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

Engineering
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I struggled with this type of problem when I was in physics last quarter :-(
did you find the acceleration... since they're roped together you assume that the acceleration of each is the same

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if ou have acceleration you'll have two unknowns that can be related to each other
Yes, I tried finding it. I got a=4.70 m/s^2
So for A, I did sum of the forces in x (x-axis 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.
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I just made my x-axis 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)
\[sin(40)=\frac{W_x}{W}\] \[W_x=mgsin(40)\] \[W_y=mgcos(40)\] \[\sum F_y=N-W\] \[N=W\] \[\sum F_{x1}=m_1a_{1x}=T-W_x-f_k=T-mgsin(40)-(\mu_s)mgcos(40)\] \
where is your weight x component?
ha, maybe that's the problem!
that must've been where I goofed :-p. Just silly mistakes I guess.
\[m_1a_{1x}=T-m_1gsin(40)-\mu_s(m_1gcos(40))\]
I didn't bother to label the acceleration since it will be the same. Would that be correct?
\[m_2a_{2y}=T-m_2g \] \[a_{2y}=-a_{1x}\]
is the pulley frictionless?
yes it is
\[T=-m_2a_{1x}-m_2g=-m_2(a_{1x}+g)\] \[T=m_1a_{1x}+m_1gsin(40)+\mu_km_1gcos(40)\]
Now that I added in the x-component of weight for Block A, my acceleration is 1.754 m/s^2
\[-m_2a_{1x}-m_2g=m_1a_{1x}+m_1gsin(40)+\mu_km_1gcos(40)\]
\[-m_2a_{1x}-m_1a_{1x}=m_2g+m_1gsin(40)+\mu_km_1gcos(40)\] \[a_{1x}(-m_2-m_1)=m_2g+m_1gsin(40)+\mu_km_1gcos(40)\] \[a_{1x}=
i think i dropped a sign somewhere let me get a paper and pen
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)
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.
how'd you define your axis for the y part?
my axis for the y-part is regular. Vertical for y, horizontal for x.
I mean for block B, that is.
isn't downwards negative for the mg?
When I took physics, my instructor told us to make the positive direction for each object the same as the direction of motion.
so for block A, up and to the right would be positive. And for block B, down would be positive.
so you're accounting for the system as a whole.
i got -1.67
that's for B it could be rounding error though
so for mine \[a_{2x}=-1.67\] \[a_{1x}=-a_{2x}=-(-1.67)=1.67\]
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
lol
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 re-read stuff in there.
so anyways now you have \[S=\frac{1}{2}(1.754)t^2+vt+S_0\] and \[V=at+v_0=1.754t+0\]
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 \[R=\frac{l}{a}\]
my friend sent me her stuff, so I'll try it out and tell you if it worked.
Thank you for taking the time to help me.
no problem if you need anymore help let me know

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