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Dyn Prob Slvg Notes

• 1. Solving Collinear Dynamics Problems CP Physics Elise Burns
• 2. The most important thing is... You must show supporting work! You must show supporting work! You must show supporting work! You must show supporting work! You must show supporting work!
• 3. We’ll do an example... Here’s a typical problem: A rope exerts a 280 N force while pulling an 80 kg water skier. A 110 N friction force opposes the skier’s motion. If the skier starts at rest, determine his acceleration. We’ll go through each step, page by page.
• 4. STEP 1: Draw a picture. A rope exerts a 280 N force while pulling an 80 kg water skier. A 110 N friction force opposes the skier’s motion. If the skier starts at rest, determine his acceleration.
• 5. STEP 2: Label all known information T = 280-N m = 80-kg f = 110-N vi = 0 A rope exerts a 280 N force while pulling an 80 kg water skier. A 110 N friction force opposes the skier’s motion. If the skier starts at rest, determine his acceleration.
• 6. STEP 3: Label the unknown information T = 280-N m = 80-kg f = 110-N vi = 0 a = ??? A rope exerts a 280 N force while pulling an 80 kg water skier. A 110 N friction force opposes the skier’s motion. If the skier starts at rest, determine his acceleration.
• 7. STEP 4: Draw a Motion Diagram. v v v a T = 280-N m = 80-kg f = 110-N vi = 0 a = ??? A rope exerts a 280 N force while pulling an 80 kg water skier. A 110 N friction force opposes the skier’s motion. If the skier starts at rest, determine his acceleration.
• 8. STEP 5: Draw a Free Body Diagram. (Don’t forget the axis!) v v v +y a +x T = 280-N N m = 80-kg T f f = 110-N w vi = 0 a = ??? A rope exerts a 280 N force while pulling an 80 kg water skier. A 110 N friction force opposes the skier’s motion. If the skier starts at rest, determine his acceleration.
• 9. STEP 6: WRITE OUT THE COMPONENTS OF NEWTON’S SECOND LAW. v v v +y a +x N T = 280-N m = 80-kg f T f = 110-N Σ Fx = max Σ Fy = may w vi = 0 a = ??? A rope exerts a 280 N force while pulling an 80 kg water skier. A 110 N friction force opposes the skier’s motion. If the skier starts at rest, determine his acceleration.
• 10. STEP 7: Substitute in the specific forces for “Net Force”. v v v +y a +x N T = 280-N m = 80-kg f T f = 110-N ΣFx = max ΣFy = may w vi = 0 T - f = max N - w = may a = ??? A rope exerts a 280 N force while pulling an 80 kg water skier. A 110 N friction force opposes the skier’s motion. If the skier starts at rest, determine his acceleration.
• 11. STEP 8: Plug in all known information into the equation. v v v +y a +x N T = 280-N m = 80-kg f T f = 110-N ΣFx = max w vi = 0 T - f = max Σ Fy= may a = ??? 280 - 110 = (80)ax N - w = may A rope exerts a 280 N force while pulling an 80 kg water skier. A 110 N friction force opposes the skier’s motion. If the skier starts at rest, determine his acceleration.
• 12. STEP 9: If necessary, solve for weight... This is not needed here. v v v +y a +x N T = 280-N m = 80-kg f T f = 110-N ΣFx = max w vi = 0 T - f = max Σ Fy= may a = ??? 280 - 110 = (80)ax N - w = may A rope exerts a 280 N force while pulling an 80 kg water skier. A 110 N friction force opposes the skier’s motion. If the skier starts at rest, determine his acceleration.
• 13. STEP 10: Solve the problem. Put units on your answer! T = 280-N v v v m = 80-kg a +y f = 110-N N +x vi = 0 T a = ??? Σ Fx = max f T - f = max w 280 - 110 = (80)ax ΣFy = may 170 = (80)ax N - w = may 2.1 m/s2 = ax A rope exerts a 280 N force while pulling an 80 kg water skier. A 110 N friction force opposes the skier’s motion. If the skier starts at rest, determine his acceleration.
• 14. Question: How did I know whether to use the x or y axis equation? They both have acceleration in the equation. There are several hints: 1. The forces given (T and f) are both in the x-axis equation. 2. The object is not moving vertically, so that ay is equal to zero anyway. 3. You aren’t given Normal force, so you can’t solve the y-axis equation.
• 15. Question: How did I know whether the person or the boat was the system? There are several hints: 1. The mass of the person is given and the mass of the boat was not. 2. The question asks for the acceleration of the person. 3. The friction force opposes the person’s motion.
• 16. Question: Could I have drawn a sketch where the person moved from right to left, instead of left to right? YES! You would get the same answer. The same exact answer! (The sign may be different if you don’t also change your +x axis to go in the direction of motion.)
• 17. Question: Could I have used an axis that showed the +x pointing opposite the direction of motion? YES! You would get the same answer. The sign would be opposite, but as long as you show your work, that’s not a problem.
• 18. Question: Do I really have to write every single step out? It’s so annoying. Why are you being so picky about this? YES! You absolutely must show all of your work. I have 3 reasons. 1. Science is about solving problems. Solving problems is a process. I am checking to make sure that you learn the process. 2. I cannot read your mind. The only way you can communicate your understanding of this material is by putting pen to paper and showing me. 3. If you want partial credit when you err, you must show all of your steps so I can give you points for what you did correctly.
• 19. Want to see another solved problem? A 60-kg person falls from a storage building and lands stiff-legged in some mud. The person stops with an acceleration of 490 m/s2, after sinking into the mud. Determine the average force of the mud in stopping the person.
• 20. A 60-kg person falls from a storage building and lands stiff-legged in some mud. The person stops with an acceleration of 490 m/s2, after sinking into the mud. Determine the average force of the mud in stopping the person. m = 60-kg V N +y a = 490 m/s2 V a +x vf = 0 V w Nmud = ?? ΣFx = max Σ Fy = may N - w = may N - mg = may Notice that in this problem, you must solve for weight, N - (60)(10) = (60)(490) using w = mg. N - 600 = 29,400 N = 30,000 N
• 21. Want to see one more solved problem? 1000-kg out of control car initially moving at 20 m/s enters a horizontal escape surface made of loose gravel. The force of friction is 2500-N. Determine the acceleration of the car while coming to a stop.
• 22. 1000-kg out of control car initially moving at 20 m/s enters a horizontal escape surface made of loose gravel. The force of friction is 2500-N. Determine the acceleration of the car while coming to a stop. N +y m = 1000-kg +x f = 2500-N f vf = 0 w v v v vi = 20 m/s a Σ Fx = max Σ Fy= may a = ??? - f = max N - w = may - 2500 = (1000)ax -2.5 m/s2 = ax
• 23. Referring to this same situation, how far does the car travel while stopping? How would you approach this? +y N m = 1000-kg +x f = 2500-N f vf = 0 w v v v vi = 20 m/s a a = ??? Σ Fx = max Σ Fy= may - f = max N - w = may - 2500 = (1000)ax -2.5 m/s2 = ax
• 24. Use kinematics to find out how far the car travels while stopping. You’ve got initial and final velocities given, plus you just solved for acceleration. All that’s left to do is pick the equation. +y N m = 1000-kg +x f = 2500-N f vf = 0 w v v v vi = 20 m/s a -2.5 m/s2 = ax vf2 = vi2 + 2ad You are not 02 = (20)2 + 2(-2.5)d allowed to 0 = 400 +(-5)d -400 =(-5)d forget 80m = d kinematics!
• 25. Now it’s your turn... Solve the 6 problems on the worksheet. SHOW ALL STEPS! Don’t skip any. Use the list on the notes page to assist you!
• 26. Other Physics News... Complete Dynamics Worksheet #1 - 6 Inspect Lab Journals... Post-Lab Quiz on this week’s lab will be on Monday! Test on Tuesday on Newton’s First and Second Laws. You may do the Newton’s Laws Internet Activity for Extra Credit. It must be handed in by Tuesday. Project Analysis should be complete up through Step J.
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