#$&*
course Phy 121
12:45 pm July 2
A rubber band has no tension until it reaches a length of 7.5 cm. Beyond that length its tension increases by .7 Newtons for every additional centimeter of length. •What will be its tension if its endpoints are at the points (5 cm, 9 cm) and (10 cm, 17 cm) as measured on an x-y coordinate system?
answer/question/discussion:
Draw diagram…legs of triangle are 5 cm and 8 cm. (10-5=5 and 17-9=8)
Then take the square root of (5^2 + 8^2)= square root of 80 cm^2= 9.4
9.4 cm - 7.5 cm = 1.9 cm
tension= .7 N * 1.9 cm = 1.3 N
#$&*
• What is the vector from the first point to the second?
answer/question/discussion:
The vector (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
#$&*
• What is the magnitude of this vector?
answer/question/discussion:
The magnitude is sqrt(5^2 + 8^2 ) = sqrt( 89^2) = 9.4 cm
(8 /5 ) = (1.6) = 58 degrees
#$&*
• What vector do you get when you divide this vector by its magnitude? (Specify the x and y components of the resulting vector).
answer/question/discussion:
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>
#$&*
What are the x and y components of the new vector?
answer/question/discussion:
The tension is about 1.3 N. Multiplying the vector <.53, .83> by 1.3 N we obtain the new vector
<.53, .83> * 1.3 N = <.7 N, 1.1 N>
(1.1 N / (.7 N) ) = 58 degrees
#$&*
This new vector is called the tension vector. It is a force vector which represents the tension. A force vector can be specified by its components, or equivalently by its magnitude and direction.
________________________________________
Your instructor is trying to gauge the typical time spent by students on these questions. Please answer the following question as accurately as you can, understanding that your answer will be used only for the stated purpose and has no bearing on your grades:
• Approximately how much time did you spend on this question?
30 min
"
&#This looks very good. Let me know if you have any questions. &#