carlj0nes
  • carlj0nes
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
Mathematics
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
carlj0nes
  • carlj0nes
acxbox22
  • acxbox22
@dan815
TheSmartOne
  • TheSmartOne
@dan815

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TheSmartOne
  • TheSmartOne
@Michele_Laino (when you come online later on)
dan815
  • dan815
okay , so do you want me to show you another way to do it? or check your work?
carlj0nes
  • 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.
carlj0nes
  • carlj0nes
So im assuming it is to tweak the current question to add acceleration
carlj0nes
  • carlj0nes
out of interest was does Fnet refer to
dan815
  • dan815
|dw:1435246603669:dw|
dan815
  • dan815
|dw:1435246874207:dw|
dan815
  • dan815
that says Fapp = P
carlj0nes
  • 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.
carlj0nes
  • 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
carlj0nes
  • 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
dan815
  • dan815
okay right, that is what we are doing
dan815
  • dan815
add all the forces
dan815
  • dan815
oh actuallly there was something that wasnt considered properly
dan815
  • dan815
once the block starts moving up, the force of friction is not acting the other way so, we have to subtract that
dan815
  • dan815
besides that the equations are same for all forces
carlj0nes
  • carlj0nes
N=187.836 Weight is 16.62Kg therefore mg =163.042 u=0.4 P=119.772
dan815
  • dan815
let me check what i get
dan815
  • dan815
|dw:1435260170676:dw|
dan815
  • dan815
sub the values in see if that gives the same answer too
dan815
  • dan815
just make sure your signs are right for all of them
carlj0nes
  • carlj0nes
Angle=0 (using zero for ease on here) =14.5 pcos0=115.95 mgsin0=40.822 uN=0.4x187.836=75.1344
carlj0nes
  • carlj0nes
what do we use for v. is that the acceleration.
carlj0nes
  • carlj0nes
@dan815 would you be able to check please.
Michele_Laino
  • 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|
Michele_Laino
  • Michele_Laino
R is the friction force
Michele_Laino
  • Michele_Laino
If we rewrite that vector equation, using components, we get these subsequent scalars equations: \[\Large \left\{ \begin{gathered} - mg\sin \theta + P\cos \theta + \mu mg = 0 \hfill \\ N - mg\cos \theta - P\sin \theta = 0 \hfill \\ \end{gathered} \right.\]
Michele_Laino
  • Michele_Laino
which you can solve for P and N
Michele_Laino
  • Michele_Laino
that above is the case without acceleration
Michele_Laino
  • Michele_Laino
If we want to include an acceleration say a, then the new vector equation is: \[\Large {\mathbf{N}} + {\mathbf{R}} + {\mathbf{P}} + {\mathbf{W}} = m{\mathbf{a}}\]
carlj0nes
  • carlj0nes
but are some of these opposing forces
Michele_Laino
  • Michele_Laino
So, if the vector a is directed at the top of the inclined plane, namely: |dw:1435262337778:dw|
Michele_Laino
  • Michele_Laino
then the new scalars equations are: \[\Large \left\{ \begin{gathered} - mg\sin \theta + P\cos \theta + \mu mg = ma \hfill \\ N - mg\cos \theta - P\sin \theta = 0 \hfill \\ \end{gathered} \right.\]
Michele_Laino
  • Michele_Laino
all those forces are the external forces acting on our box of mass m
carlj0nes
  • 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.
Michele_Laino
  • Michele_Laino
what is the value of \theta?
carlj0nes
  • carlj0nes
angle is14.5 deg
Michele_Laino
  • 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?
carlj0nes
  • 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
Michele_Laino
  • Michele_Laino
ok! Then we have to substitute P, \mu and mg, into the first equation. After a simple computation I get this: \[\Large ma = 140.43\]
Michele_Laino
  • Michele_Laino
are your forces measured in Newtons
carlj0nes
  • carlj0nes
yes
carlj0nes
  • carlj0nes
I also got that answer on the picture I uploaded a few minutes ago
Michele_Laino
  • Michele_Laino
then we have this value: \[\Large a = \frac{{140.43}}{{16.62}} \cong 8.45\;{\text{m/se}}{{\text{c}}^{\text{2}}}\]
carlj0nes
  • carlj0nes
so what you are saying the acceleration speed is 8.45m/sec^2
Michele_Laino
  • 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
carlj0nes
  • 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.
Michele_Laino
  • Michele_Laino
:)

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