Inclined Planes

Inclined Planes On an inclined plane the normal force is not opposite the weight Therefore it is necessary to break one of them into its x and y components so you can find the net force N θ mg

Inclined Planes Since the box is speeding up down the incline, the acceleration is in that direction, and therefore the net force is in that direction as well N a θ mg It is easier to break the weight into components because the acceleration is down the incline which is perpendicular to the normal force

Inclined Planes Now some trigonometry y N x a θ 90- θ mg θ

Inclined Planes Replace the force of gravity with its components. Perpendicular to the ramp will be Fg cosƟ. Parallel to the ramp will be Fg sinƟ. N a mg sinθ mg cosθ θ θ mg

Inclined Planes mg sinθ mg cosθ mg Use Newton’s second law for both the parallel and perpendicular directions x N y a mg sinθ mg cosθ θ θ mg

Example A person pushes a 30-kg shopping cart up a 10 degree incline with a force of 85 N. Calculate the coefficient of friction if the cart is pushed at a constant speed. Fa Fn q Ff mg 0.117 q

Find the acceleration of a 1000 kg roller coaster down an incline of 45o if µ=.12

Example What is the ms required to prevent a sled from slipping down a hill of slope 30 degrees? ms = 0.577

Example - a block of weight 180 N is pulled along a horizontal surface by a force of 80 N acting at an angle 30 with the horizontal at a constant speed. What is the normal force equal to?

Example The man pushes/pulls with a force of 200 N. The child and sled combo has a mass of 30 kg and the coefficient of kinetic friction is 0.15. For each case: What is the frictional force opposing his efforts? What is the acceleration of the child? f=59 N, a=3.80 m/s2 / f=29.1 N, a=4.8 m/s2