Lect05 FBD, ropes,springs

Lect05 FBD, ropes,springs - Physics 211: Lecture 5 Todays...

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Physics 211: Lecture 5, Pg 1 Physics 211: Lecture 5 Physics 211: Lecture 5 Today’s Agenda Today’s Agenda More discussion of dynamics Recap Newton's Laws The Free Body Diagram Free Body Diagram tension ) Hooke’s Law ( springs ) Original: Mats A.Selen Modified: W.Geist
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Physics 211: Lecture 5, Pg 2 Review: Newton's Laws Review: Newton's Laws Law 1 : An object subject to no external forces is at rest or moves with a constant velocity if viewed from an inertial reference frame. Law 2 : For any object, F NET = m a Where F NET = Σ F Law 3 : Forces occur in action-reaction action-reaction pairs, F A ,B = - F B ,A . Where F A ,B is the force acting on object A due to its interaction with object B and vice-versa. m is “mass” of object
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Physics 211: Lecture 5, Pg 3 Gravity: Gravity: Mass vs Weight Mass vs Weight What is the force of gravity exerted by the earth on a typical physics student? Typical student mass m = 55kg g = 9.81 m/s 2 . F g = mg = (55 kg)x(9.81 m/s 2 ) F g = 540 N = WEIGHT F E,S = - = - m g g F S,E = F = F g = = m g g
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Physics 211: Lecture 5, Pg 4 Lecture 5, Lecture 5, Act 1 Act 1 Mass vs. Weight Mass vs. Weight An astronaut on Earth kicks a bowling ball straight ahead and hurts his foot. A year later, the same astronaut kicks a bowling ball on the moon in the same manner with the same force. His foot hurts. .. (a) more (b) less (c) the same
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Physics 211: Lecture 5, Pg 5 Lecture 5, Lecture 5, Act 1 Act 1 Solution Solution The masses of both the bowling ball and the astronaut remain the same, so his foot will feel the same resistance and hurt the same as before. Ouch.
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Physics 211: Lecture 5, Pg 6 Lecture 5, Lecture 5, Act 1 Act 1 Solution Solution However the weights of the bowling ball and the astronaut are less: Thus it would be easier for the astronaut to pick up the bowling ball on the Moon than on the Earth. W = m g Moon g Moon < g Earth
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Physics 211: Lecture 5, Pg 7 The Free Body Diagram The Free Body Diagram Newton’s 2nd Law says that for an object F = m a . Key phrase here is for an object for an object . Object has mass and experiences forces Object has mass and experiences forces So before we can apply F = m a to any given object we isolate the forces acting on this object:
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Physics 211: Lecture 5, Pg 8 The Free Body Diagram. .. The Free Body Diagram.
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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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Lect05 FBD, ropes,springs - Physics 211: Lecture 5 Todays...

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