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Work problem. Please see attached. I'm having trouble with part (b)
 one year ago
 one year ago
Work problem. Please see attached. I'm having trouble with part (b)
 one year ago
 one year ago

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

Outkast3r09Best ResponseYou've already chosen the best response.1
did you find the acceleration... since they're roped together you assume that the acceleration of each is the same
 one year ago

Outkast3r09Best ResponseYou've already chosen the best response.1
if ou have acceleration you'll have two unknowns that can be related to each other
 one year ago

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

brinetheryBest ResponseYou've already chosen the best response.0
So 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.
 one year ago

Outkast3r09Best ResponseYou've already chosen the best response.1
dw:1359083523070:dw
 one year ago

Outkast3r09Best ResponseYou've already chosen the best response.1
dw:1359083623616:dw
 one year ago

Outkast3r09Best ResponseYou've already chosen the best response.1
dw:1359083742313:dw
 one year ago

brinetheryBest ResponseYou've already chosen the best response.0
I 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)
 one year ago

Outkast3r09Best ResponseYou've already chosen the best response.1
\[sin(40)=\frac{W_x}{W}\] \[W_x=mgsin(40)\] \[W_y=mgcos(40)\] \[\sum F_y=NW\] \[N=W\] \[\sum F_{x1}=m_1a_{1x}=TW_xf_k=Tmgsin(40)(\mu_s)mgcos(40)\] \
 one year ago

Outkast3r09Best ResponseYou've already chosen the best response.1
where is your weight x component?
 one year ago

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

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

Outkast3r09Best ResponseYou've already chosen the best response.1
\[m_1a_{1x}=Tm_1gsin(40)\mu_s(m_1gcos(40))\]
 one year ago

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

Outkast3r09Best ResponseYou've already chosen the best response.1
\[m_2a_{2y}=Tm_2g \] \[a_{2y}=a_{1x}\]
 one year ago

Outkast3r09Best ResponseYou've already chosen the best response.1
is the pulley frictionless?
 one year ago

Outkast3r09Best ResponseYou've already chosen the best response.1
\[T=m_2a_{1x}m_2g=m_2(a_{1x}+g)\] \[T=m_1a_{1x}+m_1gsin(40)+\mu_km_1gcos(40)\]
 one year ago

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

Outkast3r09Best ResponseYou've already chosen the best response.1
\[m_2a_{1x}m_2g=m_1a_{1x}+m_1gsin(40)+\mu_km_1gcos(40)\]
 one year ago

Outkast3r09Best 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}=
 one year ago

Outkast3r09Best ResponseYou've already chosen the best response.1
i think i dropped a sign somewhere let me get a paper and pen
 one year ago

brinetheryBest ResponseYou've already chosen the best response.0
What 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)
 one year ago

brinetheryBest ResponseYou've already chosen the best response.0
I 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.
 one year ago

Outkast3r09Best ResponseYou've already chosen the best response.1
how'd you define your axis for the y part?
 one year ago

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

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

Outkast3r09Best ResponseYou've already chosen the best response.1
isn't downwards negative for the mg?
 one year ago

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

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

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

Outkast3r09Best ResponseYou've already chosen the best response.1
that's for B it could be rounding error though
 one year ago

Outkast3r09Best ResponseYou've already chosen the best response.1
so for mine \[a_{2x}=1.67\] \[a_{1x}=a_{2x}=(1.67)=1.67\]
 one year ago

Outkast3r09Best ResponseYou've already chosen the best response.1
if 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
 one year ago

brinetheryBest ResponseYou've already chosen the best response.0
We'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.
 one year ago

Outkast3r09Best ResponseYou've already chosen the best response.1
so anyways now you have \[S=\frac{1}{2}(1.754)t^2+vt+S_0\] and \[V=at+v_0=1.754t+0\]
 one year ago

Outkast3r09Best ResponseYou've already chosen the best response.1
so 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 \[R=\frac{l}{a}\]
 one year ago

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

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

Outkast3r09Best ResponseYou've already chosen the best response.1
no problem if you need anymore help let me know
 one year ago
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