangular block is on a horizontal surface. • Reaction is always perpendicular to the floor. • Action and reaction pairs in Newton’s third law

  13. pulling force reaction pulling force friction weight Back to Force diagrams 2 Force 2 Click Back to • A rectangular block is pulled on a horizontal rough surface.

  14. reaction weight Back to Force diagrams 2 Force 2 Click Back to • A rectangular block is sliding downwards along an inclined plane.

  15. tension pulling force pulling force weight Back to Force diagrams 2 Force 2 Click Back to • A rectangular block, which is suspended by a string, is pulled by a pulling force.

  16. Reaction pulling force 6 N pulling force 6 N Weight = 20 N smooth surface Back to Force diagrams 3 Force 2 Click Back to • A pulling force 6 N is used to pull an object (mass : 2 kg) on a horizontal surface. Find the resultant force and acceleration.

  17. Reaction pulling force 6 N pulling force 6 N friction = 2 N rough surface Weight = 20 N Back to Force diagrams 3 Force 2 Click Back to • A pulling force 6 N is used to pull an object (mass : 2 kg) on a rough surface (friction : 2 N). Find the resultant force and acceleration.

  18. Tension 30° pulling force 10 N pulling force 10 N Weight Force diagrams 3 Force 2 Next Slide • A block which is suspended by a string is maintained at rest by a pulling force 10 N. The string makes an angle 30°with the horizontal. Find the tension and the weight of the block.

  19. T sin30° Tcos30° pulling force 10 N Weight Back to Force diagrams 3 Force 2 Click Back to • The acceleration of the block is zero (no motion) means that the resultant force is also zero (Why?!). • Tension inside the string is 11.5 N and the weight of the block is 5.8 N

  20. R : reaction 30° 30° Weight = 40 N Force diagrams 3 Force 2 Next Slide • An object (mass 4 kg) is placed on a smooth inclined plane which makes an angle 30°with the horizontal. Find the resultant force and the acceleration when it slides downwards along the inclined plane.

  21. R : reaction 40 sin 30°N 40 cos 30°N Back to Force diagrams 3 Force 2 Click Back to • Since the direction of acceleration and net force is pointing along the inclined plane (consider the direction of motion), the components of the weight is resolved along the inclined plane and perpendicular to the inclined plane.

  22. Block B Pulling force 10 N Block A String Smooth horizontal surface Force diagrams 3 Force 2 Next Slide • Two blocks A (mass 2 kg) and B (mass 3 kg) are connected together by a string on a smooth horizontal surface as shown. Block A is now pulled by a force 10 N. Find the acceleration of the system and the tension inside the string.

  23. Reaction = 30 N (Why?) Reaction = 20 N (Why?) T T Pulling force 10 N Weight = 30 N Weight = 20 N Block B Block A Force diagrams 3 Force 2 Next Slide • Since we have two objects, we must draw 2 force diagrams. Let T be the tension in the string and a be the acceleration

  24. Back to Force diagrams 3 Force 2 Click Back to • Since both blocks have the same acceleration, we can consider their net force. • We have simultaneous equations with two unknowns. Therefore, we can solve for T and a.

  25. (a) (b) (c) 3 m 2 m 2 m 3 m 4 m 4 m Pressure 1 Force 2 Next Slide • A rectangular block (2 m x 3 m x 4 m) is placed on a horizontal surface as shown. The mass of the block is 3 kg (What is the weight?!) Find the pressure acting on the floor in the following cases.

  26. Back to Pressure 1 Force 2 Click Back to • Even for the same block, different contact surface areas give different values of pressure.

  27. 2 m pivot 2 N Turning effect of a force 2 Force 2 Next Slide • The following diagram shows clockwise moment 4 N m • It is a clockwise moment since the force tends to rotate the rod clockwisely.

  28. 3 m pivot 2 N Back to Turning effect of a force 2 Force 2 Click Back to • The following diagram shows anti-clockwise moment 6 N m • It is an anti-clockwise moment since the force tends to rotate the rod anti-clockwisely.

  29. A (4 kg) B (2 kg) pivot 1.5 m 3 m pivot 1.5 m 3 m 20 N 40 N Turning effect of a force 2 Force 2 Next Slide • Two blocks A and B are placed on a see-saw as shown. The see-saw remains horizontal (no rotation).

  30. pivot 1.5 m 3 m 20 N 40 N Back to Turning effect of a force 2 Force 2 Click Back to • Take moment about the pivot • Clockwise moment = Anti-clockwise moment means no rotating effect

  31. Back to Turning effect of a force 2 Force 2 Click Back to • Daily Examples: (Turning effect of a door)

  32. A (3 kg) B pivot 2 m 6 m Turning effect of a force 3 Force 2 Next Slide • Two blocks A and B are placed on a see-saw as shown. The see-saw remains horizontal (no rotation). The weight of the see-saw is negligible. Find the weight of object B and the reaction force acting on the see-saw by the pivot.

  33. R: Reaction from the pivot 2 m 6 m W = Weight of B 60 N Turning effect of a force 3 Force 2 Next Slide • Take moment about the pivot

  34. R: Reaction from the pivot 2 m 6 m W = 20 N 60 N Turning effect of a force 3 Force 2 Next Slide • Consider the resultant force acting on the see-saw

  35. A (6 kg) Pivot P Pivot Q 0.4 m 0.6 m Turning effect of a force 3 Force 2 Next Slide • A block A (mass 6 kg) is placed on a rod (mass 4 kg) as shown. The rod is supported at both ends by two pivots P and Q. The rod is 1 m in length and the weight of the rod could be considered as concentrated at the mid-point of the rod. Find the reaction acting on the rod by P and Q respectively.

  36. RQ RP 0.5 m 0.5 m Pivot P Pivot Q 0.4 m 0.6 m 60 N 40 N Turning effect of a force 3 Force 2 Next Slide • Let RP and RQ be the reactions from pivot P and Q respectively • Take moment about P

  37. RQ = 44 N RP 0.5 m 0.5 m Pivot P Pivot Q 0.4 m 0.6 m 60 N 40 N Back to Turning effect of a force 3 Force 2 Click Back to • The resultant force acting on the rod is zero since it is in equilibrium.

  38. Turning effect of a force 3 Force 2 Next Slide • Daily Example:

  39. Back to Turning effect of a force 3 Force 2 Click Back to • Daily Example:

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