1. Inclined plane introduction
And you…

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

3. 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 - - -

4. 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 = 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

5. 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?

6. 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

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

8. 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
6(?a) = sin 32 (6)(9.8)
6(?a) =
a = 1.52 m/s/s

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

10. 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

11. 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 - - -

12. ∑ 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) (cos 38) 9.8 = 4.33 m/s2 !!!
SICK!!!