A community for students.
Here's the question you clicked on:
 0 viewing
brinethery
 2 years ago
Work problem. Please see attached. I'm having trouble with part (b)
brinethery
 2 years ago
Work problem. Please see attached. I'm having trouble with part (b)

This Question is Closed

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

Outkast3r09
 2 years 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
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.0Yes, I tried finding it. I got a=4.70 m/s^2

brinethery
 2 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.

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

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

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

brinethery
 2 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)

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

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

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

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

brinethery
 2 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?

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

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

Outkast3r09
 2 years 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
 2 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

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

brinethery
 2 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)

brinethery
 2 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.

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

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

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

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

brinethery
 2 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.

brinethery
 2 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.

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

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

Outkast3r09
 2 years 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
 2 years 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
 2 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.

Outkast3r09
 2 years 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
 2 years 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
 2 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.

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

Outkast3r09
 2 years ago
Best ResponseYou've already chosen the best response.1no problem if you need anymore help let me know
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.