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brinethery
 one year ago
Best ResponseYou've already chosen the best response.0I struggled with this type of problem when I was in physics last quarter :(

Outkast3r09
 one year ago
Best ResponseYou've already chosen the best response.1did you find the acceleration... since they're roped together you assume that the acceleration of each is the same

Outkast3r09
 one year ago
Best ResponseYou've already chosen the best response.1if ou have acceleration you'll have two unknowns that can be related to each other

brinethery
 one year ago
Best ResponseYou've already chosen the best response.0Yes, I tried finding it. I got a=4.70 m/s^2

brinethery
 one year ago
Best ResponseYou've already chosen the best response.0So for A, I did sum of the forces in x (xaxis being parallel to the inclined plane) and then for B, I did sum of the forces in y. I made sure to make the direction of motion positive. For A, there's a +T in the equation and B, there's a T in the equation. So I added both equations so that T would cancel out and I could solve for a.

Outkast3r09
 one year ago
Best ResponseYou've already chosen the best response.1dw:1359083523070:dw

Outkast3r09
 one year ago
Best ResponseYou've already chosen the best response.1dw:1359083623616:dw

Outkast3r09
 one year ago
Best ResponseYou've already chosen the best response.1dw:1359083742313:dw

brinethery
 one year ago
Best ResponseYou've already chosen the best response.0I just made my xaxis parallel to the plane. So my equation was m1(a) = (mew_k)(m1)(g)(cos(40)) +T And then for block B, I did m2(a) = (m2)(g)  T And then I added both equations. a(m1+m2) = (mew_k)(m1)(g)(cos(40)) + (m2)(g) a = (mew_k)(m1)(g)(cos(40)) + (m2)(g)/(m1+m2)

Outkast3r09
 one year ago
Best ResponseYou've already chosen the best response.1\[sin(40)=\frac{W_x}{W}\] \[W_x=mgsin(40)\] \[W_y=mgcos(40)\] \[\sum F_y=NW\] \[N=W\] \[\sum F_{x1}=m_1a_{1x}=TW_xf_k=Tmgsin(40)(\mu_s)mgcos(40)\] \

Outkast3r09
 one year ago
Best ResponseYou've already chosen the best response.1where is your weight x component?

brinethery
 one year ago
Best ResponseYou've already chosen the best response.0ha, maybe that's the problem!

brinethery
 one year ago
Best ResponseYou've already chosen the best response.0that must've been where I goofed :p. Just silly mistakes I guess.

Outkast3r09
 one year ago
Best ResponseYou've already chosen the best response.1\[m_1a_{1x}=Tm_1gsin(40)\mu_s(m_1gcos(40))\]

brinethery
 one year ago
Best ResponseYou've already chosen the best response.0I didn't bother to label the acceleration since it will be the same. Would that be correct?

Outkast3r09
 one year ago
Best ResponseYou've already chosen the best response.1\[m_2a_{2y}=Tm_2g \] \[a_{2y}=a_{1x}\]

Outkast3r09
 one year ago
Best ResponseYou've already chosen the best response.1is the pulley frictionless?

Outkast3r09
 one year ago
Best ResponseYou've already chosen the best response.1\[T=m_2a_{1x}m_2g=m_2(a_{1x}+g)\] \[T=m_1a_{1x}+m_1gsin(40)+\mu_km_1gcos(40)\]

brinethery
 one year ago
Best ResponseYou've already chosen the best response.0Now that I added in the xcomponent of weight for Block A, my acceleration is 1.754 m/s^2

Outkast3r09
 one year ago
Best ResponseYou've already chosen the best response.1\[m_2a_{1x}m_2g=m_1a_{1x}+m_1gsin(40)+\mu_km_1gcos(40)\]

Outkast3r09
 one year ago
Best ResponseYou've already chosen the best response.1\[m_2a_{1x}m_1a_{1x}=m_2g+m_1gsin(40)+\mu_km_1gcos(40)\] \[a_{1x}(m_2m_1)=m_2g+m_1gsin(40)+\mu_km_1gcos(40)\] \[a_{1x}=

Outkast3r09
 one year ago
Best ResponseYou've already chosen the best response.1i think i dropped a sign somewhere let me get a paper and pen

brinethery
 one year ago
Best ResponseYou've already chosen the best response.0What was your acceleration? m1(a) = (mew_k)(m1)(g)(cos(40)) (m1)gsin(40) +T m2(a) = m2(a) = (m2)(g)  T a(m1+m2) = (mew_k)(m1)(g)(cos(40)) (m1)gsin(40) + (m2)(g) a = ((mew_k)(m1)(g)(cos(40)) (m1)gsin(40) + (m2)(g))/(m1+m2)

brinethery
 one year ago
Best ResponseYou've already chosen the best response.0I have a pretty good idea on the setup of the equations, so you can just tell me if I completely missed something. And also, what your acceleration is.

Outkast3r09
 one year ago
Best ResponseYou've already chosen the best response.1how'd you define your axis for the y part?

brinethery
 one year ago
Best ResponseYou've already chosen the best response.0my axis for the ypart is regular. Vertical for y, horizontal for x.

brinethery
 one year ago
Best ResponseYou've already chosen the best response.0I mean for block B, that is.

Outkast3r09
 one year ago
Best ResponseYou've already chosen the best response.1isn't downwards negative for the mg?

brinethery
 one year ago
Best ResponseYou've already chosen the best response.0When I took physics, my instructor told us to make the positive direction for each object the same as the direction of motion.

brinethery
 one year ago
Best ResponseYou've already chosen the best response.0so for block A, up and to the right would be positive. And for block B, down would be positive.

brinethery
 one year ago
Best ResponseYou've already chosen the best response.0so you're accounting for the system as a whole.

Outkast3r09
 one year ago
Best ResponseYou've already chosen the best response.1that's for B it could be rounding error though

Outkast3r09
 one year ago
Best ResponseYou've already chosen the best response.1so for mine \[a_{2x}=1.67\] \[a_{1x}=a_{2x}=(1.67)=1.67\]

Outkast3r09
 one year ago
Best ResponseYou've already chosen the best response.1if you didn't round use your answer... min was rounded.... a lot . my graphing calc is out in the car and it's freezing lol. Funny thing, i just got back from physics 2

brinethery
 one year ago
Best ResponseYou've already chosen the best response.0We're working out of Hibbeler's book for dynamics and the chapter on work is confusing. I mean, I was already confused with these types of problems in physics I but now I just feel even more lost. This book doesn't explain things very well. You really have to read and reread stuff in there.

Outkast3r09
 one year ago
Best ResponseYou've already chosen the best response.1so anyways now you have \[S=\frac{1}{2}(1.754)t^2+vt+S_0\] and \[V=at+v_0=1.754t+0\]

Outkast3r09
 one year ago
Best ResponseYou've already chosen the best response.1so you can just sub in =]. Never heard of hibbeler. however this book from about 10 years ago im using is confusing. It uses greek lettering for everything and my teacher uses easy things like \[R=\frac{l}{a}\]

brinethery
 one year ago
Best ResponseYou've already chosen the best response.0my friend sent me her stuff, so I'll try it out and tell you if it worked.

brinethery
 one year ago
Best ResponseYou've already chosen the best response.0Thank you for taking the time to help me.

Outkast3r09
 one year ago
Best ResponseYou've already chosen the best response.1no problem if you need anymore help let me know
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