Forces and Motion beyond 1 D PHYSICS 220 Lecture 05 Forces and Motion beyond 1 D Textbook Sections 3.7, 4.1 Lecture 5 Purdue University, Physics 220

Purdue University, Physics 220 Apparent Weight Recall: F = m a Consider person accelerating up in an elevator Draw FBD Apply Newton’s second law N – mg = ma N = m(g+a) Apparent weight is equal to normal force from scale or floor Note: in free fall a=-g so N=0 y x N mg Lecture 5 Purdue University, Physics 220

Purdue University, Physics 220 iClicker You are traveling up on an elevator to the 30th floor of the Sears tower. As it nears the 30th floor, your weight appears to be A) heavier B) the same C) lighter N SF = ma N – mg = ma N = m(g+a) a < 0 so N < mg mg Lecture 5 Purdue University, Physics 220

Purdue University, Physics 220 Exercise A person has a mass of 50 kg. What is their apparent weight when they are riding on an elevator Going up with constant speed 9.8 m/s Going down with constant speed 9.8 m/s Accelerating up at a rate of 9.8 m/s2 Accelerating down at a rate of 9.8 m/s2 N = m(g+a) a = 0 so N= mg = 490 Newtons a = 0 so N= mg = 490 Newtons a = +9.8 m/s2 so N= 2 mg = 980 Newtons a = -9.8 m/s2 so N= 0 mg = 0 Newtons Lecture 5 Purdue University, Physics 220

Purdue University, Physics 220 Exercise You are standing on a scale inside an elevator. You weigh 125 pounds, but the scale reads 140 pounds. The elevator is going (up down can’t tell) The elevator is accelerating (up down can’t tell) y x A B C A B C N = m(g+a) Weight increases when accelerating up Weight decreases when accelerating down. Lecture 5 Purdue University, Physics 220

Purdue University, Physics 220 Exercise Fred throws a ball 30 m/s vertically upward. How long does it take to hit the ground 2 meters below where he released it? y = y0 + vy0t - 0.5 g t2 y-y0 - vy0t + 0.5g t2 = 0 -2 – 30 t + 0.5*9.8 t2 = 0 y = y0 + vy0t - 1/2 gt2 vy = vy0 - gt vy2 = vy02 - 2g(y-y0) t = 6.19 s or -.06 s Lecture 5 Purdue University, Physics 220

Purdue University, Physics 220 Exercise Fred throws a ball 30 m/s vertically upward. What is the maximum height the ball reaches? How long does it take to reach this height? v2-vo2 = 2 a Dy Dy = (v2-vo2 )/ (2 a) = -302 / (2 * (-9.8)) = 46 m v = v0 + a t t = (v-v0) / a = (0 – 30 m/s )/ (-9.8 m/s2) = 3.1 seconds Have students make plot? Lecture 5 Purdue University, Physics 220

Drag and Terminal Velocity Drag force = air resistance It is proportional to the square of the speed Fd W It opposes the direction of motion At the terminal velocity the drag force is equal to weight Lecture 5 Purdue University, Physics 220

Drag and Terminal Velocity Terminal velocity depends on size, mass, and shape of the object Lecture 5 Purdue University, Physics 220

Purdue University, Physics 220 Equilibrium Net Force: If are all the forces acting on an object When an object is in equilibrium, the net force acting on it is zero Lecture 5 Purdue University, Physics 220

Purdue University, Physics 220 Question Compare the net force on the two books A) Fphysics > Fbiology B) Fphysics = Fbiology C) Fphysics < Fbiology Net force is zero on both books! The normal force from table exactly cancels downward forces. Physics Biology Lecture 5 Purdue University, Physics 220

Book Pushed Across Table Calculate force of hand to keep the book sliding at a constant velocity, if the mass of the book is 1 kg, ms = .84 and mk=.75. Constant velocity  SF=0 x-direction: SF=0 Fhand-Ffriction = 0 Fhand=Ffriction Fhand=mk FNormal Combine: Fhand=mk FNormal Fhand=0.759.8 N Fhand=7.3 N y-direction: SF=0 FNormal-FGravity = 0 FNormal = FGravity FNormal =19.8=9.8 N hand Gravity Normal friction y x Key idea here is we can do each direction independently! Physics Lecture 5 Purdue University, Physics 220

Forces in 2 Dimensions: Ramp Calculate tension in the rope necessary to keep the 5 kg block from sliding down a frictionless incline of 20 degrees. 1) Draw free-body diagram 2) Label axes Weight is not in x or y direction! Need to decompose it! N Tell them any choice is OK, but this way only have 1 force to decompose T y x W q Lecture 5 Purdue University, Physics 220

Forces in 2 Dimensions: Ramp Calculate force necessary to keep the 5 kg block from sliding down a frictionless incline of 20 degrees. x- direction: S F = 0 W sinq – T = 0 W sin q W q W cos q y x q Lecture 5 Purdue University, Physics 220

Forces in 2 Dimensions: Ramp Calculate force necessary to keep the 5 kg block from sliding down a frictionless incline of 20 degrees. x- direction: W sinq – T = 0 T = W sinq = m g sinq = 16.8 Newtons N Tell them any choice is OK, but this way only have 1 force to decompose T y x W q Lecture 5 Purdue University, Physics 220

Purdue University, Physics 220 iClicker What is the normal force of ramp on block? A) N > mg B) N = mg C) N < mg In “y” direction: SF = ma N – W cos q = 0 N = W cos q T N W W q W sin q W cos q y x N = m g cos q q Lecture 5 Purdue University, Physics 220

Purdue University, Physics 220 Inclined Plane Ff N N = m g cos m g sin - Ff = 0 Ff = N m g sin -  m g cos = 0  mg Special case 1: Start with  at zero and slowly increase . Just before it slides s m g cos = m g sin tan = s Special case 2: Object is sliding down at constant velocity, that is a = 0 k m g cos = m g sin tan = k Lecture 5 Purdue University, Physics 220

Purdue University, Physics 220 Force at Angle Example A person is pushing a 15 kg block across a floor with mk= 0.4 at a constant speed. If she is pushing down at an angle of 25 degrees, what is the magnitude of her force on the block? x- direction: SFx = 0 Fpush cosq – Ffriction = 0 Fpush cosq – m FNormal = 0 FNormal = Fpush cosq / m Combine: ( Fpush cosq / m)–mg – FPush sinq = 0 Fpush ( cosq / m - sinq) = mg Fpush = m g / ( cosq/m – sinq) y- direction: SFy = 0 FNormal –Fweight – FPush sinq = 0 FNormal –mg – FPush sinq = 0 Fpush = 80 N Normal Could do this on over head to help people work through it. Pushing y x q q Friction Weight Lecture 5 Purdue University, Physics 220