Lect56short examples - Physics 211 week4 Todays Agenda q...

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Physics 211: Lecture 5, Pg 1 Physics 211: week4 Physics 211: week4 Today’s Agenda Today’s Agenda More discussion of dynamics: Tension, Springs, Pulleys, Incline, Blocks,Circular motion.
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Physics 211: Lecture 5, Pg 2 Tools: Ropes & Strings Tools: Ropes & Strings Can be used to pull from a distance. Tension Tension ( T ) at a certain position in a rope is the magnitude of the force acting across a cross-section of the rope at that position. The force you would feel if you cut the rope and grabbed the ends. An action-reaction pair. cut T T T Tension doesn’t have a direction. When you hook up a wire to an object the direction is determined by geometry of the hook up.
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Physics 211: Lecture 5, Pg 3 Tools: Ropes & Strings. .. Tools: Ropes & Strings. .. Consider a horizontal segment of rope having mass m : Draw a free-body diagram (ignore gravity). Using Newton’s 2nd law (in x direction): F NET = T 2 - T 1 = m a So if m = 0 (i.e. the rope is light) then T 1 = T 2 T is constant anywhere in a rope or string as long as its massless T 1 T 2 m a x x
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Physics 211: Lecture 5, Pg 4 Tools: Ropes & Strings. .. Tools: Ropes & Strings. .. An ideal (massless) rope has constant tension along the rope. If a rope has mass, the tension can vary along the rope For example, a heavy rope hanging from the ceiling. .. We will deal mostly with ideal massless ropes. T = T g T = 0 T T 2 skateboards
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Physics 211: Lecture 5, Pg 5 Tools: Ropes & Strings. .. Tools: Ropes & Strings. .. What is force acting on box by the rope in the picture below? (always assume rope is massless unless told different) mg T m Since a y = 0 (box not moving), T = mg
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Physics 211: Lecture 5, Pg 6 Lecture 5, Lecture 5, Act 3 Act 3 Force and acceleration Force and acceleration A fish is being yanked upward out of the water using a fishing line that breaks when the tension reaches 180 N . The string snaps when the acceleration of the fish is observed to be is 12.2 m/s 2 . What is the mass of the fish? m = ? a = 12.2 m/s 2 snap ! (a) 14.8 kg (b) 18.4 kg (c) 8.2 kg
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Physics 211: Lecture 5, Pg 7 Lecture 5, Lecture 5, Act 3 Act 3 Solution: Solution: Draw a Free Body Diagram!! T mg m = ? a = 12.2 m/s 2 Use Newton’s 2nd law in the upward direction: F TOT = m a T - mg = m a T = m a + mg = m ( g + a ) m T g a = + ( 29 kg 2 8 s m 2 12 8 9 N 180 m 2 .
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This note was uploaded on 12/05/2009 for the course PHYSICS PHYS taught by Professor Phys during the Spring '09 term at Abu Dhabi University.

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Lect56short examples - Physics 211 week4 Todays Agenda q...

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