Inclined Planes Lecture and Lab!!

Inclined Planes and Gravitational Force To analyze the forces acting on an object on an inclined plane (a tilted surface), the weight of the object, F g, is resolved into 2 perpendicular components. One component is parallel to the incline, F- parallel, and the other is perpendicular to the surface, F-perpendicular

Continued!! F-Parallel θ F-perpendicular θ F g ** Due to similar triangles the two angles θ are equal** Also note that W is the hypotenuse!

Example # 1 A book weighing 55N is resting on an inclined plane tilted at 30° from the horizontal. Find the F-parallel and F- perpendicular for this weight.

Solution F-parallel 30° F-perpendicular F g = 55N F-parallel = opposite side F-perpendicular = adjacent side F g = hypotenuse Angle = 30° Therefore: sin 30° = F-parallel = F-parallel = 27.5N (F-parallel) F g 55N

Solution Part 2 cos 30 ° = F-perpendicular = F-perpendicular = 47.6N (F-perpendicular) F g 55N The perpendicular component is 47.6N, and that the parallel component is 27.5N

Inclined Planes What other forces are acting on the block besides gravity? F N (normal force) is pushing upward perpendicular to the surface. Since the book is not moving downwards(through the surface), F net = 0.

Inclined Planes If there is no friction, any motion down the plane is caused by F- parallel (note that F- parallel is F net ) Therefore: a = F-parallel = F g sin θ m m Since F g = mg, then a = g sin θ

Example # 1 Continues Find the acceleration and direction of the book’s motion. F g = 55N Angle = 30° Force causing motion is F-parallel (assume the surface is frictionless)

Example # 1 Continued a = g sin θ = (-9.8m/s 2 )(sin 30°) = -4.9m/s 2 (downward) How does friction fit into this? Static friction is overcome when an angle is reached that causes an object to move down the plane.

Example # 2 A key is placed on a book cover and begins to move when the cover is tilted at 40° from the horizontal. What is μ (coefficient of friction) between the key and the book?

Solution Angle = 40° F f = μ F N F N = F-perpendicular F f = F- parallel (since friction is parallel to the surface) F-perpendicular = F g cos θ F-parallel = F g sin θ

Solution continued.. Therefore: F f = μ F N F- parallel = μ F- perpendicular F g sin θ = μ F g cos θ Sin θ = μ cos θ Therefore μ = tan θ = tan 40° = 0.84

Inclined Planes Get into your lab groups and make your way to the back!!!!