1 / 10

UNIT 4 Forces and the Laws of Motion

UNIT 4 Forces and 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. In the collision between the car and the truck, which has the greater acceleration?.

brent
Download Presentation

UNIT 4 Forces and the Laws of Motion

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. UNIT 4Forces and the Laws of Motion

  2. 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?

  3. 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.

  4. T1 sin(60) T1 T2 T1 cos(60) mg Everyday Forces q Find Tensions T1 and T2 m q = 60o m = 20 kg 24

  5. 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

  6. 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

  7. 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

  8. 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

  9. 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

  10. END

More Related