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» 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 » 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 » Now some trigonometry y N x a θ 90- θ θ mg » 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 θ θ mg » Use Newton’s second law for both the parallel (y) and perpendicular (x) directions x y N a θ θ mg 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 Ff mg sin Fa Fn Ff k FN Fa k FN mg sin FN mg cos Fa k mg cos mg sin Fa mg sin k mg cos mg cos Ff mg mg sin Fa mg sin k mg cos 85 (30)(9.8)(sin10) k (30)(9.8)(cos10) 0.117 What is the s required to prevent a sled from slipping down a hill of slope 30 degrees? s = 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? 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