Day 31 Ppt A Fish On Two Hooks
- 1. A fish ontwo hooks
- 2. Rules of This Activity#1 All students must perform all of the work of the activity in their science journals, then when you are satisfied that it is correct, transfer it to the mini white board!#2 Hold your mini white board up when the assigned work is completed.#3 Do not erase your mini white board until it has been checked and approved.#4 Find another student in the room who is having difficulty and sit with them to help them through the correct process.
- 3. The Problem
- 4. Working The ProblemT2 = 6.52NT130*910 g
- 5. Working the ProblemStep #1 Draw a force diagram of the fish suspended from the two fishing lines. Remember to use the four stepsC of M Long term forcesContact forces Labeled axes
- 6. The Force Diagram yT26.52NT1 30* xFg8.92N
- 7. Working the ProblemStep #1 Draw a force diagram of the fish suspended from the two fishing lines. Step #2 Use SOH-CAH-TOA, to find the magnitude of the T2x , and T2y components. Show all of your work.
- 8. SOH – CAH - TOACOS θ = T2x/T2 COS 30* = T2x/T2 (6.52N)(0.87) = T2x 5.67N = T2x SIN θ = T2y/T2 (6.52N)(0.50) = T2y 3.26N = T2y
- 9. Working the ProblemStep #2 Use SOH-CAH-TOA, to find the magnitude of the T2x and T2y components. Show all of your work.Step #3 Draw a “clean” force diagram, breaking Fg, T1 and T2 into their x and y components. Again remember to use the four steps (C of M, long term forces, contact forces, labeled axes) in creating a force diagram!
- 10. The “Clean” Force Diagram yT1y T2y 3.26N x T2x 5.67N T1x Fg 8.92N
- 11. Working the ProblemStep #3 Draw a “clean” force diagram, breaking Fg, T1 and T2 into their x and y components. Step #4 State all of your “knowns” then using SOH-CAH-TOA once again, calculate the angle of the dock fisherman’s line. When your work is completed, consult with other students to decide if the final answer is thought to be correct before submitting it for approval.
- 12. SOH - CAH - TOAFg = 8.92Nand T1y + T2y = Fgand T2y = 3.26N and T2x = 5.67N and T1x = T2x and T1 = 8NSO… T1x = 5.67N and T1y = 8.92N – 3.26N = 5.66NIf you used the “x” direction COS θ = T1x/T1 COS θ = 5.67N/8N COS θ = .709 (45*)If you used the “y” direction SIN θ = T1y/T1 SIN θ = 5.66N/8N SIN θ = .708 (45*)