course Phy 201
I did the preceding introductory problems on vectors and had no trouble whatsoever, and I watched the videos on the disk. But none of that seems to relate to this seed question. If you will point me in the right direction for the second question I will redo all of it and resubmit.
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?
d=sqrt((10-5)^2 + (17-9)^2)=9.43cm
9.43cm-7.5cm = 1.93cm
1.93cm*.7N=1.35N of tension
Good start.
What is the vector from the first point to the second?
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 is the magnitude of this vector?
9.43cm
What vector do you get when you divide this vector by its magnitude?
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?
answer/question/discussion:
The tension is about 1.35 N. Multiplying the vector <.53, .83> by 1.35 N we obtain the new vector <.53, .83> * 1.35 N = <.71 N, 1.13 N>, approximately.
This represents a force vector with x and y components .71 N and 1.1 N, respectively.
The magnitude of this vector is the original 1.35 N, and it is directed at angle arcTan (1.13 N / (.71 N) ) = 58 degrees, approximately.
What are the x and y coordinates of the new vector?
answer/question/discussion:
See my notes and be sure you understand. If not please submit questions.