capa 2 - Phys 1120, CAPA #2 solutions. NOTE: Here, I will...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Phys 1120, CAPA #2 solutions. NOTE: Here, I will provide "generic" solution. Since the numbers and some details vary a little, your problems may look a wee bit different, but this solution should tell you enough to solve your problem) The answers to your particular questions are available through CAPA (on the login page, click on "view previous set". Just be sure to fill in the set number first, before clicking! COMMENT: You should ALWAYS try to solve problems using symbols only, even ( especially ) on "number crunching " problems. There are lots of reasons - it's easier to check your work, it's easier to see if the units are correct, it's easier for a TA to see if you've made a mistake, it's easier to see if various limits make sense, it's easier to see the PHYSICS when you have a formula instead of plugging numbers in right away. It's how any professional, in any field, will solve real life problems, even numerical ones. Get in the habit of this! We will ask exam questions where we won't give any numbers, you will have to work on a symbolic level. 1) Particle 1 has a given mass and acceleration, so Newton-II says F(on 1) = m(1)*a(1) is given to us. But Newton-III says F(on 2) is the same magnitude! N-II says for object 2, F(on 2) = m(2)*a(2). The problem asks for m(2): m(2) = F(on 2) / a(2) (I'm considering magnitude only) = F(on 1) / a(2) = m(1)*a(1)/a(2). All the numbers on the right are given, just plug in. 2) Now use F = k Q1 Q2 / R^2. The problem says Q1=Q2, call it Q, and we know F from the previous problem. R was given. (Don't forget to convert cm to m first!) So Q^2 = F*R^2/ k, or Q = Sqrt[ F*R^2/k] = Sqrt[ m(1)*a(1) * R^2 / k] Plug in the numbers now. 3) A force diagram is needed here. Let me describe it, I have three force arrow on the charge: F = qE straight to the LEFT (from electricity), F = mg straight DOWN (from gravity), and tension T along the string direction. The question asks for how much charge is needed). What I'm thinking is that the object is in equilbrium. That means the NET FORCE on it is zero. So if I write down Newton's law (F(net)=ma), with a=0 (that's what equilbrium implies), then I will have TWO equations to help me, the x- and y- components. There are TWO unknowns (the Tension ithe rope, and the charge on the ball.) So 2 equations, 2 unknowns, we should be ok! Components of Fnet=ma: Do x- and y- separately. F_netx = -qE + Tsin(theta) = 0 (Do you see why I use sin(theta) for the x component! Look carefully at the picture in CAPA, and which angle is named theta.)
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
F_nety = -mg + Tcos(theta) = 0 (I've chosen up =+, and right = +) Look again at this equation, and the previous one, and make sure YOU SEE why the signs are what they are. So the first one gives Tsin(theta) = qE, and the second gives Tcos(theta) = mg.
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 04/28/2008 for the course PHYS 1120 taught by Professor Rogers during the Fall '08 term at Colorado.

Page1 / 4

capa 2 - Phys 1120, CAPA #2 solutions. NOTE: Here, I will...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online