#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.

An early morning fishing trip has a 910 gram fish hooked by two fishermen

both of whom are using an 8N test line.

One of the fishermen is on the dock. The other fisherman is in a boat and has 6.52N on his line at

an angle of 30*. Neither is able to move the fish closer to them and neither fishing line snaps.

At what angle is the dock fisherman’s line if his fishing line is currently just

under the breaking point?

910 g

T1

T2 = 6.52N

30*

Step #1 Draw a force diagram of the fish suspended from the two fishing lines. Remember to use the four steps

C of M

Long term forces

Contact forces

Labeled axes

y

T2 6.52N T1

30* x

Fg 8.92N

Step #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.

COS 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

Step #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 T2into 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!

y

T1y

T2y 3.26N x

T2x 5.67N T1x

Fg 8.92N

Step #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.

Fg = 8.92N and T1y + T2y = Fg and T2y = 3.26N and T2x = 5.67N and T1x = T2x and T1 = 8N

SO… T1x = 5.67N and T1y = 8.92N – 3.26N = 5.66N

If 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*)