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

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