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CONNECTED PARTICLES

CONNECTED PARTICLES. LESSON THREE By Ronald Ddungu ronaldddungu@yahoo.com. MOTION ON A ROUGH INCLINED PLANE.

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CONNECTED PARTICLES

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  1. CONNECTED PARTICLES LESSON THREE By Ronald Ddungu ronaldddungu@yahoo.com

  2. MOTION ON A ROUGH INCLINED PLANE A particle of mass 4kg rests on the surface of a rough plane which is inclined at 300 to the horizontal. It is connected by a light inelastic string passing over a light smooth pulley at the top of the plane, to a particle of mass 6kg which is hanging freely. If the coefficient of friction between the 4kg mass and the plane is 0.45, find the acceleration of the system when it is released from rest and the tension in the string. Find also the force exerted by the string on the pulley.

  3. DIAGRAM

  4. SOLUTION(All forces acting are shown) a ms-2 a ms-2 Fr

  5. Equations of motion Consider the 4kg mass; There is no motion perpendicular to the plane. Thus resolving perpendicular to the plane gives; R – 4gcos300 = 0 R = 4g x √3/2 R= 2g√3N……(i) There is an acceleration ams-2 up the plane. The resultant force parallel to the plane is; T – 4gsin300 –Fr . Thus T – 4gsin300 –Fr = 4a..(ii) But Fr = μR = 0.45(2g√3)N = 9g√3N ……………(iii) 10

  6. Calculations Using (iii) in (ii) T – 4gsin300 –9g√3 = 4a 10 T – 2g – 9g√3 = 4a…….(iv) 10 For the 6kg mass assumed to have a resultant force vertically downwards; 6g –T = 6a…………(v) Adding (iv) and (v) T – 2g – 9g√3 = 4a 10 + 6g –T = 6a g(4 - 0.9x√3) = 10a 10a = 2.4411x 9.8 a = 2.3923ms-2

  7. Finding the tension 6g –T = 6a T = 6g – 6a T = 6x(9.8 - 2.3923) T = 44.4462N Therefore the acceleration of the system is 2.3923ms-2 and the tension in the string is 44.4462N Force exerted on the pulley is through the tension as shown below 300

  8. Finding the resultant force Resolving the forces; • Horizontally( ); -Tcos300 = - 44.4462cos300 = - 38.49263N • Vertically( ); -Tsin300 - T = - 1.5T = -1.5 X44.4462 = - 66.6693N • Resultant force R 66.6693N 38.49263N R2 = 38.492632 + 66.66932 R2 = 5926.478127 R = √(5926.478127) R = 76.9836N Inclined at an angle Tan-1 (66.6693/38.49263) = 60.00 . The force exerted by the string onto the pulley is 76.9836N inclined at 600 to the horizontal.

  9. HOME WORK DIAGRAM

  10. QUESTION A particle of mass M kg rests on the surface of a rough plane which is inclined at 300 to the horizontal. It is connected by a light inelastic string passing over a light smooth pulley at the top of the plane, to a particle of mass N kg which is hanging freely. If the coefficient of friction between the M kg mass and the plane is 0.35, find the acceleration of the system when it is released from rest and the tension in the string. Find also the force exerted by the string on the pulley in the cases below. (a) When M = 5 and N = 7 (b) When M = 8 and N = 5

  11. RESEARCH • http://www.youtube.com/watch?v=jHVRF-5c6I4(Watch this video) • http://www.mei.chosenhill.org/Solutions/mechanics1/Revision%20Connected%20Particles.pdf (Down load this document and bring it with you to school) -END-

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