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PHYSICS 220 Lecture 05 Forces and Motion beyond 1 D Textbook Sections 3.7, 4.1 Lecture 5 Purdue University, Physics 220 1 •UNIMPORTABLE: •#00E9BF56,2.00,2.00 •#09B63F80,6.00,6.00 •#0E133429,2.00,2.00 •#225BEB92,4.00,4.00 •#80330CBF,6.00,6.00 •#803AF64C,6.00,6.00 •#809B0813,4.00,4.00 • #80E6096F,6.00,6.00 •#810C8D00,2.00,2.00 •#81709E6F,6.00,6.00 •#817EE11E,6.00,6.00 •#8279AE55,6.00,6.00 •#831033A0,4.00,4.00 •#8341E92B,4.00,4.00 •#834C955A,6.00,6.00 •#83CA7831,6.00,6.00 Lecture 5 Purdue University, Physics 220 2 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 y = y0 + vy0t + 1/2 gt2 vy = vy0 + gt vy2 = vy02 + 2g(y-y0) -2 – 30 t + 0.5*9.8 t2 = 0 ax bx c 0 2 b b2 4ac x 2a Lecture 5 2 30 30 4 9.8 2 x 9.8 2 t = 6.19 s or -.06 s Purdue University, Physics 220 3 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 v = v0 + a t Dy = (v2-vo2 )/ (2 a) t = (v-v0) / a = -302 / (2 * (-9.8)) = (0 – 30 m/s )/ (-9.8 m/s2) = 46 m = 3.1 seconds Lecture 5 Purdue University, Physics 220 4 Contact Force: Spring • Force exerted by a spring is directly proportional to the amount by which it is stretched or compressed. Fspring = k x always trying to restore its original length • Example: When a 5 kg mass is suspended from a spring, the spring stretches 8 cm. Determine the spring constant. Fspring- Fgravity = 0 k =mg/x Fspring = Fgravity = (5 kg) x (9.8 m/s2) /(0.08 m) kx=mg = 612 N/m Fspring Fgravity y x Lecture 4 Purdue University, Physics 220 5 Contact Force: Tension • Tension – A force transmitted by a rope, cord, cable or the like which transmits a force from one end to an object attached at the other end • Ideal string (or cord, rope, etc.): – Always maintains constant tension everywhere. – Has a zero mass. – Tension is parallel to the string Lecture 4 Purdue University, Physics 220 Pulley changes the direction of the force associated with the tension in the rope. 6 Pulley Example Two boxes are connected by a string over a frictionless pulley. Box 1 has mass 1.5 kg, box 2 has a mass of 2.5 kg. Box 2 starts from rest 0.8 meters above the table, how long does it take to hit the table. • Compare the acceleration of boxes 1 and 2 A) |a1| > |a2| B) |a1| = |a2| 1) T - m1 g = m1 a1 2) T - m2 g = -m2 a1 C) |a1| < |a2| y using a1 = -a2 x T 2) T = m2 g -m2 a1 1) m2 g -m2 a1 - m1 g = m1 a1 a1 = (m2 – m1)g / (m1+m2) Lecture 4 Purdue University, Physics 220 1 1 T 2 2 m1g m2g 7 Pulley Example Two boxes are connected by a string over a frictionless pulley. Box 1 has mass 1.5 kg, box 2 has a mass of 2.5 kg. Box 2 starts from rest 0.8 meters above the table, how long does it take to hit the table. a1 = (m2 – m1)g / (m1+m2) a = 2.45 m/s2 Dx = v0t + ½ a t2 Dx = ½ a t = sqrt(2 Dx/a) t = 0.81 seconds y x t2 Lecture 4 T 1 T 2 2 m1g Purdue University, Physics 220 1 m2g 8 iClicker • Two boxes are connected by a string over a frictionless pulley. In equilibrium, box 2 is lower than box 1. Compare the weight of the two boxes. A) They have the same weight B) Box 1 is heavier C) Box 2 is heavier SF = 0 1) T - m1 g = 0 2) T – m2 g = 0 => m1 = m2 Lecture 4 T 1 T 1 m1g 2 2 m2g Purdue University, Physics 220 9 Tension Example • Determine the force exerted by the hand to suspend the 45 kg mass as shown in the picture. 1) 220 N 2) 440 N 3) 660 N 4) 880 N 5) 1100 N SF = 0 T+T–W=0 2T=W T=mg/2 = (45 kg x (9.8 m/s2)/ 2 = 220 N y T T x W Remember the magnitude of the tension is the same everywhere along the rope! Lecture 4 Purdue University, Physics 220 10 Question • What is the force on the ceiling? y A) 220 N B) 440 N D) 880 N C) 660 N x E) 1100 N SF = 0 Fc Fc -T - T – T = 0 Fc = 3 T Fc = 3 x 220 N = 660 N Lecture 4 Purdue University, Physics 220 T 11 Question • What does the scale read? A) 225 N B) 550 N C) 1100 N The sum of the forces here will be 0. Thus, the force tension will equal the force gravity. Since the force of gravity is 550N, the force of tension (which is measured by the scale) will also be 550N. Lecture 4 Purdue University, Physics 220 12 30 Two blocks one sliding one hanging •A block of mass m1=3kg rests on a frictionless horizontal surface. A second block of mass m2=2kg hangs from an ideal cord of negligible mass who runs over an ideal pulley. Block 2 starts from rest 0.8 meters above the floor, how long does it take to hit the floor? a a a1 = (m2 )g / (m1+m2) a = 3.9 m/s2 Dy = v0t + ½ a t2= ½ a t2 t = sqrt(2 Dy/a) t = 0.64 seconds Lecture 4 Purdue University, Physics 220 13 Friction • Magnitude of frictional force is proportional to the normal force. Fkinetic = mk N Fstatic ms N mk coefficient of kinetic friction ms coefficient of static friction Be Careful! •Static friction , can be any value up to msN •Direction always opposes motion Lecture 4 F Ff Purdue University, Physics 220 14 Inclined Plane Ff N N = m g cos m g sin - Ff = 0 Ff = mN m g sin - m m g cos = 0 mg • Special case 1: Start with at zero and slowly increase . Just before it slides ms m g cos = m g sin tan = ms • Special case 2: Object is sliding down at constant velocity, that is a = 0 mk m g cos = m g sin Lecture 5 tan = mk Purdue University, Physics 220 15 iClicker What is the normal force of ramp on block? A) N > mg B) N = mg In “y” direction: SF = ma N – mg cos = 0 N = mg cos N T Lecture 5 C) N < mg W N = m g cos Purdue University, Physics 220 W 16 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 cos – Ffriction = 0 Fpush cos – m FNormal = 0 FNormal = Fpush cos / m y- direction: SFy = 0 FNormal –Fweight – FPush sin = 0 FNormal –mg – FPush sin = 0 Combine: ( Fpush cos / m)–mg – FPush sin = 0 Fpush ( cos / m - sin ) = mg Fpush = m g / ( cos/m – sin) Fpush = 80 N Normal Pushing y x Friction Weight Lecture 5 Purdue University, Physics 220 17