Calculating Force Direction of a Water-Skier Tow Rope

AI Thread Summary
To calculate the force direction of the tow rope for a water-skier, the tension in the rope can be resolved into vertical and horizontal components using trigonometric functions. The forward force component is determined by multiplying the tension (T) by the cosine of the angle (14.5 degrees), while the sideward force component is found by multiplying T by the sine of the angle. The length of the rope does not provide direct information about the tension, which remains unspecified. The discussion emphasizes the need to understand the relationship between the angle and the forces involved in the skier's motion. Further clarification on the tension value is necessary to complete the calculations.
DOMINGO79
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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?
 
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DOMINGO79 said:
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?

You can resolve that into vertical and horizontal components

The one perpendicular to motion wouild be Tension in the rope * sin 14.5

the one parallel ( question 1) Would be T cos 14.5

I am not sure about that T actually is . The length of the rope doesn't help.

I hope that give you a start

Sid
 

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