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 The force that act on the object are balanced in all direction.  The force cancel each other, so that the resultant force or net force is zero.  Newton’s.

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Presentation on theme: " The force that act on the object are balanced in all direction.  The force cancel each other, so that the resultant force or net force is zero.  Newton’s."— Presentation transcript:

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2  The force that act on the object are balanced in all direction.  The force cancel each other, so that the resultant force or net force is zero.  Newton’s Third Law of motion states that to every action there is an equal but opposite reaction.

3  Is a single force that represents the combined effect of 2 or more forces in magnitude and direction.

4 1. 2 forces that act along the same direction. =  Resultant force,F = F1 + F forces that act in opposite direction. =  Resultant force, F = F1 – F2 F1 = 8 N F2 = 3 N F = 11 N F1 = 8 N F = 5 N F2 = 3 N

5 3. 2 forces acting at a point at an angle to each other. a) 2 non-parallel force i.Parallelogram law  The tails of 3 forces F, F1 and F2 must originate from the same point. ii.The triangle method (tail to tip method) θ θ θ F2 F1 F2 F1 F2 F1 F θ F2 F1 F2 F The tail of F1connecting to the tip of F2 F2 F1 θ F2 F1 θ θ

6 b) 2 perpendicular forces i.Using parallelogram law i.Using Pythagoras Theorem › Resultant force, F = √(F1)² + (F2)² › tan θ = F2 F1 F2 F1 F2 F1 F2 F θ

7 1. Find the resultant force for the two forces as shown in figure 1. Answer : F = 10.6 N at angle of 41° 120° 8 N 12 N Figure 1

8 2. Samy and Heng Gee pull a crate with force of 70 N and 90 N respectively. Find the resultant force on the crate due to these two forces. Answer :  Resultant force, F = 114 N  tan θ =  θ = 37.9° Samy F2 = 70 Ns Heng Gee F1 = 90 N θ 70 N 90 N F

9  A single force can be resolved into 2 perpendicular components.  Fx = F cos θ ( Fx / F = cos θ)  Fy = F sin θ ( Fy / F = sin θ ) Fx FyF θ θ Fy = F sin θ Fx = F cos θ

10 1. A tourist is pulling a bag with a force of 12 N at an angle 60° to the horizontal floor. What is the horizontal and vertical components of the force?  Answer :  Fx = 6 N  Fy = 10.4 N Fy Fx 12 N

11 2. Santhiran pulls a 5 kg crate on the floor with a force of 35 N.  Find the horizontal component of the force.  If the crate is moving with constant velocity, what is the friction against the crate.  If the friction against the box is 8 N, what is the acceleration of the crate?  Answer : › Fx = N › Friction : N › a = 4.7 ms-2 25° 35 N

12  2 types: A. Object in equilibrium on an inclined plane. I. If the object is at rest. II. If the object is moving on a smooth inclined plane. B. Three forces in equilibrium.

13 A. Object in equilibrium on an inclined plane. I. If the object is at rest.  The net force perpendicular to the plane = 0  Normal reaction – mg cos  = 0  F normal = mg cos   The net force parallel to the plane = 0  Frictional force – mg sin  = 0  F friction = mg sin  friction mg cos θ Normal reaction mg sin θ mg

14 II. If the object is moving on a smooth inclined plane, vertical component of the forces are balanced but the force down the plane is not balanced.  F net = ma  mg sin  = ma  a = g sin  Normal reaction mg sinθ mg mg cos θ Motion a friction

15  A carton of mass 5 kg is at rest on an inclined plane making an angle of 15° with the horizontal. Find the frictional force and the normal force acting on the carton.  Answer :  F friction = 12.9 N  Normal force = 48.3 N Normal reaction mg sin θ mg mg cos θ 15° friction

16 B. Three forces in equilibrium  Problem involving 3 forces in equilibrium can be solving using : i. Resolution of forces  Total force to the left = total force to the right  Total force upward = total force downward ii. Drawing a closed tail- to –tip ( triangle method) A B C

17  A 12 kg mass is suspended from a hook in the ceiling. The object is pulled aside by a horizontal string and makes an angle of 45° with the horizontal. Find the tension in both string.  Answer :  T1 = N  T2 = 120 N 12 kg 45° T1 T1 sin θ T1 cos θT2 mg

18 1. Lift ( refer the notes) 2. Pulley system 2) Pulley system A frictionless pulley serves to change the direction of a force Have 2 types : i. A force pulling a mass over a pulley In this situation, the tension, T is equal to the pulling force, F even if the rope is slanting. ii. A pulley with 2 masses. The heavier mass will accelerate downwards while the lighter one will accelerate upwards at the same rate. The tension is not equal to the weight of either mass.

19  A 5 kg mass is used to accelerate a 3 kg block along a table as shown in figure1. The friction between the table and the block is 10 N. Assuming that the pulley is smooth and the string is of negligible mass, find › The acceleration of the system › The tension in the string  Answer :  a = 5 ms -2  T = 25 N 3kg 5 kg a a


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