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

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- carlj0nes

##### 2 Attachments

- acxbox22

@dan815

- TheSmartOne

@dan815

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## More answers

- TheSmartOne

@Michele_Laino (when you come online later on)

- dan815

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

- 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

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

- carlj0nes

out of interest was does Fnet refer to

- dan815

|dw:1435246603669:dw|

- dan815

|dw:1435246874207:dw|

- dan815

that says Fapp = P

- 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

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

@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

okay right, that is what we are doing

- dan815

add all the forces

- dan815

oh actuallly there was something that wasnt considered properly

- dan815

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

- dan815

besides that the equations are same for all forces

- carlj0nes

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

- dan815

let me check what i get

- dan815

|dw:1435260170676:dw|

- dan815

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

- dan815

just make sure your signs are right for all of them

- carlj0nes

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

- carlj0nes

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

- carlj0nes

@dan815 would you be able to check please.

##### 1 Attachment

- 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

R is the friction force

- 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

which you can solve for P and N

- Michele_Laino

that above is the case without acceleration

- 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

but are some of these opposing forces

- Michele_Laino

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

- 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

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

- 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.

##### 1 Attachment

- Michele_Laino

what is the value of \theta?

- carlj0nes

angle is14.5 deg

- 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

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

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

are your forces measured in Newtons

- carlj0nes

yes

- carlj0nes

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

- 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

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

- 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

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

:)

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