## 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 Group Title

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2. milliex51 Group Title

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 Group Title

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 Group Title

Someone? Help? :(

5. elica85 Group Title

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

6. milliex51 Group Title

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

7. elica85 Group Title

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 Group Title

ohh.. Tension force.

9. elica85 Group Title

is this moving to the left?

10. milliex51 Group Title

to the right.

11. elica85 Group Title

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

12. milliex51 Group Title

really? ohh.. okay.

13. elica85 Group Title

gravity causes the masses to move

14. milliex51 Group Title

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

15. milliex51 Group Title

yup.

16. elica85 Group Title

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

17. milliex51 Group Title

alright.

18. elica85 Group Title

now the resistant force is already given in the problem

19. milliex51 Group Title

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

20. elica85 Group Title

what are you using for m2?

21. milliex51 Group Title

2.0 kg

22. milliex51 Group Title

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

23. elica85 Group Title

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 Group Title

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 Group Title

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

26. milliex51 Group Title

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

27. elica85 Group Title

my final is 2.9m/s^2

28. milliex51 Group Title

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

29. elica85 Group Title

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

30. elica85 Group Title

your formula seems a little off but i will double check

31. milliex51 Group Title

ohh.. okay.

32. elica85 Group Title

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 Group Title

ohh..

34. elica85 Group Title

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 Group Title

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

36. milliex51 Group Title

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

37. elica85 Group Title

np

38. milliex51 Group Title

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

39. elica85 Group Title

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

40. elica85 Group Title

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

41. milliex51 Group Title

ohhh, so it's like m2g?

42. elica85 Group Title

yes

43. milliex51 Group Title

ohh, okay. Thank you!

44. elica85 Group Title

np

45. milliex51 Group Title

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

46. elica85 Group Title

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 Group Title

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

48. elica85 Group Title

yup