PHYS 1443 – Section 501Lecture #7AnnouncementsMaximum Range and HeightKinetic Equations of Motion in a Straight Line Under Constant AccelerationNewton’s First Law and Inertial FramesNewton’s Second Law of MotionExample of Newton’s 2nd Law of MotionExample for Newton’s 2nd Law of MotionGravitational Force and WeightNewton’s Third Law (Law of Action and Reaction)Example of Newton’s 3rd LawPHYS 1443 – Section 501Lecture #7Wednesday, Feb. 11, 2004Dr. Andrew Brandt•Newton’s Laws of Motion•The Force of GravityWednesday, Feb. 11, 2004 PHYS 1443-501, Spring 2004Dr. Andrew Brandt1Announcements• HW #2 on Ch. 3 due tonight, Feb. 11 at midnight• Beware negative points• HW#3 on Ch. 4 due Weds Feb 18, HW#4 on Ch. 5will be due Mon Feb. 23• HW should help your grade, if it’s not doing that start earlier, ask questions, physics clinic--RM 10• Remember the 1stmidterm, Wednesday, Feb. 25,– Covers chapters 1-5– Don’t get behind!Wednesday, Feb. 11, 2004 PHYS 1443-501, Spring 2004Dr. Andrew Brandt2Maximum Range and Height• What are the conditions that give maximum height and range of a projectile motion?Wednesday, Feb. 11, 2004 PHYS 1443-501, Spring 2004Dr. Andrew Brandt3⎟⎟⎠⎞⎜⎜⎝⎛=gvhii2sin22θThis formula tells us that the maximum height can be achieved when θi=90o!!!⎟⎟⎠⎞⎜⎜⎝⎛=gvRiiθ2sin2This formula tells us that the maximum range can be achieved when 2θi=90o, i.e., θi=45o!!!Kinetic Equations of Motion in a Straight Line Under Constant Acceleration()tavtv xxixf +=Velocity as a function of time()12fxixfxxvt vvt+−= = xiDisplacement as a function of velocity and time221tatvxxxxiif += +Displacement as a function of time, velocity, and acceleration()222ffixxi xvv axx=+ −Velocity as a function of displacement and accelerationWednesday, Feb. 11, 2004 PHYS 1443-501, Spring 2004Dr. Andrew Brandt4You may use different forms of Kinetic equations, depending on the information given to you for a specific problem!!Example+quizNewton’s First Law and Inertial FramesGalileo motivated Newton’s 1stlaw of motion (Law of Inertia): In the absence of external forces, an object at rest remains at rest and an object in motion continues in motion with a constant velocity. What does this statement tell us? • When no force is exerted on an object, the acceleration of the object is 0. • Any isolated object, an object that does not interact with its surrounding, is either at rest or moving at a constant velocity.• An object maintains its current state of motion, as long as there is no force that interferes with the motion. This tendency is called the Inertia.A frame of reference that is moving at constant velocity is called an Inertial FrameWednesday, Feb. 11, 2004 PHYS 1443-501, Spring 2004Dr. Andrew Brandt5Newton’s Second Law of MotionamFii=∑The acceleration of an object is directly proportional to the net force exerted on it and is inversely proportional to the object’s mass. How do we write the above statement in a mathematical expression?xiixmaF =∑Since it’s a vector expression, each component should also satisfy:From the above vector expression, what do you conclude the dimension and unit of force are?=]][[ amyiiymaF =∑ziizmaF =∑Mass does notequal weight!The dimension of force is ]][[2−LTM===−]][[]][[][2LTMamForce2/ smkg⋅The unit of force in SI is Wednesday, Feb. 11, 2004 PHYS 1443-501, Spring 2004Dr. Andrew Brandt6For ease of use, we define a new derived unit called, a Newton (N) lbssmkgN41/112≈⋅≡Wednesday, Feb. 11, 2004 PHYS 1443-501, Spring 2004Dr. Andrew Brandt7Acceleration Example of Newton’s 2ndLaw of MotionWhat constant net force is required to bring a 1500kg car to rest from a speed of 100km/h within a distance of 55m?This is a one dimensional motion. Which kinetic formula do we use to find acceleration?What is given?Thus, the force needed to stop the car is=xF==hkmvxi/100()ifxxixfxxavv −+= 222What do we need to know to figure out the force? Acceleration!! Initial speed:sm /28Displacement:mxxxif55=−=∆Final speed:smvxf/0=()=−−=ifxixfxxxvva222()()22/1.7552/28smmsm−=−()Nsmkgmax42101.1/1.71500 ×−=−×=How are stopping distance and force related?=−=−=∆xxixfifavvxxx222()=−xxixfmavvm222()22xixmvF−•Linearly proportional to the mass of the car•Squarely proportional to the speed of the car•Inversely proportional to the force by the brakeExample for Newton’s 2ndLaw of MotionDetermine the magnitude and direction of acceleration of an object whose mass is 0.30kg and is being pulled by two forces, F1 and F2, as shown in the picture, with magnitudes of the forces 8.0 N and 5.0 N, respectively.()NF 0.460cos0.8cos1=×=1oθ()NF 9.660sin0.8sin1=×=1oθ()NF 7.420cos0.5cos22=−×=oθ()NF 7.120sin0.5sin22−=−×=oθ=xF1Components ofF1F2F1θ1=60οθ2=−20ο=yF1=xF2=yF2Components ofF2Components oftotal force F=xF==+=+NFFxx7.87.40.421 xma==−=+NFFyy2.57.19.621 yma=yF2/293.07.8smmFaxx===Wednesday, Feb. 11, 2004 PHYS 1443-501, Spring 2004Dr. Andrew Brandt8Acceleration Vector a2/173.02.5 smmFayy===() ()222/341729 sma =+=D302917tantan 11=⎟⎠⎞⎜⎝⎛=⎟⎟⎠⎞⎜⎜⎝⎛=−−xyaaθ2/1729 smjijaiaayx⎟⎠⎞⎜⎝⎛+=+=∧∧∧∧Magnitude and direction of acceleration aGravitational Force and WeightThe attractive force exerted on an object by the Earth Gravitational Force, Fggm== amFGMggMFWG===Weight of an object with mass M isSince weight depends on the magnitude of gravitational acceleration, g, it varies depending on geographical location.By measuring the forces one can determine masses. This is why you can measure mass using spring scale.Wednesday, Feb. 11, 2004 PHYS 1443-501, Spring 2004Dr. Andrew Brandt9Newton’s Third Law (Law of Action and Reaction)If two objects interact, the force, F12, exerted on object 2 by object 1 is equal in magnitude and opposite in direction to the force, F21, exerted on object 1 by object 2. F12F212112FF −=The action force is equal in magnitude to the reaction force but in the opposite direction. These two forces always act on different objects. What is the reaction force to the force of an object in free fall ?The force exerted on the Earth do the gravitational field. A stationary object on top of a table has a reaction force (normal force) from the table to balance the action force, the gravitational force.Wednesday, Feb. 11, 2004 PHYS 1443-501, Spring 2004Dr. Andrew Brandt10Example of Newton’s 3rdLaw A large man and a
View Full Document
Unlocking...