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Old Oct 22nd 2018, 04:37 AM   #1
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Join Date: Jun 2018
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Question Questions about forces and tensions

I have difficulty with the following question:

An angler of mass 80 kg standing on a riverbank is slowly reeling in a 15 kg fish at the end of a line. The tip of the fishing rod is 8m above the water level, and there are 17 m of line between the tip of the rod and the hook. As the fish come in at constant speed there is a horizontal resistive force from the water or 105 N.

Calculate the tension in the line.


The answer to the above is, according to the answer section provided in the textbook, 119 N. But I cannot figure out how I can get this number.
Obviously, this number of 119 includes the resistive force of 105 N, but how can we get the additional 14 N?

I would much appreciate it if somone can help me with this.

Also the question goes on to say:

Draw diagrams to show the forces on

a) the fish
b) the angler and rod (considered as a single object)
Find the magnitude of each force. The mass of the rod can be neglected.


The text says that the answer to a) above is:

Weight 150 N, tension 119 N, resistance 105 N, buoyancy 94 N.

But I have no idea whatsoever how to calculate "buoyancy" as this is the first time for me to come across it in this texbook let alone it does not say anywhere at all how to work it out.


So I would also appreciate it if somone can tell me how to work out "buoyancy".


Further, the text says that the answer to b) above is:

Weight 800 N, tension 119 N, normal contact force 856 N, friction 105 N.


But I don't understand why "normal contact force" is 856 N, not 800 N, because the weight of the angler is 80 kg, not 85.6 kg. So where does this additional 56 N come from?


I have attached my diagrams so please let me know if there is anything wrong with it.


Thank you.
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Masaru is offline   Reply With Quote
Old Oct 22nd 2018, 01:00 PM   #2
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First draw a diagram. "The tip of the fishing rod is 8m above the water level, and there are 17 m of line between the tip of the rod and the hook" so we have a right triangle with hypotenuse of length 17 m, one leg of length 8 m, and the other leg of length $\displaystyle \sqrt{17^2- 8^2}= 15$ m.

The reason for the "additional 14 N" is that the fish is being reeled in horizontally but the line is the hypotenuse of the triangle. Find the total force so that the horizontal component is 105 N. The ratio of hypotenuse to horizontal side of the right triangle is 17/15 so the tension in the line is (17/15)(105)= 119.
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Old Oct 22nd 2018, 07:01 PM   #3
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Thank you for your help.

So the upward component X would be (8/15)*105 = 56.

Let B be Buoyancy, then X + B = 150


56 + B = 150


So B = 94N


So Normal contact force on the man & rod would be 800 + 56 = 856 N
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