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10 2012 ppt force review

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  • 1. Know Definitions of Key Terms & Symbols
  • 2. Focus during the entire Power Point activity. Solidify your studying skills during this class period. Perform your work in your science journal so you have created a study guide for the test. Call me over if you are having difficulty getting started. If your answer is confirmed as correct, become a student/teacher and help someone in class who does not understand the method used to solve the problem.
  • 3. Batfink, who has a mass of 50 kg is placed in a 25 kg stationary barrel. What is the Fg on Batfink and the barrel?
  • 4. SOLUTION: Force of gravity on Batfink and the barrel. mbf = 50 kg mb = 25 kg g = -9.8 m/s2 Fg Fg= -735 N
  • 5. Hugo Ago-go pushes the barrel with Batfink in it towards the end of the cliff with a 85 N force over a distance of 12 m before the barrel leaves the cliff. The force of friction is 2.75 N. Draw a force diagram of the situation.
  • 6. yy Fs = 735 N Ff = -2.75 N x Fa = 85 N Fg = -735 N
  • 7. Calculate the acceleration of the barrel in the x axis.
  • 8. SOLUTION: Acceleration of Batfink and the barrel. mbf = 5o kg mb = 25 kg g = -9.8 m/s2 Fa = 85 N Ff = 2.75 N xi = 0 m x = 12 m a a = 1.1 m/s2
  • 9. What is the Vf of the barrel just before it falls off the cliff?
  • 10. SOLUTION: Final velocity of Batfink and the barrel while still on the cliff. vi = 0 m/s a = 1.1 m/s2 xi = 0 m xf = 12 m ti = 0 s vf tf v = 5.14 m/s
  • 11. Batfink and the barrel are raised at 1.25 m/s2. What is the force of support acting on Batfink and the Barrel?
  • 12. SOLUTION: Force of support on Batfink and the barrel. mbf = 75 kg mb = 25 kg g = -9.8 m/s2 ay = 1.25 m/s2 Fs Fs= 828.75 N
  • 13. Suddenly, Batfink and the barrel are lowered at .75 2 m/s . What is the force of support acting on Batfink and the Barrel?
  • 14. SOLUTION: Force of support on Batfink and the barrel. mbf = 75 kg mb = 25 kg g = -9.8 m/s2 a = -.75 m/s2 Fs Fs= 678.75 N
  • 15. The Incredible Hulk is hanging motionless off the ground by chains attached separately to his wrists from two different walls. The Hulk has a mass of 355 kg. The chain on his right wrist (T1) forms an angle of 26˚ relative to the floor, and the chain from his left wrist (T 2) forms an angle of 32˚ relative to the floor. T2 has 2500 N acting on it. Draw a force diagram of the situation.
  • 16. y T1 T2 = 2500 N 26° Fg -3479 N 32°
  • 17. I dare you to determine the tension in T2X.
  • 18. SOLUTION: Tension in T2x. T2 = 2500 N θ = 32° m = 355 kg g = -9.8 m/s2 Fg = -3479 N T2x T2x = 2120 N
  • 19. Now determine the tension in T1!
  • 20. SOLUTION: Tension in chain #1. T2 = 2500 N θ = 26° m = 355 kg g = -9.8 m/s2 Fg = -3479 N T2x = 2120 N T1 T1 = 2358.2 N
  • 21. Determine the force, velocity and displacement for a 2.75 kg cart starting 6.2 meters left of the reference point, while traveling at 1.47 m/s for 4.63 seconds.
  • 22. SOLUTION: Rate of acceleration, force, velocity, and final displacement. a = 2 m/s2 F= m = 2.75 kg Vf = t = 4.63 s Xf = Vi = 1.47 m/s Xi = -6.2 m F = 5.5 N Vf = 10.73 m/s X = 22.04 m
  • 23. Determine the force, velocity and displacement for a 2.75 kg cart starting 6.2 meters left of the reference point, while traveling at 1.47 m/s for 4.63 seconds.
  • 24. SOLUTION: Rate of acceleration, force, velocity, and final displacement. Xi = -6.2 m F= Vi = 1.47 m/s a = m = 2.75 kg Vf = t = 4.63 s Xf = a = .33 m/s2 F = .908 N Vf = 2.99 m/s
  • 25. Create a motion map, properly labeled x-t, v-t, and a-t graphs, and a force diagram based on the actual F-m graph assuming a force of friction of 0.75 N. What would the applied force have to be to attain this rate of acceleration?
  • 26. SOLUTION: 4.14 6.2 max = Fa + Ff (2.75)(.33) = Fa+ -0.75 N Fa = 1.66 N 2.99 Fs = 26.95 N 1.47 Ff = -0.75 N Fa = 1.66 N .33 Fg = -26.95 N