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Welcome back to Physics 215 Today’s agenda: • More on free-body diagrams and force components • Applying Newton’s laws Physics 215 – Fall 2014 Lecture 05-2 1 Current homework assignment • HW4: – Knight textbook -- Ch.5: 40, 42, 44, 56 – Ch.6: 26, 28 – Due Friday, Sept. 26th in recitation Physics 215 – Fall 2014 Lecture 05-2 2 Newton’s Laws First Law: In the absence of external forces, an object at rest remains at rest and an object in motion continues in motion with constant velocity. Second Law: Fnet = ∑Fon object = m a Third Law: Physics 215 – Fall 2014 FAB = - FBA (“action = reaction”) [regardless of type of force and of motion of objects in question] Lecture 05-2 3 Consider a person sitting on a chair. We can conclude that the downward weight force on the person (by the Earth) and the upward normal force on the person (by the chair) are equal and in opposite direction, because 1. the net force on the person must be zero 2. the two forces form a Newton’s third-law pair 3. neither of the above explanations 4. both of the above explanations Physics 215 – Fall 2014 Lecture 05-2 4 Free-body diagrams: Person sitting on chair Chair Person Earth (incomplete) Physics 215 – Fall 2014 Lecture 05-2 5 There are two people facing each other, each on a separate cart. If person A pushes on person B, while person B does nothing, what will be the resulting motion of the carts? 1. Cart A doesn’t move and Cart B moves backwards 2. Cart B doesn’t move and Cart A moves backwards 3. Both carts move in opposite directions 4. Neither cart moves Physics 215 – Fall 2014 Lecture 05-2 6 A force P is applied by a hand to two blocks which are in contact on a frictionless, horizontal table as shown in the figure. The blocks accelerate to the right. Block A has a smaller mass than block B. (a) Draw free body diagrams for each block. (b) Which block experiences the larger net force? (c) Suppose that initially the mass of block A were half that of block B. If in a subsequent experiment the mass of block A were doubled, by what factor would the acceleration change assuming the pushing force remained constant? B Physics 215 – Fall 2014 Lecture 05-2 A 7 A block is held in place on a frictionless incline by a massless string, as shown. The acceleration of the block is 1. 2. 3. 4. zero straight down down and to the left (along the incline) not zero, but neither of 2 or 3 Physics 215 – Fall 2014 Lecture 05-2 8 A block is held in place on a frictionless incline by a massless string, as shown. The force on the block by the incline is 1. 2. 3. 4. a normal force given by vector A. a normal force given by vector B. the force given by vector A, but it’s not a normal force. the force given by vector B, but it’s not a normal force. Physics 215 – Fall 2014 Lecture 05-2 9 Normal force • Always perpendicular to surface of contact • Generic name given to contact force between 2 objects Physics 215 – Fall 2014 Lecture 05-2 10 Other forces Besides normal force what other forces are present for block on inclined plane? Physics 215 – Fall 2014 Lecture 05-2 11 Free-body diagram: Block on frictionless incline • Show all forces exerted on the block. • Do not show forces exerted by the block on anything else. Physics 215 – Fall 2014 Lecture 05-2 12 Geometry… y vertical component of NBP is NBPcos(q) horizontal is NBPsin(q) N q x T q Physics 215 – Fall 2014 W Lecture 05-2 13 Force components: Block on frictionless incline F = 0 implies all components of F are zero! Horizontal and vertical Physics 215 – Fall 2014 Lecture 05-2 14 When the block is held in place as shown, is the magnitude of the normal force exerted on the block by the incline 1. greater than the magnitude of the weight force (on the block by the Earth), 2. equal to the magnitude of the weight, or 3. less than the magnitude of the weight. 4. Can’t tell. Physics 215 – Fall 2014 Lecture 05-2 15 Solving system • Vertical equilibrium: NBPcos(q) + WBE = 0 • Horizontal equilibrium: NBPsin(q) + TBR = 0 2 equations solve for NBP and TBE in terms of WBE and q Physics 215 – Fall 2014 Lecture 05-2 16 What have we learned? • If at rest – net force = 0, since a = 0 ! • Write down FBD – identify all forces present • Take force components in 2 (in 2D) directions to find unknowns • Don’t plug in numbers until end Physics 215 – Fall 2014 Lecture 05-2 17 The string that holds the block breaks, so there is no more tension force exerted on the block. Will the magnitudes of either of the other two forces on the block (i.e., the weight and the normal force) change? 1. 2. 3. 4. Both weight and normal force will change. Only the weight force will change. Only the normal force will change. Neither one of the two forces will change. Physics 215 – Fall 2014 Lecture 05-2 18 Discussion • Say normal does not change. What will be the new net force? Same as old, but “minus” the tension, i.e. straight to the left (and not zero). • So what would the block do? Accelerate. OK -- Straight to the left? NO, can’t be correct answer. • Acceleration is down the incline, so normal gets smaller. Physics 215 – Fall 2014 Lecture 05-2 19 Geometry (2) N q q Physics 215 – Fall 2014 component of WBE at 900 to plane is WBEcos(q) component along plane is WBEsin(q) Wq Lecture 05-2 20 Force components: Block on frictionless incline • The string that holds the block breaks. • What does happen? • Take components now along and perpendicular to incline Physics 215 – Fall 2014 Lecture 05-2 21 Solving for second case • Since acceleration is down incline, component of net force at 90 degrees to slope is zero. NBP + WBEcos(q) = 0 • Using second law applied to components along incline: a = WBEsin(q)/mblock Physics 215 – Fall 2014 Lecture 05-2 22 Conclusion • Normal forces can change when small changes are made to the situation. • Can choose any 2 directions to find force components, but it pays to pick ones that simplify equations Physics 215 – Fall 2014 Lecture 05-2 23 Suppose we chose to resolve forces vertically and horizontally? • Vertically: (1) Ncos(q) + W = -masin(q) • Horizontally: (2) Nsin(q) = ma cos(q) • Multiply (1) by cos(q), (2) by sin(q) and add: N + Wcos(q) = 0 as before!! Physics 215 – Fall 2014 Lecture 05-2 24 If the block lies 0.5 m up such a plane inclined at an angle of 30 how long will it take for the block to reach the ground? (take g = 10 m/s2) Physics 215 – Fall 2014 Lecture 05-2 25 “real” inclined plane • For a real incline, we expect for small angles that the block still does not move when string breaks – why? • Friction – force up slope that opposes motion – more later… Physics 215 – Fall 2014 Lecture 05-2 26 Summary • To solve problems in mechanics, identify all forces and draw free body diagrams for all objects • If more than one object, use Newton’s Third law to reduce number of independent forces • Use Newton’s Second law for all components of net force on each object • Choose component directions to simplify equations Physics 215 – Fall 2014 Lecture 05-2 27 Weight, mass, and acceleration • What does a weighing scale ``weigh’’? • Does it depend on your frame of reference? • Consider elevators…. Physics 215 – Fall 2014 Lecture 05-2 28 A person is standing on a bathroom scale while riding an express elevator in a tall office building. When the elevator is at rest, the scale reads about 160 lbs. While the elevator is moving, the reading is frequently changing, with values ranging anywhere from about 120 lbs to about 200 lbs. At a moment when the scale shows the maximum reading (i.e., 200 lbs) the elevator 1. 2. 3. 4. must be going up must be going down could be going up or going down I’m not sure. Physics 215 – Fall 2014 Lecture 05-2 29 Motion of elevator Motion of elevator (if a < 0 ) (if a > 0 ) • Moving upward and slowing down, • Moving upward and speeding up, OR OR • Moving downward and speeding up. Physics 215 – Fall 2014 • Moving downward and slowing down. Lecture 05-2 30 A person is standing on a bathroom scale while riding an express elevator in a tall office building. When the elevator is at rest, the scale reads about 160 lbs. While the elevator is accelerating, a different reading is observed, with values ranging anywhere from about 120 lbs to about 200 lbs. At a moment when the scale shows the maximum reading (i.e., 200 lbs) the acceleration of the elevator is approximately 1. 2. 3. 4. 1 m/s2 2.5 m/s2 5 m/s2 12.5 m/s2 Physics 215 – Fall 2014 Lecture 05-2 31 Conclusions • Scale reads magnitude of normal force |NPS| • Reading on scale does not depend on velocity (principle of relativity again!) • Depends on acceleration only * a > 0 normal force bigger * a < 0 normal force smaller Physics 215 – Fall 2014 Lecture 05-2 32 Reminder of free-fall experiment • Objects fall even when there is no atmosphere (i.e., weight force is not due to air pressure). • When there is no “air drag” things fall “equally fast.” i.e. same acceleration • From Newton’s 2nd law, a = W/m is independent of m -- means W = mg • The weight (i.e., the force that makes objects in free fall accelerate downward) is proportional to mass. Physics 215 – Fall 2014 Lecture 05-2 33 Inertial and gravitational mass • Newton’s second law: F = mI a • For an object in “free fall” W = mG g • If a independent of mI, must have mI = mG Principle of equivalence Physics 215 – Fall 2014 Lecture 05-2 34 Reading assignment • Forces, Newton’s Laws of Motion, Weight & Friction • Ch. 5-7 in textbook Physics 215 – Fall 2014 Lecture 05-2 35