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anonymous
 3 years ago
Work problem. Please see attached. I'm having trouble with part (b)
anonymous
 3 years ago
Work problem. Please see attached. I'm having trouble with part (b)

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0did you find the acceleration... since they're roped together you assume that the acceleration of each is the same

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0if ou have acceleration you'll have two unknowns that can be related to each other

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

anonymous
 3 years 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1359083523070:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1359083623616:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1359083742313:dw

anonymous
 3 years 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)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[sin(40)=\frac{W_x}{W}\] \[W_x=mgsin(40)\] \[W_y=mgcos(40)\] \[\sum F_y=NW\] \[N=W\] \[\sum F_{x1}=m_1a_{1x}=TW_xf_k=Tmgsin(40)(\mu_s)mgcos(40)\] \

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0where is your weight x component?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ha, maybe that's the problem!

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[m_1a_{1x}=Tm_1gsin(40)\mu_s(m_1gcos(40))\]

anonymous
 3 years 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?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[m_2a_{2y}=Tm_2g \] \[a_{2y}=a_{1x}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0is the pulley frictionless?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[T=m_2a_{1x}m_2g=m_2(a_{1x}+g)\] \[T=m_1a_{1x}+m_1gsin(40)+\mu_km_1gcos(40)\]

anonymous
 3 years 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[m_2a_{1x}m_2g=m_1a_{1x}+m_1gsin(40)+\mu_km_1gcos(40)\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[m_2a_{1x}m_1a_{1x}=m_2g+m_1gsin(40)+\mu_km_1gcos(40)\] \[a_{1x}(m_2m_1)=m_2g+m_1gsin(40)+\mu_km_1gcos(40)\] \[a_{1x}=

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i think i dropped a sign somewhere let me get a paper and pen

anonymous
 3 years 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)

anonymous
 3 years 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0how'd you define your axis for the y part?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0my axis for the ypart is regular. Vertical for y, horizontal for x.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I mean for block B, that is.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0isn't downwards negative for the mg?

anonymous
 3 years 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.

anonymous
 3 years 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so you're accounting for the system as a whole.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0that's for B it could be rounding error though

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so for mine \[a_{2x}=1.67\] \[a_{1x}=a_{2x}=(1.67)=1.67\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0if 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

anonymous
 3 years 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.

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so 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 \[R=\frac{l}{a}\]

anonymous
 3 years 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Thank you for taking the time to help me.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0no problem if you need anymore help let me know
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