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Newton’s third law of motion 1 Force 2

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Presentation on theme: "Newton’s third law of motion 1 Force 2"— Presentation transcript:

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 Free body diagram (Labeled force diagram)
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 Free body diagram (Labeled force diagram)
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 Free body diagram (Labeled force diagram)
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 Force acting at right angles to a unit area
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 Newton’s third law of motion 2 Force 2
Back to Newton’s third law of motion 2 Force 2 Click Back to A rectangular block is suspended by a string. tension Tension is always along the string. Action and reaction pairs in Newton’s third law

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

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

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

15 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. pulling force tension weight pulling force

16 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. Reaction Weight = 20 N pulling force 6 N smooth surface pulling force 6 N

17 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. Reaction Weight = 20 N pulling force 6 N friction = 2 N rough surface pulling force 6 N

18 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. pulling force 10 N Tension Weight pulling force 10 N 30°

19 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?!). T sin30° Weight pulling force 10 N T cos30° Tension inside the string is 11.5 N and the weight of the block is 5.8 N

20 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. R : reaction Weight = 40 N 30° 30°

21 Back to Force diagrams 3 Force 2 Click Back to R : reaction
40 cos 30°N 40 sin 30°N 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 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. Block B Block A String Pulling force 10 N Smooth horizontal surface

23 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 T Reaction = 30 N (Why?) Block B Weight = 30 N Block A Pulling force 10 N T Reaction = 20 N (Why?) Weight = 20 N

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 Pressure 1 Force 2 Next Slide (a) 3 m 2 m (b) 4 m 2 m (c) 4 m 3 m
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. (a) 3 m 2 m (b) 4 m 2 m (c) 4 m 3 m

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

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

29 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). A (4 kg) B (2 kg) 1.5 m 3 m pivot 40 N 20 N 1.5 m 3 m pivot

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

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

32 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. A (3 kg) B 2 m 6 m pivot

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

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

35 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. A (6 kg) Pivot Q 0.4 m 0.6 m Pivot P

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

37 Turning effect of a force 3 Force 2
Back to Turning effect of a force 3 Force 2 Click Back to Pivot Q 0.4 m 0.6 m Pivot P 0.5 m 60 N 40 N RP RQ = 44 N 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 Turning effect of a force 3 Force 2
Back to Turning effect of a force 3 Force 2 Click Back to Daily Example:

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