## carlj0nes one year ago By using mathematical modelling wit vectors, explain how the acceleration of the object can be included to determine the value of the force P. I will attach my answer for the question without acceleration, but i need to modify it to explain how to include acceleration

1. carlj0nes

2. acxbox22

@dan815

3. TheSmartOne

@dan815

4. TheSmartOne

@Michele_Laino (when you come online later on)

5. dan815

okay , so do you want me to show you another way to do it? or check your work?

6. carlj0nes

my work is correct for non acceleration but for the same question I have been asked to By using mathematical modelling with vectors, explain how the acceleration of the object can be included to determine the value of the force P.

7. carlj0nes

So im assuming it is to tweak the current question to add acceleration

8. carlj0nes

out of interest was does Fnet refer to

9. dan815

|dw:1435246603669:dw|

10. dan815

|dw:1435246874207:dw|

11. dan815

that says Fapp = P

12. carlj0nes

Cheers @dan815 i was watching as you was doing it. I have just wrote it up neat and handed it in. Hopefully i should get feedback today.

13. carlj0nes

I will be asking another question to the qualified help soon which is linked with this post. posted at the bottom of this http://openstudy.com/study#/updates/557d6ee3e4b0e4e582aa72cd

14. carlj0nes

@dan815 I have just had feedback and apparently it isn't what he is after.|dw:1435259141823:dw| I have attached a drawing of what he wants and an equation

15. dan815

okay right, that is what we are doing

16. dan815

17. dan815

oh actuallly there was something that wasnt considered properly

18. dan815

once the block starts moving up, the force of friction is not acting the other way so, we have to subtract that

19. dan815

besides that the equations are same for all forces

20. carlj0nes

N=187.836 Weight is 16.62Kg therefore mg =163.042 u=0.4 P=119.772

21. dan815

let me check what i get

22. dan815

|dw:1435260170676:dw|

23. dan815

sub the values in see if that gives the same answer too

24. dan815

just make sure your signs are right for all of them

25. carlj0nes

Angle=0 (using zero for ease on here) =14.5 pcos0=115.95 mgsin0=40.822 uN=0.4x187.836=75.1344

26. carlj0nes

what do we use for v. is that the acceleration.

27. carlj0nes

@dan815 would you be able to check please.

28. Michele_Laino

in your first problem we have to apply this vector formula: $\Large {\mathbf{N}} + {\mathbf{R}} + {\mathbf{P}} + {\mathbf{W}} = {\mathbf{0}}$ |dw:1435261769664:dw|

29. Michele_Laino

R is the friction force

30. Michele_Laino

If we rewrite that vector equation, using components, we get these subsequent scalars equations: $\Large \left\{ \begin{gathered} - mg\sin \theta + P\cos \theta + \mu mg = 0 \hfill \\ N - mg\cos \theta - P\sin \theta = 0 \hfill \\ \end{gathered} \right.$

31. Michele_Laino

which you can solve for P and N

32. Michele_Laino

that above is the case without acceleration

33. Michele_Laino

If we want to include an acceleration say a, then the new vector equation is: $\Large {\mathbf{N}} + {\mathbf{R}} + {\mathbf{P}} + {\mathbf{W}} = m{\mathbf{a}}$

34. carlj0nes

but are some of these opposing forces

35. Michele_Laino

So, if the vector a is directed at the top of the inclined plane, namely: |dw:1435262337778:dw|

36. Michele_Laino

then the new scalars equations are: $\Large \left\{ \begin{gathered} - mg\sin \theta + P\cos \theta + \mu mg = ma \hfill \\ N - mg\cos \theta - P\sin \theta = 0 \hfill \\ \end{gathered} \right.$

37. Michele_Laino

all those forces are the external forces acting on our box of mass m

38. carlj0nes

@Michele_Laino would you be able to check to see if I have understood what you have done. Sorry about the quality but I will re write it once I know its correct.

39. Michele_Laino

what is the value of \theta?

40. carlj0nes

angle is14.5 deg

41. Michele_Laino

ok! your equations are right. So substituting the value of \theta, we get: $\Large \left\{ \begin{gathered} - 0.250mg + 0.968P + \mu mg = ma \hfill \\ N - 0.968mg - 0.250P = 0 \hfill \\ \end{gathered} \right.$ what is the value of a?

42. carlj0nes

I don't have a value for acceleration as it is to produce the equation that can be used. mg=163.042 p=119.772 u=0.4 and N =187.836

43. Michele_Laino

ok! Then we have to substitute P, \mu and mg, into the first equation. After a simple computation I get this: $\Large ma = 140.43$

44. Michele_Laino

are your forces measured in Newtons

45. carlj0nes

yes

46. carlj0nes

I also got that answer on the picture I uploaded a few minutes ago

47. Michele_Laino

then we have this value: $\Large a = \frac{{140.43}}{{16.62}} \cong 8.45\;{\text{m/se}}{{\text{c}}^{\text{2}}}$

48. carlj0nes

so what you are saying the acceleration speed is 8.45m/sec^2

49. Michele_Laino

I'm saying that the acceleration is toward the top of the inclined plane, and its magnitude is: 8.45 m/sec^2

50. carlj0nes

thank you for your help. Im now going to sleep as I have an early start as im helping the school take children to the science museum. I will try and write this up neat tomorrow then.

51. Michele_Laino

:)