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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
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

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TheSmartOne
 one year ago
Best ResponseYou've already chosen the best response.0@Michele_Laino (when you come online later on)

dan815
 one year ago
Best ResponseYou've already chosen the best response.2okay , so do you want me to show you another way to do it? or check your work?

carlj0nes
 one year ago
Best ResponseYou've already chosen the best response.0my 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
 one year ago
Best ResponseYou've already chosen the best response.0So im assuming it is to tweak the current question to add acceleration

carlj0nes
 one year ago
Best ResponseYou've already chosen the best response.0out of interest was does Fnet refer to

carlj0nes
 one year ago
Best ResponseYou've already chosen the best response.0Cheers @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
 one year ago
Best ResponseYou've already chosen the best response.0I 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
 one year ago
Best ResponseYou've already chosen the best response.0@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
 one year ago
Best ResponseYou've already chosen the best response.2okay right, that is what we are doing

dan815
 one year ago
Best ResponseYou've already chosen the best response.2oh actuallly there was something that wasnt considered properly

dan815
 one year ago
Best ResponseYou've already chosen the best response.2once the block starts moving up, the force of friction is not acting the other way so, we have to subtract that

dan815
 one year ago
Best ResponseYou've already chosen the best response.2besides that the equations are same for all forces

carlj0nes
 one year ago
Best ResponseYou've already chosen the best response.0N=187.836 Weight is 16.62Kg therefore mg =163.042 u=0.4 P=119.772

dan815
 one year ago
Best ResponseYou've already chosen the best response.2let me check what i get

dan815
 one year ago
Best ResponseYou've already chosen the best response.2sub the values in see if that gives the same answer too

dan815
 one year ago
Best ResponseYou've already chosen the best response.2just make sure your signs are right for all of them

carlj0nes
 one year ago
Best ResponseYou've already chosen the best response.0Angle=0 (using zero for ease on here) =14.5 pcos0=115.95 mgsin0=40.822 uN=0.4x187.836=75.1344

carlj0nes
 one year ago
Best ResponseYou've already chosen the best response.0what do we use for v. is that the acceleration.

carlj0nes
 one year ago
Best ResponseYou've already chosen the best response.0@dan815 would you be able to check please.

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2in 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
 one year ago
Best ResponseYou've already chosen the best response.2R is the friction force

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2If 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
 one year ago
Best ResponseYou've already chosen the best response.2which you can solve for P and N

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2that above is the case without acceleration

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2If 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
 one year ago
Best ResponseYou've already chosen the best response.0but are some of these opposing forces

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2So, if the vector a is directed at the top of the inclined plane, namely: dw:1435262337778:dw

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2then 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
 one year ago
Best ResponseYou've already chosen the best response.2all those forces are the external forces acting on our box of mass m

carlj0nes
 one year ago
Best ResponseYou've already chosen the best response.0@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
 one year ago
Best ResponseYou've already chosen the best response.2what is the value of \theta?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2ok! 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
 one year ago
Best ResponseYou've already chosen the best response.0I 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
 one year ago
Best ResponseYou've already chosen the best response.2ok! 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
 one year ago
Best ResponseYou've already chosen the best response.2are your forces measured in Newtons

carlj0nes
 one year ago
Best ResponseYou've already chosen the best response.0I also got that answer on the picture I uploaded a few minutes ago

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2then we have this value: \[\Large a = \frac{{140.43}}{{16.62}} \cong 8.45\;{\text{m/se}}{{\text{c}}^{\text{2}}}\]

carlj0nes
 one year ago
Best ResponseYou've already chosen the best response.0so what you are saying the acceleration speed is 8.45m/sec^2

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2I'm saying that the acceleration is toward the top of the inclined plane, and its magnitude is: 8.45 m/sec^2

carlj0nes
 one year ago
Best ResponseYou've already chosen the best response.0thank 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.
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