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A rubber band has no tension until it reaches a length of 7.5 cm. Beyond that length its tension increases by .7 Newtons for every additional centimeter of length.

What will be its tension if its endpoints are at the points (5 cm, 9 cm) and (10 cm, 17 cm) as measured on an x-y coordinate system?

We are given a way to find the tension of the rubber band from its length.

The length runs from one of the given points to the other. So wefind the distance between the two given points:

• A right triangle can be constructed from the first point to the second, with a horizontal leg equal to the 'run' between the points and a vertical leg equal to the 'rise'.
• The legs of the triangle are therefore 10 cm - 5 cm = 5 cm, and 17 cm - 9 cm = 8 cm.
• The hypotenuse of this triangle is the distance between the points, which is therefore the length of the rubber band.
• The hypotenuse is easily found from the legs using the Pythagorean Theorem.
• The length of the rubber band is therefore sqrt( (5 cm)^2 + (8 cm)^2 ) = sqrt( 89 cm^2) ) = 9.4 cm, approx.

Having found the length we use the given information to find the tension:

• Tension increases by .7 N for every cm beyond the 7.5 cm length.
• The rubber band is about 9.4 cm long, which is 9.4 cm - 7.5 cm = 1.9 cm beyond the 7.5 cm length.
• The tension is therefore .7 N * 1.9 cm = 1.3 N, approx.

What is the vector from the first point to the second?

What is the magnitude of this vector?

A vector has magnitude and direction. The vector from (5 cm, 9 cm) to (10 cm, 17 cm) is <5 cm, 8 cm>, indicating a vector with components 5 cm in the x direction and 8 cm in the y direction.

The magnitude of this vector is sqrt( (5 cm)^2 + (8 cm)^2 ) = sqrt( 89 cm^2) = 9.4 cm, approx..

Its angle is arctan(8 cm / (5 cm) ) = arctan(1.6) = 58 degrees, approx..

What vector do you get when you divide this vector by its magnitude? (Specify the x and y components of the resulting vector).

If we divide the vector <5 cm, 8 cm> by its magnitude we get

<5 cm, 8 cm> / sqrt(89 cm^2) =

< 5 cm / sqrt(89 cm^2), 8 cm / sqrt(89 cm^2) > =

<.53, .83>, approximately.

That is, we get a vector with x component .53 and y component .83. Note that both these components are unitless, since dividing cm by sqrt(cm^2) yields cm/cm so that the units cancel out.

This vector is 1 unit in length (therefore called a unit vector), and directed at the same angle as the original vector.

What vector do you get when you multiply this new vector by the tension?

What are the x and y coordinates of the new vector?

The tension is about 1.3 N. Multiplying the vector <.53, .83> by 1.3 N we obtain the new vector

<.53, .83> * 1.3 N = <.7 N, 1.1 N>, approximately.

This represents a force vector with x and y components .7 N and 1.1 N, respectively. The magnitude of this vector is the original 1.3 N, and it is directed at angle arcTan (1.1 N / (.7 N) ) = 58 degrees, approximately.