1 / 11

UNIT 4 Forces and the Laws of Motion

UNIT 4 Forces and the Laws of Motion. N. T. m 1. T. m 1 g. m 2. m 2 g. Force Lab Notes. Forces on m 1. m 1 a = T = m 2 g – m 2 a. Forces on m 2. 23. Everyday Forces. A 75.0-kg box slides down a 25.0˚ ramp with an acceleration of 3.60 m/s 2.

rae
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. N T m1 T m1g m2 m2g Force Lab Notes Forces on m1 m1a = T = m2g – m2a Forces on m2 23

  3. Everyday Forces A 75.0-kg box slides down a 25.0˚ ramp with an acceleration of 3.60 m/s2. a) Find the μk between the box and the ramp. What acceleration would a 175 kg box have on this ramp? 25

  4. A 75.0-kg box slides down a 25.0˚ ramp with an acceleration of 3.60 m/s2. ΣFy = 0 a) Find the μ between the box and the ramp. FN = mgcos(25°) = 667 N ΣFx ≠ 0 FNET = ma = mgsin(25°) - Ff FN FNET = 270 N = 311- Ff Ff Ff = µFN = µ(667N) = 41N mgcos(25°)  µ = .0614 mg mgsin(25°)  26

  5. A 75.0-kg box slides down a 25.0˚ ramp with an acceleration of 3.60 m/s2. ΣFx ≠ 0 b) What acceleration would a 175 kg box have on this ramp? Ff = µFN FN FNET = ma ma = mgsin(25°) - Ff Ff ma = mgsin(25°) – μmgcos(25˚) mgcos(25°) mass does not matter, the acceleration is the same!! mg mgsin(25°) 27

  6. Everyday Forces A 75.0-kg box is pushed with a 90.0N exerted downward at a 30˚ below the horizontal. If the coefficient of kinetic friction between box and the floor is 0.057, how long does it take to move the box 4.00m, starting from rest? A 75.0-kg box is pushed with a 90.0N exerted downward at a 30˚ below the horizontal. If the coefficient of kinetic friction between box and the floor is 0.057, how long does it take to move the box 4.00m, starting from rest? FN t = ? vi = 0 F  4.00 m Ffk Fg 28

  7. 90.0N  77.9 N 45.0 N FN  = 30˚ 735.8 N 1. Draw a free-body diagram to find the net force. 2. Convert all force vectors into x- and y- components. Ffk 29

  8. = 781 N 90.0N  77.9 N 45.0 N FN 735.8 N  = 30˚ 3. Is this an equilibrium or net force type of problem? Net force ! 4. The sum of all forces in the y-axis equals zero. Ffk FN = 45.0 + 735.8 N 5. Solve for the normal force. FN = 781 N 30

  9. = 781 N 90.0N  77.9 N 45.0 N 44.5 N FN 735.8 N  = 30˚ 6. Given the μk = 0.057, find the frictional force. μkFN = Ff (0.057) 781 N = 44.5 N Ff = 44.5 N Ffk 7. Given this is a net force problem, net force equals m times a. 77.9 N – 44.5 N = (75 kg) a a = .445 m/s2 31

  10. = 781 N 90.0N  77.9 N a = .445 m/s2 45.0 N 44.5 N FN 735.8 N  = 30˚ 8. Which constant acceleration equation has a, vi, x, and t? t = 4.24 s Ffk 32

  11. END

More Related