anonymous
  • anonymous
Work problem. Please see attached. I'm having trouble with part (b)
Engineering
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
anonymous
  • anonymous
I struggled with this type of problem when I was in physics last quarter :-(
anonymous
  • anonymous
did you find the acceleration... since they're roped together you assume that the acceleration of each is the same

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anonymous
  • anonymous
if ou have acceleration you'll have two unknowns that can be related to each other
anonymous
  • anonymous
Yes, I tried finding it. I got a=4.70 m/s^2
anonymous
  • anonymous
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.
anonymous
  • anonymous
|dw:1359083523070:dw|
anonymous
  • anonymous
|dw:1359083623616:dw|
anonymous
  • anonymous
|dw:1359083742313:dw|
anonymous
  • anonymous
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)
anonymous
  • anonymous
\[sin(40)=\frac{W_x}{W}\] \[W_x=mgsin(40)\] \[W_y=mgcos(40)\] \[\sum F_y=N-W\] \[N=W\] \[\sum F_{x1}=m_1a_{1x}=T-W_x-f_k=T-mgsin(40)-(\mu_s)mgcos(40)\] \
anonymous
  • anonymous
where is your weight x component?
anonymous
  • anonymous
ha, maybe that's the problem!
anonymous
  • anonymous
that must've been where I goofed :-p. Just silly mistakes I guess.
anonymous
  • anonymous
\[m_1a_{1x}=T-m_1gsin(40)-\mu_s(m_1gcos(40))\]
anonymous
  • anonymous
I didn't bother to label the acceleration since it will be the same. Would that be correct?
anonymous
  • anonymous
\[m_2a_{2y}=T-m_2g \] \[a_{2y}=-a_{1x}\]
anonymous
  • anonymous
is the pulley frictionless?
anonymous
  • anonymous
yes it is
anonymous
  • anonymous
\[T=-m_2a_{1x}-m_2g=-m_2(a_{1x}+g)\] \[T=m_1a_{1x}+m_1gsin(40)+\mu_km_1gcos(40)\]
anonymous
  • anonymous
Now that I added in the x-component of weight for Block A, my acceleration is 1.754 m/s^2
anonymous
  • anonymous
\[-m_2a_{1x}-m_2g=m_1a_{1x}+m_1gsin(40)+\mu_km_1gcos(40)\]
anonymous
  • anonymous
\[-m_2a_{1x}-m_1a_{1x}=m_2g+m_1gsin(40)+\mu_km_1gcos(40)\] \[a_{1x}(-m_2-m_1)=m_2g+m_1gsin(40)+\mu_km_1gcos(40)\] \[a_{1x}=
anonymous
  • anonymous
i think i dropped a sign somewhere let me get a paper and pen
anonymous
  • anonymous
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)
anonymous
  • anonymous
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.
anonymous
  • anonymous
how'd you define your axis for the y part?
anonymous
  • anonymous
my axis for the y-part is regular. Vertical for y, horizontal for x.
anonymous
  • anonymous
I mean for block B, that is.
anonymous
  • anonymous
isn't downwards negative for the mg?
anonymous
  • anonymous
When I took physics, my instructor told us to make the positive direction for each object the same as the direction of motion.
anonymous
  • anonymous
so for block A, up and to the right would be positive. And for block B, down would be positive.
anonymous
  • anonymous
so you're accounting for the system as a whole.
anonymous
  • anonymous
i got -1.67
anonymous
  • anonymous
that's for B it could be rounding error though
anonymous
  • anonymous
so for mine \[a_{2x}=-1.67\] \[a_{1x}=-a_{2x}=-(-1.67)=1.67\]
anonymous
  • anonymous
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
anonymous
  • anonymous
lol
anonymous
  • anonymous
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.
anonymous
  • anonymous
so anyways now you have \[S=\frac{1}{2}(1.754)t^2+vt+S_0\] and \[V=at+v_0=1.754t+0\]
anonymous
  • anonymous
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 \[R=\frac{l}{a}\]
anonymous
  • anonymous
my friend sent me her stuff, so I'll try it out and tell you if it worked.
anonymous
  • anonymous
Thank you for taking the time to help me.
anonymous
  • anonymous
no problem if you need anymore help let me know

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