Question

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A water-skier is pulled behind a boat by a rope that is at an angle of 13° and has a tension of 490 N. The water-skier has a mass of 49 kg.

What are the magnitudes of the x- and y- components of the tension?

What is the normal force acting on the skier?

Answer 1

Answer:

normal force= 480.2 x component=477.4  y component= 110.4

Explanation:

I'm sorry I just did this today in class but i think normal force has to be equal to mg or the mass * gravity of the person and that would be 480.2 newtons. I multiplied 9.8 (gravity) by the mass of 49g

I think for the x and y components you can make a triangle with the angled string and then use sohcahtoa to solve for the tension. I used cos(13) = x/490 to solve for x component and then I used the pythagoran theroem to get the remaining side which would be : a^2 + 477.4^2 = 490^2

hope this can help and that it is correct. good luck

Answer 2

The magnitudes of the x- and y- components of the tension is 477.44N and 110.23N

The normal force acting on the skier is 480.2N

The magnitude of the x- and y- components of the tension is expressed as:

• x = Tcos 13°
• y = T sin  13°

Given the following

Tension T = 490N

x = 490 cos13° = 477.44N

y = 490sin13° =  110.23N

Hence the magnitudes of the x- and y- components of the tension is 477.44N and 110.23N

The normal force acting on the skier is equal to the weight.

N = W = mg

N = W = 49 * 9.8

N = W = 480.2N

Hence the normal force acting on the skier is 480.2N

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