Report

UNIT 3 Forces and the Laws of Motion Tuesday October 18th FORCES & THE LAWS OF MOTION 2 ConcepTest 4.9b Collision Course II In the collision between 1) the car the car and the truck, 2) the truck which has the greater 3) both the same acceleration? 4) it depends on the velocity of each 5) it depends on the mass of each ConcepTest 4.9b Collision Course II In the collision between 1) the car the car and the truck, 2) the truck which has the greater 3) both the same acceleration? 4) it depends on the velocity of each 5) it depends on the mass of each We have seen that both vehicles experience the same magnitude of force. But the acceleration is given by F/m so the car has the larger acceleration, since it has the smaller mass. TODAY’S AGENDA Tuesday, October 18 Laws of Motion Mini-Lesson: Everyday Forces (1st Law Problems) Hw: Practice D (All) p139 UPCOMING… Wed: Even More Everyday Forces Quiz #1 1st Law Problem Thurs: NO SCHOOL (Teacher In-Service Day) Fri: NO SCHOOL (Teacher In-Service Day) Mon: Lab Review: LAWS OF FORCE Lab Everyday Forces q Find Tensions T1 and T2 Fy 0 T1 sin(60) mg T1 sinq T1 mg 20(9.8) T1 226 N sinq sin60 Fx 0 T2 T1 cosq T1 cos(60) T2 mg m q = 60o m = 20 kg 24 Everyday Forces F3 A lantern of mass m is suspended by a string that is tied to two other strings, as shown to the right. The free-body diagram shows the forces exerted by the three strings on the knot. F2 q1 In terms of F1, F2, and F3, what is the net force acting on the knot? q2 F1 F3 Fnet = F1+F2+F3 = 0 F1 F2 Everyday Forces A lantern of mass m is suspended by a string that is tied to two other strings, as shown to the right. The free-body diagram shows the forces exerted by the three strings on the knot. F3 F2 Find the magnitudes of the x and y components for each force acting on the knot. q1 q2 F1 Everyday Forces F3 Find the magnitudes of the x and y components for each force acting on the knot. X component q1 q2 y component F1 0.0 F2 -F2cosӨ1 +F2sinӨ1 F3 +F3cosӨ2 +F3sinӨ2 Fx net = F3cosӨ2 – F2cosӨ1 = 0 F2 -mg Fy net = F2sinӨ1 + F3sinӨ2 - mg = 0 F1 Everyday Forces F3 F2 Assume that Ө1= 30°, Ө2= 60°, and the mass of the lantern is 2.1 kg. Find F1, F2, and F3. Fx net = F3cosӨ2 – F2cosӨ1 = 0 q1 Fy net = F2sinӨ1 + F3sinӨ2 - mg = 0 F1 Find F3 in terms of F2. F3cosӨ2 = F2cosӨ1 q2 F3 = F2 cos 1 cos 2 F3 = F2 cos 30 cos 60 F3 = F2(1.73) Substitute for F3 with F2(1.73). mg = 20.6N F2sinӨ1 + F3sinӨ2 - mg =0 F2sin(30) + F2(1.73)sin(60) – 20.6 N =0 F2(.5) + F2(1.73)(.866) = 20.6 N F2(.5) + F2(1.50) = 20.6 N F2(.5 + 1.50) = 20.6 N F2(2.00) = 20.6 N F2 = 10.3 N Everyday Forces F3 Assume that Ө1= 30°, Ө2= 60°, and the mass of the lantern is 2.1 kg. Find F1, F2, and F3. F1 = -mg = -20.6 N F3 = F2(1.73) F3 = 17.8 N F2 = 10.3 N F3 = 10.3N(1.73) F2 q1 q2 F1 END 12