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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 Third Law of motion states that to every action there is an equal but opposite reaction.

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

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1. 2 forces that act along the same direction. = Resultant force,F = F1 + F2 2. 2 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

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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

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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

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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

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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 = 0.7778 θ = 37.9° Samy F2 = 70 Ns Heng Gee F1 = 90 N θ 70 N 90 N F

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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

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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

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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 = 31.72 N › Friction : 31.72 N › a = 4.7 ms-2 25° 35 N

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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.

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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

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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

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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

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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

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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 = 169.7 N T2 = 120 N 12 kg 45° T1 T1 sin θ T1 cos θT2 mg

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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.

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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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Newton’s Laws of Motion. HFinks '072 6/2/2015 Basic Concepts Force – push or pull on an object - Vector quantity Mass – amount of matter in a body.

Newton’s Laws of Motion. HFinks '072 6/2/2015 Basic Concepts Force – push or pull on an object - Vector quantity Mass – amount of matter in a body.

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