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Forces On An Inclined Plane

Forces On An Inclined Plane. F N. Think about the forces as she sleds down the hill in the laundry basket (if it is not a non-frictionless surface). Can you label them? . F f. 30°. What is the formula used to calculate F g ?. F g. = mg. How many other forces are there? (Don’t make any up!).

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Forces On An Inclined Plane

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  1. Forces On An Inclined Plane

  2. FN Think about the forces as she sleds down the hill in the laundry basket (if it is not a non-frictionless surface). Can you label them? Ff 30° What is the formula used to calculate Fg? Fg = mg How many other forces are there? (Don’t make any up!) Two; the normal force (FN) and the force of friction (Ff).

  3. The force due to gravity can be broken up into vectors using the tip to tail method. They are Fgll and Fg┴. Fgll = Fgsin θ 30° The formulas used to calculate the components are: Fg┴ = Fgcos θ Fg One way to remember when to use cosine or sine is to look at the formulas on your reference table. Ax = A cos θ (horizontal) and Ay = A sin θ (vertical). Just use the opposite function for that direction when on an incline. Fg┴ = Fgcos θ (vertical) and Fgll = Fgsin θ(horizontal).

  4. Other formulas you need to know: • Ff= μFN (μ = coefficient of friction) • ΣFx = Fnet = ma = Fgll – Ff (ΣFx = sum of the forces in the x- direction)

  5. Example: If a 60 kg person is sledding down a hill on waxed skis (yes, it’s a laundry basket, but just pretend!) at a 30o incline at a constant velocity, calculate Fnet, FN, Fgll, Fg┴ , Fg, and Ff. The easiest thing to find first is the weight (Fg). Fg= mg Fg= (60kg)(9.81m/s2) Fg= 588.6N Fgll = 294.3N Use the weight to find it’s vectors. 30° Fg┴ = 509.73N = 588.6N Fg Fg┴ = Fgcos θ Fgll = Fgsin θ Fgll = (588.6N)sin(30°) Fgll = (588.6N)(0.500) Fgll = 294.3N Fg┴ = (588.6N)cos(30°) Fg┴ = (588.6N)(0.866) Fg┴ = 509.73N

  6. What is the normal force equal to? (Remember, it has to be equal and opposite of another force.) It is equal to Fg┴. The normal force is pushing down with 509.73N of force while the ground is pushing up with an equal amount of force. FN = 509.73N = 25.49N Ff To find the force of friction, plug in the normal force and the coefficient of friction that says waxed ski on snow. Fgll 30° This would be ________ friction because it is SLIDING. kinetic Fg┴ Fg Ff = μFN *Since coefficients of friction don’t have units, keep the unit of a Newton.* Ff = (0.05)(509.73N) Ff = 25.49N

  7. The last thing to figure out with all of the forces is the net force. Since she is going down the hill at a constant velocity, what does that mean? Ff = 25.49N ***ZERO acceleration.*** Fgll= 294.3N Keep that in mind! 0 30° ΣFx = Fnet = ma = Fgll – Ff Fnet = Fgll– Ff Subtract the force of friction because it is in the opposite direction of motion. Fnet = 294.3N – 25.49N Fnet= 268.81N

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