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Transcript
Ch. 7 Forces and Motion in Two Dimensions Milbank High School Sec. 7.1 Forces in Two Dimensions • Objectives – Determine the force that produces equilibrium when three forces act on an object – Analyze the motion of an object on an inclined plane with and without friction What is meant by two dimensions? • Consider a golf ball being hit out of a sand trap – It has a horizontal force AND a vertical force – We can solve for many different things using a combination of forces and vectors • Height of the ball • Time in the air • Velocity when it hits the ground Equilibrant • A force exerted on an object to produce equilibrium • Same magnitude as the resultant force but opposite in direction Solving problems in two dimensions • Draw it out! • Rearrange vectors to form a triangle if possible • Solve for the resultant vector – Opposite in direction – Example Problem Pg. 151 Sec. 7.2 Projectile Motion • Objectives – Recognize that the vertical and horizontal motions of a projectile are independent – Relate the height, time in the air, and the initial velocity of a projectile using its vertical motion, then determine the range. – Explain how the shape of the trajectory of a moving object depends upon the frame of reference from which it is observed. Projectiles have independent motions! • Projectiles have two velocities, one in the “x” direction, and one in the “y” direction • x is always constant • y will be changing due to the acceleration due to gravity Displacement • y displacement y = yo - 1/2gt2 • x displacement x = vxot v = 25m/s Velocity of projectiles launched horizontally • vx = initial velocity • vy = (-g)t • v = resultant velocity vector • Example Pg. 157 Effects of air resistance • We ignore the effects of air resistance for these problems • Sometimes it would make a large difference, other times it wouldn’t • Many projectiles modified so that they reduce air resistance Projectiles launched at an Angle • • • • Usually given angle of launch and velocity What do we have to find? Maximum height Range – Horizontal distance • Flight time – hang time Projectiles Launched at an Angle • Two initial velocity components • vxo • vyo How do we find these? vx = vo(cosθ) vy = vo(sinθ) Projectiles Launched at an Angle • tup = vyo/g • ttotal = 2(tup) • Peak Height y = vyot - ½gt2 • Range R = vxot Projectiles launched at an Angle • The Flight of a Ball • Example Problem Pg. 159 Sec. 7.3 Circular Motion • Objectives – Explain the acceleration of an object moving in a circle at constant speed – Describe how centripetal acceleration depends upon the object’s speed and the radius of the circle – Recognize the direction of the force that causes centripetal acceleration – Explain how the rate of circular motion is changed by exerting torque on it. Uniform Circular Motion • Movement of an object at constant speed around a circle with a fixed radius • Merry-go-round Circumference = 2*pi*Radius Vectors Acceleration • Which direction? • Always towards the center Centripetal Force • “Center seeking” • Net force towards the center that causes the object to try to seek the center • What force is pulling it in? • As a bucket of water is tied to a string and spun in a circle, the force of tension acting upon the bucket provides the centripetal force required for circular motion. Net Force • Example Problem Pg. 165