Inclined plane introduction And you…

Start with a good diagram!!!! Axis labels, signs, forces,… Y FN + X - Ffr Fmg y θ + θ Fmg Fmg x - - -

Break things into x and y components ∑ Fx = ma = Fmgx – Ffr Fmgx = sin θ mg Y FN + X - Ffr Fmg y θ + θ Fmg Fmg x - - -

So what is the rate of acceleration of a 10 kg mass on an 12O inclined plane that has no friction? ∑ Fx = ma = Fmgx – Ffr Therefore: ma = and recall that Fmgx = So substitute in and solve: m a = sin θ m g 10 a = sin 12 (10) 9.8 a = 2.037 m/s2 *Take note that you can also cancel the mass (10) on either side as well prior to finding the answer. Fmgx sin θ mg

You try one… make the whole diagram as well please. What is the rate of acceleration if a 16 kg mass is on an inclined plane that is at an 67O angle and there is no friction?

So what is the rate of acceleration of a 10 kg mass on an 12O inclined plane that has no friction? ∑ Fx = ma = Fmgx – Ffr Therefore: ma = and recall that Fmgx = So substitute in and solve: m a = sin θ m g 16 a = sin 67 (16) 9.8 a = 9.02 m/s2 *Take note that your calculated rate of acceleration on a plane will never surpass 9.8 m/s2 (g). Fmgx sin θ mg

Well what if there is friction? Just include the force of friction into your calculations like this…

So what is the rate of acceleration if the mass = 6kg, θ = 32O and the Ffr = 22N? ∑ Fx = ma = Fmgx - Ffr 6(?a) = sinθ mg - 22 6(?a) = sin 32 (6)(9.8) - 22 6(?a) = 31.159 - 22 a = 1.52 m/s/s

But you will not always be given the Ffr You may need to calculate it based on the ∑Fy equation

Diagram with signs, Define: Fgy, Fgx, ∑Fy, ∑Fx THE QUESTION: What is the rate of acceleration of an object on a ramp if: θ = 38O , μ = .22, ????????????? Diagram with signs, Define: Fgy, Fgx, ∑Fy, ∑Fx

Break things into x and y components ∑ Fx = ma = Fmgx – Ffr Fmgx = sin θ mg ∑ Fy = ma = 0 = Fn – Fmgy Fmgy = cos θ mg ∴ Fn = Fmgy Recall: Ffr = μFn θ = 38O , μ = .22 Y FN + X - Ffr Fmg y + Θ = 38 Θ = 38 Fmg Fmg x - - -

∑ Fx = ma = Fmgx – Ffr θ = 38O , μ = .22 Fmgx = sin θ mg ∑ Fy = ma = 0 = Fn – Fmgy Fmgy = cos θ mg ∴ Fn = Fmgy Recall: Ffr = μFn TIE EVERYTHING TOGETHER ∑ Fx = ma = Fmgx – Ffr ∑ Fx = ma = sin θ mg – Ffr ma = sin θ mg – μFn ma = sin θ mg – μ Fmgy ma = sin θ mg – μ cos θ mg 2 unknown variables: m and a So kick some mass! a = sin 38 (9.8) - .22 (cos 38) 9.8 = 4.33 m/s2 !!! SICK!!!