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Aim: How can we explain forces at an angle? Do Now: Solve for the x and y components: x=5N x 30° 10 N y x = 8.7 N Why do crazy football coaches say to get low? Demo http://www.youtube.com/watch?v=eE6x1 rfogIU&feature=related If force is at an angle, some component of the force is in the x direction and some is in the y Forces at an Angle A 12 kg box is pulled across a table with a force of 50 N at an angle of 40o above the horizontal. Draw a free body diagram Write a net force equation in the x and y direction What is the normal force? If the box moves at a constant velocity, what is the force of friction? What is the coefficient of friction? 12 kg 40o Question 2 A small child pulls a 25 kg wagon with 30 N of force over a frictionless surface. The angle that the handle makes with the ground is 27o What is the horizontal component of the force? What is the vertical component of the force? Write a net force equation in the x and y direction What is the acceleration of the wagon If the wagon starts from rest, how far does the child go in 4 seconds? Normal Force Do you remember where the normal force is directed? What if the plane is inclined? On an incline: The problem with an incline: The object moves along the plane y y x x Fg Everything would need to be resolved into x and y components That’s a lot of sins and cosines Solution – rotate the x and y axis This way, only Fg needs to be resolved into x and y components - F║ and F┴ FN How can we solve for F║ and F┴ FII Mathematically, these θ’s are equal θ F θ Fg = mg At rest or moving with a constant velocity: FF = ? FF = F║ FF = Fgsinθ FF = mgsinθ FN FN = ? FN = F┴ FN = Fgcosθ FN = mgcosθ FF FII θ F θ Fg FN On a frictionless incline: As θ increases, FII increases, so Fnet increases, and the object accelerates faster FNet = ma FF F║ = ma FII Fgsinθ = ma mgsinθ = ma θ F gsinθ = a θ Fg On a surface with friction… The force of friction decreases as the angle increases since: Remember: cos(90o) = 0 and cos(0o) = 1 The higher the angle, the lower the value of cosine Ex: A 50 kg object rests on a table that is inclined 25o from the horizontal. (a) Determine the components of gravity acting on the object. (b) What is the Normal force? FII = Fgsinθ FII = mgsinθ FII = (50 kg)(9.8 m/s2)sin(25o) FII = 207 N F = Fgcosθ F = mgcosθ F = (50 kg)(9.8 m/s2)cos(25o) F = 444 N FN = F┴ = 444 N What is the force of friction if the object is at rest? FF = F║ FF = 207 N Assume the incline is now frictionless What is the acceleration down the incline? FNet = ma F║ = ma 207 N = (50 kg)a a = 4.14 m/s2 What is the coefficient of static friction? FF = µFN 207 N = µ(444 N) µ = 0.47 How far will the object travel in 5 s? d = vit + ½at2 d = (0 m/s)(5 s) + ½(4.14 m/s2)(5 s)2 d = 51.75 m