Lecture10-2A

# Lecture10-2A - Today Today ’ ’ s Lecture s Lecture...

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Unformatted text preview: Today Today ’ ’ s Lecture s Lecture Lecture 10:Chapter 6, Using Newton’s Laws Lots of Examples First Example: Skier First Example: Skier A skier with mass m = 65kg slides down A slope with an incline angle of q = 32 o . Find (a) skier’s acceleration and (b) the magnitude of the force that the snow (slope) exerts on the skier. We now have a choice of coordinate system. Do we choose the x axis to be horizontal or parallel to the slope? Since the acceleration is parallel to the slope and there is no acceleration perpendicular to the slope, it is more convenient to choose our x axis to be parallel to the slope and subsequently the y axis perpendicular to the slope . For this problem we choose to ignore any effects of friction and our Free-Body consists of only two forces, the force of gravity, F g , and the normal force, N , of the snow on the skier. First Example: Skier First Example: Skier A skier with mass m = 65kg slides down A slope with an incline angle of q = 32 o . Find (a) skier’s acceleration and (b) the magnitude of the force that the snow (slope) exerts on the skier. With this choice of coordinates the Free-Body diagram takes on the form shown in the figure. With the use of a little trigonometry the two EOM’s ( x and y directions) are F g sin mg sin ma x and N − F g cos N − mg cos ma y First Example: Skier First Example: Skier A skier with mass m = 65kg slides down A slope with an incline angle of q = 32 o . Find (a) skier’s acceleration and (b) the magnitude of the force that the snow (slope) exerts on the skier. Recognizing that there is no acceleration in the y directon, a y = 0 , the two EOM’s, yield the solutions: Like a falling body, the acceleration down the hill is independent of mass! F g sin mg sin ma x and (a) a x g sin 9.8sin32 ∘ 5.19 m / s 2 (b) N mg cos 65 9.8 cos32 ∘ 540 N N − F g cos N − mg cos First Example: Skier First Example: Skier It is worthwhile to reemphasize that there is only one vector equation here, namely: We could have solved this problem with a coordinate system in which the x axis is horizontal and the y axis vertical. Now the gravitational force simplifies but both The normal force and acceleration have components in both directions: Our original choice of coordinates clearly simplified the problem, N x and a y were zero!...
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Lecture10-2A - Today Today ’ ’ s Lecture s Lecture...

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