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Transcript
More of Newton’s Laws Announcements: • • • • • Tutorial Assignments due tomorrow. Pages 19-21, 23, 24 (not 22,25) Note Long Answer HW due this week. CAPA due on Friday. Have added together the clicker scores so far and will try to have results placed on D2L (Desire to Learn – replacement for CULEARN). There are still several clickers not registered. • Exam – Feb. 16 – can use calculator and 1 sheet (8.5”x11”) either side – not xeroxed. 1 Web page: http://www.colorado.edu/physics/phys1110/phys1110_sp12/ Force review Force is a vector so superposition of forces is true (can combine all forces into one net force using vector addition). We draw a free body diagram to figure out the effect of the various forces and to calculate the net force. Newton’s 1st law: A body will stay at a constant velocity unless acted upon by a net force. Newton’s 2nd law: . Can split: & A net force causes an acceleration which is inversely proportional to the mass of the object 2 Clicker question 1 Set frequency to BA Q. An astronaut in intergalactic space is twirling a rock on a string. Suddenly the string breaks when the rock is at the point shown. Which path does the rock follow after the string breaks? Before the string breaks, the astronaut must be exerting a force on the rock causing the centripetal acceleration Snap! A B D C After the string breaks, no net force is acting so the rock travels at constant velocity which was horizontal to the right (tangent to circle) at the moment the string broke. 3 Force of gravity We all feel the force of gravity. It keeps the Earth together and keeps us from floating off. We have also learned that if there is no opposing force (free fall) the acceleration due to gravity is Thus, from Newton’s 2nd law we can deduce that the force of gravity is We call the force due to gravity the weight: 4 Clicker question 2 Set frequency to BA Q. An object is being lowered on a cord at a constant speed. How does the tension T in the cord compare to the weight mg of the object? A. T > mg B. T = mg constant velocity C. T < mg D. Impossible to tell Constant speed in a straight line means acceleration is 0 so and therefore 5 Clicker question 3 Set frequency to BA Q. An object is being lowered on a cord at a speed which is decreasing. The acceleration of the object is A. Up B. Down C. 0 velocity down, D. Impossible to tell speed decreasing If the speed is decreasing the acceleration direction must be opposite to the velocity direction. If v1 and v2 are positive and v2 is less than v1 (slowing down) then Δv is negative (opposite direction from v1 and v2) 6 Clicker question 4 Set frequency to BA Q. An object is being lowered on a cord at a speed which is decreasing. How does the tension T in the cord compare to the weight mg of the object? A. T > mg B. T = mg velocity down, C. T < mg speed decreasing D. Impossible to tell Let up be positive. From the previous question and ay is positive and therefore T > mg 7 More on weight We will use the term weight to refer to the force on an object due to gravity alone. We may also use the term apparent weight to describe the value returned by measuring the weight in an accelerated environment. A spring scale effectively measures the tension in a string holding up the object to be measured. As we just found out, the tension may not equal the weight (mg) if there is acceleration and so an apparent weight will be measured. This applies to most scales like a bathroom scale. 8 Units of force From Newton’s 2nd law we can deduce the units of force. has units of kg·m/s2 also known as a newton (N) That is, Mass is an intrinsic property of an object independent of where it is measured. Weight is the force due to gravity and depends on the gravitational field the object is in. In SI units, kilogram is a unit of mass and newton is a unit of force (including weight) In the English system, pound measures force (including weight) and slug is the unit of mass: A mass of 1 kilogram weighs about 2.2 lbs on Earth 9 Newton’s third law Two formulations of Newton’s 3rd law. For every action there is an equal and opposite reaction. If a force is exerted on body A due to B then, there is a force which is equal in magnitude but opposite in direction on body B due to A. Fon A due to B = − Fon B due to A This is how a rocket works. Hot gas is ejected out the back which creates an equal force on the rocket in the opposite direction. € 10 Newton’s third law Imagine being on a stationary skateboard and someone throws a basketball at you from the left which you catch. What would happen to you and the skateboard? Would you move left, right, or not move at all? The basketball exerts a force on you moving you left and you exert a force on the basketball, slowing it down Imagine being on a stationary skateboard and throwing a basketball to the left. What would happen to you and the skateboard? Would you go left, right, or not move at all? You exert a force on the basketball moving it left and the basketball exerts a force on you moving you right. 11 Newton’s third law The equal and opposite forces in Newton’s third law for a particular situation are sometimes referred to as an action-reaction pair. Let’s identify action-reaction forces in 1 the diagram. For each pair, exactly one ton force acts on one object and the “reaction” force acts on the other object. Normal force pushing object up and object pushing ground down Friction pushing object right and object pushing ground left Tension pulling object left and object pulling rope right Gravity pulls object toward Earth; object pulls Earth toward object 12 Newton’s laws relationship Newton’s 1st law is basically a special case of Newton’s 2nd law when force and acceleration are 0. Newton’s 2nd law deals with individual objects. We draw free body diagrams to show the forces acting on an object. Newton’s 3rd law deals with the forces between two objects (equal and opposite). Sometimes we need this to determine the force acting on the object of interest. 13 Clicker question 5 Set frequency to BA Situation I: A car on Earth rides at a constant speed of 35 mph over the top of a round hill with a radius of curvature of 100 m. Situation II: A monorail car in space moves at constant speed of 35 mph along a round monorail, with a 100 m radius of curvature. Which car experiences the larger acceleration? A. Earth car I B. Space car C. Both cars have the II same acceleration D. Impossible to tell The speed is constant so there is no linear acceleration. But the velocity vector direction is changing so there is centripetal acceleration which is the same for both. 14 Clicker question 6 Set frequency to BA Q. A car of mass m, traveling at constant speed, rides over the top of a hill. At the top of the hill, the magnitude of the normal force N of the road on the car is… A. N = mg B. N > mg C. N < mg D. Impossible to tell Picking the up direction as the positive y-axis we get We know the acceleration is down so and therefore 15 Clicker question 7 Set frequency to BA Q. Assuming the girls have a mass of 40 kg and the men have a mass of 80 kg, what is the magnitude of the downward force on one shoulder of a skier in the bottom row? (Assume g = 10 m/s2.) No acceleration in the y direction A. 200 N so the forces have to add to 0. B. 300 N C. 400 N Weight of top girl is 400 N (W=mg), D. 800 N split equally between two middle girls. E. 0 N Total downward force of a middle row girl is 400 N of her own weight plus 200 N of the girl above weight. The middle row girl’s total downward force is 600 N which is split onto two shoulders. Each bottom row shoulder has 300 N on it. 16 A real pyramid Be thankful I didn’t ask you to calculate the forces for this pyramid! 17