Calculating Tow Rope Force for Water-Skiing

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To calculate the tow rope force for water-skiing, the force can be resolved into forward and sideward components based on the angle created by the rope. Given the rope length of 8m and the skier's lateral movement of 2m at an angle of 14.5 degrees, trigonometric functions can be used to find these components. The forward component can be calculated using cosine, while the sideward component uses sine. Since the actual tow rope force is unknown, the results will be expressed as fractions of this unknown force. Drawing a diagram and applying basic trigonometry will aid in visualizing and solving the problem.
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I am given the information: A water-skier is pulled behind a motorboat with a rope 8m long. As she is approaching a ramp, she pulls herself 2m to the side of the path of the boat. It creates an angle of 14.5 degrees.

I am asked:
1.) how much of the tow rope force is in the forward direction?
2.) how much of the tow rope force is sidewards?

How do i find that info when i am not given any additional infor? :confused:
 
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1) The tow rope force is parallell to the rope
a)Find the components of that force in the forwards and sideways directions.
b)Find the fractional amount of the total rope force each component force equals.
2). Do not double post for any reason
3) welcome to PF!
 
Since you are not given the actual tow rope force, you can only give your answer as a fraction of the (unknown) tow rope force. The tow rope force pulls in the direction of the rope (of course!). Since the rope is now at an angle, find the components of that force in the forward and sideways directions. Draw a picture and use a little trig.
 
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