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World of Physics

World of Physics. Force-vector diagram. Type # 5. +y. F N. +x. F f. F g cos . . F g sin . . F g. FVD diagram type 5 is an object sliding down an “inclined plane”. Because of the angle of the ramp, we tip our axis:. The above diagram needs to drawn for every problem.

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World of Physics

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  1. World of Physics Force-vector diagram Type # 5

  2. +y FN +x Ff Fg cos   Fg sin   Fg FVD diagram type 5 is an object sliding down an “inclined plane”. Because of the angle of the ramp, we tip our axis: The above diagram needs to drawn for every problem. Note that the force of gravity or “weight” is always drawn STRAIGHT down, and that the components are dotted with arrowheads and drawn parallel and perpendicular to the incline. THIS IS VERY IMPORTANT! Does your picture look EXACTLY like mine?

  3. Problem # 1 - Question • A box massing 5.0 kg is on a plane inclined at 55° with a 0.20 coefficient of friction. • What is the normal force applied by the inclined plane? • What is box’s acceleration? A photograph of an object on an incline plane

  4. +y FN +x Set up the two equations for each problem. Ff = .2FN 49cos55  49sin55  Fg = (5) (9.8) = 49 N Problem # 1 - Procedure Fy = FN - Fg cos(55) = 0 Fx = -Ff + Fg sin(55) = ma

  5. Problem # 1 - Workout Fy = FN - Fgcos(55) = ma FN - (5)(9.8)cos(55) = 0 FN = 28.1 N Ff = ()(FN) = (.2)(28.1) = 5.62 N Fx = -Ff + Fgsin(55) = ma (-5.62) + (5)(9.8)sin(55) = 5a a = 6.9 m/s2 Solution: a = 6.9 m/s2

  6. Problem # 2 - Question • A 2009-N skier is on a inclined plane of 55° with a 0.70 coefficient of friction. • What is the normal force applied by the inclined plane? • What is the acceleration? An image of a skier on a inclined surface.

  7. +y FN +x Again: Set up the above two equations for each problem. Ff = FN Fg cos   Fg sin   Fg = 2009 N Problem # 2 - Procedure Fy = FN - Fg cos(55) = 0 Fx = -Ff + Fg sin(55) = ma

  8. Problem # 2 - Workout Fy = FN - Fgcos(55) = ma FN - (2009)cos(55) = 0 FN = 1152.32 N Ff = ()(FN) = (.7)(1152.315) = 806.62 N Fx = Ff + Fgsin(55) = ma (-806.62) + (2009)sin(55) = 205a a = 4.09 m/s2 Solution: a = 4.1 m/s2

  9. Step-by-Step guide to solve Force Problems, Type # 5 Step 1: Draw the diagram first; label & plug in the knowns. Step 2: Write the equations, first for the y and then for the x. Step 3: Plug in the numbers and solve, first for FN, then for Ff, and finally for the acceleration.

  10. Just For Fun... What do you think about this seen? (Hint: Think in physics (force, acceleration))

  11. The End Presented By an anonymous physics student World is Full of Physics

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