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UNIT 3 Forces and the Laws of Motion. Tuesday October 18 th. FORCES & THE LAWS OF MOTION. ConcepTest 4.9b Collision Course II. 1) the car 2) the truck 3) both the same 4) it depends on the velocity of each 5) it depends on the mass of each.
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UNIT 3Forces and the Laws of Motion
Tuesday October 18th FORCES & THE LAWS OF MOTION
ConcepTest 4.9bCollision Course II 1) the car 2) the truck 3) both the same 4) it depends on the velocity of each 5) it depends on the mass of each In the collision between the car and the truck, which has the greater acceleration?
ConcepTest 4.9bCollision Course II 1) the car 2) the truck 3) both the same 4) it depends on the velocity of each 5) it depends on the mass of each In the collision between the car and the truck, which has the greater acceleration? We have seen that both vehicles experience the same magnitude of force. But the acceleration is given byF/mso thecarhas thelarger acceleration, since it has thesmaller 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
T1 sin(60) T1 T2 T1 cos(60) mg Everyday Forces q Find Tensions T1 and T2 m q = 60o m = 20 kg 24
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 In terms of F1, F2, and F3, what is the net force acting on the knot? F2 F2 q1 q2 Fnet = F1+F2+F3 = 0 F3 F1 F1
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 Find the magnitudes of the x and y components for each force acting on the knot. F2 q1 q2 F1
Everyday Forces Find the magnitudes of the x and y components for each force acting on the knot. F3 X component y component F1 0.0 -mg F2 -F2cosӨ1 +F2sinӨ1 F2 F3 q1 q2 +F3cosӨ2 +F3sinӨ2 Fx net = F3cosӨ2 – F2cosӨ1 = 0 Fy net = F2sinӨ1+F3sinӨ2- mg = 0 F1
Everyday Forces Assume that Ө1= 30°,Ө2= 60°, and the mass of the lantern is 2.1 kg. Find F1, F2, and F3. F3 Fx net = F3cosӨ2 – F2cosӨ1 = 0 Fy net = F2sinӨ1+F3sinӨ2- mg = 0 Find F3 in terms of F2. F3cosӨ2= F2cosӨ1 F3 = F2 F3 = F2 F2 q1 q2 F3 = F2(1.73) mg= 20.6N Substitute for F3 with F2(1.73). F2(.5)+F2(1.50)= 20.6 N F2sinӨ1+F3sinӨ2- mg=0 F2(.5+1.50)= 20.6 N F2sin(30)+F2(1.73)sin(60)– 20.6 N=0 F2(2.00)= 20.6 N F2(.5)+F2(1.73)(.866)= 20.6 N F1 F2= 10.3 N
Everyday Forces Assume that Ө1= 30°,Ө2= 60°, and the mass of the lantern is 2.1 kg. Find F1, F2, and F3. F3 F1 = -mg = -20.6 N F2= 10.3 N F3 = F2(1.73) F3 = 10.3N(1.73) F2 q1 q2 F3 = 17.8 N F1