Mechanics 2

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Information about Mechanics 2

Published on December 4, 2013

Author: adityaabeysinghe

Source: slideshare.net

Mechanics - 2 For more information about Newton’s laws and applications search the presentation- Mechanics-1 By Aditya Abeysinghe Mechanics-2 By Aditya Abeysinghe 1

Tension of a string Tension is the external force that act on objects connected to the string or the internal forces that develop inside the string. F (internal force to the T string) F (External force to the object) mg Mechanics-2 By Aditya Abeysinghe 2

Different types of tension 1. Tension in accelerated environments 2. Tension in non-accelerated environments a. Normal tension b. Tension in inclined planes c. Tension in pulley systems Mechanics-2 By Aditya Abeysinghe 3

Tension in non-accelerated environments Normal tensionSingle object T F T > F for the object to move Mg * Note: For the next examples, the friction on the object(s) is/are not considered. Mechanics-2 By Aditya Abeysinghe 4

Two objects For M1 , F = ma T = M1 a a T T For M2, F = ma F – T = M2a ① ② F ①+②, M1g F = (M1 + M2 )a M2g Therefore, T = M1F/ (M1 + M2 ) Three or more objects T2 T1 M3g 5 Mechanics-2 By Aditya Abeysinghe

For M1, For M2, For M3 , F = ma F = ma F = ma F – T1 = M1a ① T1 – T2 = M2a ② T2 = M3a ③ ①+②+③ F = (M1 + M2 + M3 )a Therefore, T1 = (M2 + M3 )F/ (M1 + M2 + M3 ) T2 = M3F/ (M1 + M2 + M3 ) Mechanics-2 By Aditya Abeysinghe 6

Tension in pulley systems T” a T M1g M2g For M1, F =ma M1g – T = M1a ① For M2, F = ma T– M2g= M1a ② ①+② a = (M1 - M2) g / (M1 + M2) Therefore, T = 2M1 M2 g / (M1 + M2) T” = 2T T” = 4M1 M2 g / (M1 + M2) Mechanics-2 By Aditya Abeysinghe 7

Tension in inclined planes F = ma Mg Sinθ = Ma a = g Sinθ F = ma R – Mg Cosθ = M × 0 R = Mg Cosθ θ Mg Mechanics-2 By Aditya Abeysinghe 8

Tension in accelerated environments T θ a mg For m, F = ma T Sinθ = ma ① F = ma T Cosθ –mg = m× 0 T Cosθ = mg ② ①/② , tanθ = a/g Mechanics-2 By Aditya Abeysinghe 9

T For m, F = ma T-mg = ma a mg T = m(g + a) Pulley Systemsa T m1 a T m1 g F m2 a Mg M’g

For m, F = ma T = m1a For M, F = ma T = Mg Thus, m1a = Mg Therefore, a = (M/m1)g By considering the whole system, F =ma F = (M + M’ + m1) a = (M + M’ + m1) (M/m1)g Inclined planes – a maCosθ maCosθ = mgSinθ ma θ mg mgSinθ a = g tanθ Thus, a α θ 11

Apparent weight in a lift Consider the following occasions: N 1. The lift is at rest or moving with constant velocity For man, F=ma , N – Mg = 0 OR N = Mg (Apparent weight equals true weight) 2. The lift accelerates upwards For man, F = ma , N – Mg = Ma OR N = M (g+a) (Apparent weight is more than true weight) Mg 3. The lift accelerated downwards For man, F =ma , Mg – N = Ma OR N = M(g-a) (Apparent weight is less than true weight, for g > a ) 4. The lift falls under gravity For man, F = ma , Mg – N = Mg OR N = 0 (Apparent weight is zero) Mechanics-2 By Aditya Abeysinghe 12

Contact Force (Masses in contact) a M F m For M, F = ma F – R = Ma For m, F = ma R = ma ①+② ① ② Thus,F = (M + m)a And, a = F / (M +m) M R Internal force , retards the object R m R – Contact Force Internal force, accelerat es the object Method II: By applying F = ma to the whole system, the contact force can be ignored. Thus, F = (M + m)a Mechanics-2 By Aditya Abeysinghe 13

Motion of a body on a frictionless inclined plane Mgsinθ = Ma Therefore, a = gsinθ R R = MgCosθ a θ Mg Mechanics-2 By Aditya Abeysinghe 14

A body moving down a rough inclined plane When a body is moving down a rough inclined plane, the frictional force, which opposes the motion, acts upwards. μR R a θ R = MgCosθ However, F = μR Therefore, F = μ MgCosθ By applying F = ma, MgSinθ – μMgCosθ = Ma OR a = g(Sinθ – μCosθ) Mg Mechanics-2 By Aditya Abeysinghe 15

A body moving up a rough inclined plane When a body is moving up a rough inclined plane, the frictional force,which opposes the motion, acts downwards. R a θ Mg μR R = MgCosθ However, F = μR Therefore, F = μ MgCosθ By applying F = ma, MgSinθ+ μMgCosθ = Ma OR a = g(Sinθ + μCosθ) Acceleration also lies downwards as the object retards and becomes stationary on its way up. Mechanics-2 By Aditya Abeysinghe 16

Flying of birds A bird flies by displacing air below its wings. F The force due to the change of momentum is the force given to air by the wings of the bird. If F > Mg , it flies up. Similarly, if F = Mg , it stays stationary. And, if F < mg, it flies down. A V mg If V is the speed of the bird, the distance it travels within a unit amount of time is also V. V However, force = rate of change of momentum. Therefore, F = mv /1 = ρAV × V = ρAV2 F = ρAV2 Then, the volume of air displaced by the bird within a unit time is AV. However, m = ρ × Volume (ρ – density of air). Thus, m = ρAV Mechanics-2 By Aditya Abeysinghe 17

Movement of a bicycle Movement of a bicycle is due to the movement of the body of a bicycle on wheels. When applying brakes frictional forces are arranged as follows F1 First a frictional force occurs when the whel tries to move backward relative to Earth F2 F Due to this F1 force the body moves forward which causes the front wheels to develop a F2 frictional force Mechanics-2 By Aditya Abeysinghe F 18

The momentum or the speed of an object changes only when an external unbalanced force acts on it Mass of trolley – m u v Mass of man - M F v F Although when considered as a whole, there’s no unbalanced force, the frictional force F which acts on the trolley decelarates the trolley while the frictional force which acts on the object accelerates the object Mechanics-2 By Aditya Abeysinghe Since there’s no unbalanced force within the system, Applying Principle of Conservation of Linear Momentum to the system, mu = (m + M)V 19

But if a person leaves the trolley perependicularly while the trolley is moving, trolley’s speed will not be affected, since there’s no generation of frictional forces. Rod Rod v u From Principle of Conservation of Linear Momentum, mu + Mu = Mu + mv Solving this equation we get, u=v Mechanics-2 By Aditya Abeysinghe Thus the speed is unaffected when there’re no frictional force in both the objects 20

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