## anonymous 4 years ago 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.

1. anonymous

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2. anonymous

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. anonymous

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. anonymous

Someone? Help? :(

5. anonymous

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

6. anonymous

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

7. anonymous

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. anonymous

ohh.. Tension force.

9. anonymous

is this moving to the left?

10. anonymous

to the right.

11. anonymous

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

12. anonymous

really? ohh.. okay.

13. anonymous

gravity causes the masses to move

14. anonymous

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

15. anonymous

yup.

16. anonymous

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

17. anonymous

alright.

18. anonymous

now the resistant force is already given in the problem

19. anonymous

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

20. anonymous

what are you using for m2?

21. anonymous

2.0 kg

22. anonymous

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

23. anonymous

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. anonymous

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. anonymous

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

26. anonymous

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

27. anonymous

my final is 2.9m/s^2

28. anonymous

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

29. anonymous

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

30. anonymous

your formula seems a little off but i will double check

31. anonymous

ohh.. okay.

32. anonymous

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. anonymous

ohh..

34. anonymous

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. anonymous

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

36. anonymous

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

37. anonymous

np

38. anonymous

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

39. anonymous

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

40. anonymous

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

41. anonymous

ohhh, so it's like m2g?

42. anonymous

yes

43. anonymous

ohh, okay. Thank you!

44. anonymous

np

45. anonymous

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

46. anonymous

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. anonymous

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

48. anonymous

yup