Normal Force. W = mg cos(θ) Normal force F N = mg cos(θ) F N

Transcription

1 Normal Force W = mg cos(θ) Normal force F N = mg cos(θ) Note there is no weight force parallel/down the include. The car is not pressing on anything causing a force in that direction. If there were a person holding the car from rolling or sliding down the hill in that direction then there would be weight force of the car on pressing on the person. a θ F gec F N θ W M. Herndon, Phys

2 Free Body Diagram Consider a car sliding down the road a a? F N How many forces are acting on the car? F friction from road on car Identify and draw the forces Sometimes easier if you identify all action reaction pairs, F g M. Herndon, Phys

3 Gravity Newton s 3 rd Law Action Reaction Pairs - Force of gravity of the earth on the car - Understand components: perpendicular (pressing the object into the road) and parallel a (accelerating the car down the road) - Force of gravity of the car on the earth Normal force and weight - Road pushing up (normal to) on the car - Weight of the car pushing down on the road. This is a force because the car physically pushes down the road against the frictional force. Friction and weight - Weight of the car pushing down (parallel to) the road, but only the portion pushing on the road - Friction pushing up the road F friction from road on car W, F gec M. Herndon, Phys a? F N θ Portion down the road not negated by friction is the acceleration F gce

4 Free Body Diagram Force of gravity and the Weight W - interested in the force of gravity and the weight component caused by gravity that moves the car down the hill and pushes against the friction. W dependent on F gec = mg sin(θ) - and the components that press the car into the road to compute the normal force. W perp = F gecperp = mg cos(θ) then a a? F N F friction from road on car Normal force F N = mg cos(θ) Friction F k = µ k F N =µ k mg cos(θ) The we need to check if the possible component of the weight that is pointed down the hill is larger than the static friction. W = mg sin(θ) > F s = µ s F N = µ s mg cos(θ), tan(θ)>µ s θ W, F gec Accelerating forces: F gecperp - F k M. Herndon, Phys

5 Tension: Atwood Machine Two objects are connected by a soft string around a frictionless pulley as shown. What is the tension on the string? What are the acceleration of the objects, respectively? (see board, g=9.8 m/s 2 ) Ø Ø Ø Free Body Diagrams Coordinate system Equations: +x Object 1: T - M 1 g = M 1 a 1 Object 2: M 2 g T = M 2 a 2 Ø Connection: a 1 =a 2 =a Ø Solve: a=(m 2 -M 1 )/(M 2 +M 1 )g = 2.45 m/s 2 à a 1 =2.45 m/s 2, a 2 =2.45 m/s 2 T=2M 1 M 2 /(M 1 +M 2 )g = N T M 1 g T M 2 g +x M. Herndon, Phys

6 Average Velocity An object moving uniformly around a circle of radius r has a period T. What is its average velocity over the period T? A: 0 B: 2πr/T C: r/t D: Not enough information as the mass of the object is not given. E: None of above A Average velocity is displacement divided by time interval and displacement depends only on initial and final positions. Conceptual questions like this will make up about 1/4-1/3 of the exam. M. Herndon, Phys

7 Basic Kinematic Equations A boat is traveling at 4.0 m/s as it passes the starting line of a race. If the boat accelerates at 1.0 m/s2, then the distance the boat has traveled after 6.0 seconds is: x(t) = x 0 + v 0 t + ½a 0 t 2 = 0 + 4*6.0 + ½*1.0x6.0 2 = 42 m Some problems will be basic calculation and others involve more challenging calculations. M. Herndon, Phys

8 Projectile Motion, More Complex A projectile is projected on the ground with a velocity of 45.0 m/s at an angle of 60.0 degrees above the horizontal. On its way down, it lands on a rooftop of 4m high. What is the horizontal distance between the launching and landing points? v yi =vsin(60) = 39.0 m/s y(t) = y 0 + v yi t 1/2 gt 2 solving quadratic eq. gives t=7.9 s v xi =vcos(60) = 22.5 m/s x-x 0 = v xi t = 22.5 x 7.9 = 178 m M. Herndon, Phys

9 Circular Motions and Accelerations A particle is moving a circle with radius r =2.0m. The particle experiences a uniform tangential acceleration from a velocity of 1.0m/s to 2.0m/s. Some possible question to practice What is the angular velocity at the end of the acceleration? What distance does the particle travel during the acceleration? What is the magnitude of the displacement during the acceleration? Does the centripetal acceleration increase uniformly during the acceleration. On the last: The tangential velocity is increasing linearly. The centripetal acceleration goes as v T2 /r rather than linearly M. Herndon, Phys

10 Circular Motions and Accelerations What is the angular velocity at the end of the acceleration? velocity: v T = 2.0m/s angular velocity ω = v T /r = 2.0/2.0 = 1 rad/sec What distance does the particle travel during the acceleration? average velocity 1.5 m/s. d= vt = 1.5*1 = 1.5 m What is the magnitude of the displacement during the acceleration? Assume we traveled from (r,theta) = (2,0), or (x,y) = (2,0) Traveling 1.5 m means theta = d/r = 1.5/2 = 0.75 rad Then calculate the Cartesian coordinates at (r,theta) = (2.0,0.75 rad) and find the magnitude of the displacement between the two sets of coordinates M. Herndon, Phys