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Friction Equilibrium on slopes

Friction Equilibrium on slopes. Friction: Equilibrium. KUS objectives BAT Solve equilibrium problems on slopes using coefficient of friction. Starter :. j. i, j notation. i. WB3 . Object on a slope about to go up the slope.

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Friction Equilibrium on slopes

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  1. Friction Equilibrium on slopes

  2. Friction: Equilibrium • KUS objectives BAT Solve equilibrium problems on slopes using coefficient of friction Starter:

  3. j i, j notation i WB3. Object on a slope about to go up the slope A Block with mass 3 kg is pushed by a force parallel to the direction of the slope .The block is at a tipping point where it is about to go up the slope. The slope surface has coefficient of friction of 0.3 and is at 280 to the horizontal. Find the values of friction and the Push force Push Reaction Friction 280 Weight

  4. WB3Object on a slope about to go up the slope solution Push Reaction i) R = 25.96 ii) Friction =R = 0.3 x 25.96 = 7.79 N Friction 280 280 • Push =friction + component of weight downslope Push = 7.79 + 29.4 sin 28 = 21.59 N

  5. WB4Object on a slope about to go down the slope A Block with mass 8 kg is at the point of slipping down a rough slope. The slope is at 280 to the horizontal. Find the value of the coefficient of friction of the slope Friction will be up the slope Friction Reaction 280 Weight

  6. WB4 Object on a slope about to go down the slope solution Friction Reaction i) R = 69.2 ii) Equilibrium Friction= 78.4 sin 28 = 36.8 N 78.4sin28 = 36.8 280 280 iii) Friction =R Weight = 8 x 9.8=78.4

  7. WB5. Simultaneous equations question A Block with mass 3 kg is held at rest on a rough slope by a horizontal force (Push). The slope has coefficient of friction of 1/3 and the slope is at 300 to the horizontal. Find the value of the Push force Reaction Push Friction 300 Weight

  8. j i, j notation i WB5 (cont). Simultaneous equations question solutionpart 1 Reaction 300 Fr 300 300 Weight = 3 x 9.8=29.4

  9. WB5 (cont). Simultaneous equations question solutionpart 2 29.4sin30 + Fr – Pcos30 = 0 (1) Reaction R = 29.4cos30 + Psin30 (2) Friction = R = R (3) Fr So 29.4sin30 +R – Pcos30 = 0 (1) 300 solve simultaneously gives So 29.4sin30 +– Pcos30 = 0 P = 33.2 N (and R = 42.06 N)

  10. j i, j notation i WB6. horizontal push, object is about to go down the slope A Block with Weight 10 Newton’s is held at rest on a rough slope by a horizontal force (Push). The slope has coefficient of friction of 0.3 and the slope is at 300 to the horizontal. Find the value of the Push, Reaction and friction forces Friction Reaction Push 300 Weight

  11. WB6. rough slope in equilibrium, object is about to go down the slope solution Fr R 300 300 300

  12. WB6. rough slope in equilibrium, object is about to go down the slope solution  R = 10cos30 + Psin30 (1) Fr R  Fr + Pcos30 = 10sin30 (2) 300 Fr = R = 0.3 R (3) Solve simultaneously 300 300  0.3 + Pcos30 = 10sin30 (2) 

  13. KUS objectives BAT Solve equilibrium problems on slopes using coefficient of friction self-assess One thing learned is – One thing to improve is –

  14. END

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