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

1. brinethery Group Title

2. brinethery Group Title

I struggled with this type of problem when I was in physics last quarter :-(

3. Outkast3r09 Group Title

did you find the acceleration... since they're roped together you assume that the acceleration of each is the same

4. Outkast3r09 Group Title

if ou have acceleration you'll have two unknowns that can be related to each other

5. brinethery Group Title

Yes, I tried finding it. I got a=4.70 m/s^2

6. brinethery Group Title

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.

7. Outkast3r09 Group Title

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8. Outkast3r09 Group Title

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9. Outkast3r09 Group Title

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10. brinethery Group Title

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)

11. Outkast3r09 Group Title

$sin(40)=\frac{W_x}{W}$ $W_x=mgsin(40)$ $W_y=mgcos(40)$ $\sum F_y=N-W$ $N=W$ $\sum F_{x1}=m_1a_{1x}=T-W_x-f_k=T-mgsin(40)-(\mu_s)mgcos(40)$ \

12. Outkast3r09 Group Title

where is your weight x component?

13. brinethery Group Title

ha, maybe that's the problem!

14. brinethery Group Title

that must've been where I goofed :-p. Just silly mistakes I guess.

15. Outkast3r09 Group Title

$m_1a_{1x}=T-m_1gsin(40)-\mu_s(m_1gcos(40))$

16. brinethery Group Title

I didn't bother to label the acceleration since it will be the same. Would that be correct?

17. Outkast3r09 Group Title

$m_2a_{2y}=T-m_2g$ $a_{2y}=-a_{1x}$

18. Outkast3r09 Group Title

is the pulley frictionless?

19. brinethery Group Title

yes it is

20. Outkast3r09 Group Title

$T=-m_2a_{1x}-m_2g=-m_2(a_{1x}+g)$ $T=m_1a_{1x}+m_1gsin(40)+\mu_km_1gcos(40)$

21. brinethery Group Title

Now that I added in the x-component of weight for Block A, my acceleration is 1.754 m/s^2

22. Outkast3r09 Group Title

$-m_2a_{1x}-m_2g=m_1a_{1x}+m_1gsin(40)+\mu_km_1gcos(40)$

23. Outkast3r09 Group Title

$-m_2a_{1x}-m_1a_{1x}=m_2g+m_1gsin(40)+\mu_km_1gcos(40)$ $a_{1x}(-m_2-m_1)=m_2g+m_1gsin(40)+\mu_km_1gcos(40)$ $a_{1x}= 24. Outkast3r09 Group Title i think i dropped a sign somewhere let me get a paper and pen 25. brinethery Group Title 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) 26. brinethery Group Title 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. 27. Outkast3r09 Group Title how'd you define your axis for the y part? 28. brinethery Group Title my axis for the y-part is regular. Vertical for y, horizontal for x. 29. brinethery Group Title I mean for block B, that is. 30. Outkast3r09 Group Title isn't downwards negative for the mg? 31. brinethery Group Title When I took physics, my instructor told us to make the positive direction for each object the same as the direction of motion. 32. brinethery Group Title so for block A, up and to the right would be positive. And for block B, down would be positive. 33. brinethery Group Title so you're accounting for the system as a whole. 34. Outkast3r09 Group Title i got -1.67 35. Outkast3r09 Group Title that's for B it could be rounding error though 36. Outkast3r09 Group Title so for mine \[a_{2x}=-1.67$ $a_{1x}=-a_{2x}=-(-1.67)=1.67$

37. Outkast3r09 Group Title

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

38. brinethery Group Title

lol

39. brinethery Group Title

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.

40. Outkast3r09 Group Title

so anyways now you have $S=\frac{1}{2}(1.754)t^2+vt+S_0$ and $V=at+v_0=1.754t+0$

41. Outkast3r09 Group Title

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 $R=\frac{l}{a}$

42. brinethery Group Title

my friend sent me her stuff, so I'll try it out and tell you if it worked.

43. brinethery Group Title

Thank you for taking the time to help me.

44. Outkast3r09 Group Title

no problem if you need anymore help let me know