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Published byBruce Watkins Modified over 8 years ago
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Newton’s Second Law I
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Review: Newton’s First Law IF the object experiences NO net external force…. Resting objects remain at rest. Moving objects move at a constant velocity.
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The ubiquitous block on an inclined plane. F gravity F block-on-book F friction
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Newton’s Second Law The acceleration of an object is: Directly proportional to the net external force acting on it, and… Inversely proportional to its mass
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In other words… F net = m ×a
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Let’s try a different incline: F gravity F normal F friction Ice
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Example 1: Suppose you want to push a 8 kg block up a frictionless inclined plane of 30 degrees. What would be the acceleration of the block, if a force of 50 Newtons were exerted parallel to the plane?
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Example 1 (cont.) Step 1 – Draw a picture Step 2 – Force-body diagram Y X F app. F normal FgFg
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Example 1 (cont.) Step 3 – What do we know? Mass = 8 kg Fapp.= 50 N. θ = 30˚ g = 9.8m/s 2 Y X F app. FgFg
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Example 1 (cont.) Step 4 – Apply Newton’s Second Law, in a single direction. Y X F app. FgFg
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Example 1 (cont.) Step 5 – Plug ‘n Chug Y X F app. FgFg
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On October 29, 1998, space shuttle Discovery lifted off with a total mass of 2,000,000 kg http://spaceflight.nasa.gov/gallery/images/s huttle/sts-103/html/sts103s005.html Example 2
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The Problem: What would be the acceleration if the thrust force (F thrust ) were 33,000,000 Newtons (an accurate assumption)? http://trc.dfrc.nasa.gov/Gallery/Gra phics/STS/Small/STS_launch.gif
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Example 2 (cont.) Step 1 – Picture Step 2 – Force- body diagram FgFg FtFt Y X
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Example 2 (cont.) Step 3 – What do we know? Mass = 2,000,000kg F thrust = 33MN g = 9.8m/s 2 FgFg FtFt Y X
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Example 2 (cont.) Step 4 – apply Newton’s second law in one direction (Y) FgFg FtFt Y X
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Example 2 (cont.) Step 5 – Plug ‘n Chug FgFg FtFt But…..in reality, the acceleration during the first stages of launch was closer to 14m/s 2 Why the difference?
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θ F g∙sinθ FtFt Launch video a=14 m/s 2
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