PHY 101 1st Edition Lecture 10 Outline of Last Lecture I. Why do things move?II. Aristotle vs. NewtonIII. 4.1 ForcesIV. 4.2 Newton’s First LawOutline of Current Lecture V. 4.3 Newton’s Second LawVI. Gravitational ForceVII. Gravitational Force, Mass, WeightCurrent Lecture4.3 Newton’s Second Law- First law: explains what happens to an object if there is not net force acting on it- What happens if there is a net force acting on an oobject?o The velocity of the object will change (i.e there will be acceleration)o The greater the force, the greater the accelerationo The greater the mass, the smaller the acceleration- Newton’s second law of motion:o The acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass.o A = F/m or F = mao F is the net external force acting on the object: F = vector sum of all external forceso 2nd law in dimensions: Fx = max Fy = mayo The x and y components can be treated separately and independentlyo Units: Newtons (N) 1N = 1 kg·m/s2 Meaning: when 1N of force acts on an object that has a mass of 1kg, it produces an acceleration of 1 m/s2These notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute. Projectile motion: a projectile only experiences the gravitational force, which gives the acceleration of -9.8 m/s2 along the y-direction.- Example: 1D motiono Airboat mass = 3.5 x 102 kgo Net horizontal force acting on it: 7.0 x 102 No A) find acceleration A = fnet/m A = 7.0 x 102/ 3.5 x 102 A = 2.0 m/s2o B) from rest, how long does it take to reach a speed if 12.0 ms/? V = at T = v/a = 12.0/2.0 = 6so Once it reaches 12.0 m/s, the pilot shuts off the engine and the boat drifts 50.0muntil it stops. Find the resistance forceo Analysis: to find the resistance force, we need to find acceleration due to resistance force. USE KINEMATICS** 2aΔx = v2 - v2v = 0m/s 2a(5.0) = 0 - 122v0 = 12m/s A = -144/100 = -1.44m/s2 Fresist = ma = (3.5 x 102kg)(-1.44m/s2) = -504N- Example: horse pulling a bargeo Mass of barge = 2.0 x 103 kgo Each horse exerts a force of magnitude = 6.0 x 102 No Direction of forces: Horse 1: 30° Horse 2: 45° (Use -45°)o Find the initial acceleration of the barge. (ignore water, air etc. resistance)o Draw diagram, coordinate systemo Force and acceleration are vectors, calculate x and y components separately thenfind magnitude and direction Horse 1: Fx1 = Fcos(30) = (6.0 x 102)(0.866) = 5.20 x 102N Horse 2: Fx2 = Fcos(-45) = (6.0 x 102)(0.707) = 4.24 x 102No X-component Fx = fx1 + fx2 = (5.20 x 102N) + (4.24 x 102N) = 9.44 x 102No Y-component Horse 1: Fy1 = Fsin(30) = (6.0 x 102)(0.5) = 3.0 x 102N Horse 2: Fy2 = Fsin(-45) = (6.0 x 102)(-0.707) = -4.24 x 102N Fy = Fy1 + Fy2 = (3.0 x 102N) + (-4.24 x 102N) = -1.24 x 102N Y-component of the net force is along the negative y directiono Initial acceleration: X – a = fx/m = 9.44 x 102N/2.0 x 102 kg = 0.472 m/s2 Y – a = fy/m = -1.24 x 102N/2.0 x 102 kg = -0.062 m/s2o Magnitude:  A = (ax2 + ay2)1/2 = ((0.472)2 + (-0.062)2)1/2 = 0.476 m/s2o Direction: Tan(Θ) = ay/ax = -0.062/ 0.472 = -0.131 Θ = tan-1(-0.131) = -7.46° Angle measured from the x-axis- Gravitational Forceo Why do things always fall to the ground when released?o Why does earth circle the sun?o Gravitational force Mutual attraction between any two objects Things on earth obey the same laws as things in the “heaven” **every particle in the universe attracts each other with a force that is directly proportional to the mass of the particles and inversely proportional to the square of the distance between them- Gravitational Force, Mass and Weighto Newton’s law of universal gravitation Fg = Gm1m2/r2 G = 6.67 x 10-12 N·m2kg2: the universal gravitation constanto Weight: the magnitude of the gravitational force acting on an object of mass m is called the weight w of the objecto Relation between g and the universal gravitational constant G Fg = w = GmEm/r2 = mg g = GmE/r2o Example:o It is known that the average radius of Earth is 6371 km calculate the mass of Earth. g = GmE/r2  mE= gr2/ G g = 9.8 (6371 x 103)2/ 6.67 x 10 -11 kg = 6.0 x 1024