Phy 231
Your 'cq_1_16.1' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** **
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:
Answer: 1.4N.
• What is the vector from the first point to the second?
answer/question/discussion:
Answer: v=vx+vy=5cos(theta)+8sin(theta), theta=tan^-1(8/5)=58deg.
5 cm * cos(theta) is the x component of a vector of magnitude 5 and angle theta. There is no vector of magnitude 5 in this problem.
Nor is there a vector of magnitude 8 and angle theta.
So 5cos(theta)+8sin(theta) is not a useful calculation.
The length vector v has components vx = 5 and vy = 8, so its angle is tan^-1(8/5) = 58 deg.
The magnitude of the length vector is about 9.4; you don't calculate this but your 1.4 N force is consistent with this length so I assume you have already obtained the length correctly. You document this calculation correctly below.
• What is the magnitude of this vector?
answer/question/discussion:
Answer: magnitude = sqrt(5^2 + 8^2) = sqrt(89) = 9.4cm.
• 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:
Answer: vx=(5cos(58)) / 9.4 = .28. vy=(8sin(58)) / 9.4 = .72, v=vx+vy=1
The components are 5 and 8, not 5 cos(58 deg) and 8 sin(58 deg).
A vector with components vx = .28 and vy = .72 has magnitude sqrt(.28^2 + .72^2) =.76 or so; this vector would not have magnitude 1.
vx + vy is not a calculation you would ever make. There is no point in adding the x and y components of a vector.
• What vector do you get when you multiply this new vector by the tension?
answer/question/discussion:
Answer: v=1, v*tension = 1.4N.
The angle is right.
A vector must also have a specified direction.
• What are the x and y coordinates of the new vector?
answer/question/discussion:
vx=.39, vy=1, angle=tan^-1(1/.39)=68.7deg. (7.5, 13).
Correct this after correcting your components.
** **
20 minutes
** **
When they ask: what is the vector, what they looking for? v=vx+vy=5cos(theta)+8sin(theta) or v=vx+vy=2.6+6.8, or something else? Do they really mean what are the x and y components of the vector?
You have some errors, though you are generally on the right track here.
&#Please see my notes and submit a copy of this document with revisions and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end). &#