## milliex51 Group Title HEEEELP!! How do I calculate the acceleration of the cart shown, given the following assumption: A frictional resistance of magnitude 2.0 N is acting on the cart. 2 years ago 2 years ago

1. milliex51

|dw:1324432388146:dw|

2. milliex51

How do I calculate the acceleration of the cart shown, given the following assumption: A frictional resistance of magnitude 2.0 N is acting on the cart. It first asks what the acceleration will be with no friction is acting on a cart and my formula is: T = m1a m2a = m2g - T T = m2g - m2a m1a+ m2a = m2g a = m2g / (m1 + m2)

3. milliex51

and the answer for my current problem stated above is 2.9 m/s^2... and I don't know how to solve it D;

4. milliex51

Someone? Help? :(

5. elica85

a=F/m F=driving force-resistant force what's the driving force in this case?

6. milliex51

driving force? It.. It doesn't say.

7. elica85

well i'm just asking you to think about it. it will help you in solving problems in the future. what force is causing the system to move?

8. milliex51

ohh.. Tension force.

9. elica85

is this moving to the left?

10. milliex51

to the right.

11. elica85

ok, so no it's not the tension force. it's the force from the hanging mass

12. milliex51

really? ohh.. okay.

13. elica85

gravity causes the masses to move

14. milliex51

ohh, i see it. 'Cause when the hanging mass falls the cart moves.

15. milliex51

yup.

16. elica85

so $driving force=m _{hanging}*g$

17. milliex51

alright.

18. elica85

now the resistant force is already given in the problem

19. milliex51

how? And is your equation the same as: a = m2g / (m1 + m2)?

20. elica85

what are you using for m2?

21. milliex51

2.0 kg

22. milliex51

and you said how the resistant force is already given in the problem-- how?

23. elica85

ok then yes, it will end up as that equation you have but i will break it down a=F/m resistant force = 2N F=2g-2N =19.6N-2N back to acceleration... a=F/mass of the system = (19.6N-2N)/(2kg+4kg)

24. milliex51

how did you get 2N as the resistant force? o.O 'cause the answer in the back of the text is 2.9 m/s^2

25. elica85

oops, actually different than yours but i'm confident in what i put up there

26. milliex51

do you know a way in which I can use my formula?

27. elica85

my final is 2.9m/s^2

28. milliex51

ohhh but do you know how to answer the problem with my formula? ...

29. elica85

tension cancels out for this system which is why i didn't use it in any of my equations

30. elica85

your formula seems a little off but i will double check

31. milliex51

ohh.. okay.

32. elica85

break down the system in 2 parts. hanging mass will be labeled 1. $m _{1}g-T _{1}=m _{1}a$ ^^^derived from F=ma solve for T $T _{1}=m _{1}(g-a)$ for second part... $T _{2}-m _{2}g=m _{2}a$ solve for T... $T _{2}=m _{2}(g+a)$

33. milliex51

ohh..

34. elica85

so what you have to calculate a would work if there was no friction but since there is, the best way to do it is the way i did it

35. milliex51

ohh, cos that formula I showed you was for no friction & I thought maybe to switch a few places would be okay.

36. milliex51

Well then, thank you so much elica for putting up with me!! Thank you!!

37. elica85

np

38. milliex51

btw, why is it: F=2g-2N? Why is it 2g?

39. elica85

after you calculate 2g (2kg*9.8m/s^2) it will equal to 19.6N. the unit N=kg*m/s^2

40. elica85

just remember you can't add or subtract unless they have the same units

41. milliex51

ohhh, so it's like m2g?

42. elica85

yes

43. milliex51

ohh, okay. Thank you!

44. elica85

np

45. milliex51

so F = m2g - frictional resistance? (what's the variable for it?)

46. elica85

in this case, frictional force is denoted as $F _{k}=\mu mg$ k for kinetic friction and the u-looking sign is the friction coefficient which is usually given in a problem. in this problem, the Fk was already solved (2N)

47. milliex51

ohhhh, I'm familiar with that! Oh, thank you!

48. elica85

yup