Can anyone see where I go wrong (applied maths question...)

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pazzy78

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Here is the question, sorry I posted here, couldn't find applied maths area.

Here is my solution ..

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This method has given me correct answers for other questions, I don't see where I go wrong here ... even taken friction into account..

For some background here is a worked example from the book :

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This is similar question to the one I can't get correct, this is in fact easier as it has no friction.
 
Seems the R I am using for the wedge of course is not the same R as the particle... but even using the new R value (mg) , I still don't get the answer ...
 
pazzy78 said:
This is similar question to the one I can't get correct, this is in fact easier as it has no friction.
Click to expand...
Did you try to solve Question 6 assuming there is no friction between the wedge and the table?
 
With zero friction I get 8g/21 .. still wrong ... [as in not getting the expected answer of 3g/11]
 
Last edited:
Correct answer without friction should be 8g/21 ...

answer it is looking for with friction is 3g/11.

Best I can get is 19g/63
 
Did you try to use this formula?

Npsin⁡θ−μNw=Ma1\displaystyle N_p\sin\theta - \mu N_w = Ma_1
 
mario99 said:
Did you try to use this formula?

Npsin⁡θ−μNw=Ma1\displaystyle N_p\sin\theta - \mu N_w = Ma_1
Click to expand...

Presumably this is the F = ma formula ?

and the Sin here is you are resolving the force into vectors, yes .. I did .. doing that and subbing R (reaction force) into another equation gives me the value of 19g/63...
 
mario99 said:
Wrong!

It's the normal forces on the wedge!
Click to expand...

Yes, but that's the same as the weight force ... it doesn't move in the vertical direction so up force = down force.


edit sorry, I see i originally posted that it was friction ...
 
There is also another force. When the particle push on the wedge, there is a component of NpN_p acts downward. Therefore the vertical forces on the wedge are:

∑Fy=Mg+Npcos⁡θ−Nw=0\displaystyle \sum F_y = Mg + N_p\cos\theta - N_w = 0

Nw=Mg+Npcos⁡θ\displaystyle N_w = Mg + N_p\cos\theta
 
∑Fx=Ma1\displaystyle \sum F_x = Ma_1


Npsin⁡θ−μNw=Ma1\displaystyle N_p\sin \theta - \mu N_w = Ma_1


a1=Npsin⁡θ−μNwM\displaystyle a_1 = \frac{N_p\sin \theta - \mu N_w}{M}


Npsin⁡θ=m(a2cos⁡θ−a1)\displaystyle N_p\sin\theta = m(a_2\cos\theta - a_1)


Npsin⁡θ=ma2cos⁡θ−m(Npsin⁡θ−μNwM)\displaystyle N_p\sin\theta = ma_2\cos\theta - m\left(\frac{N_p\sin \theta - \mu N_w}{M}\right)


Np=mg−ma2sin⁡θcos⁡θ\displaystyle N_p = \frac{mg - ma_2\sin\theta}{\cos\theta}


[mg−ma2sin⁡θcos⁡θ]sin⁡θ=ma2cos⁡θ−m([mg−ma2sin⁡θcos⁡θ]sin⁡θ−μNwM)\displaystyle \left[\frac{mg - ma_2\sin\theta}{\cos\theta}\right]\sin\theta = ma_2\cos\theta - m\left(\frac{\left[\frac{mg - ma_2\sin\theta}{\cos\theta}\right]\sin \theta - \mu N_w}{M}\right)


[mg−ma2sin⁡θcos⁡θ]sin⁡θ=ma2cos⁡θ−m([mg−ma2sin⁡θcos⁡θ]sin⁡θ−μ(Mg+[mg−ma2sin⁡θcos⁡θ]cos⁡θ)M)\displaystyle \left[\frac{mg - ma_2\sin\theta}{\cos\theta}\right]\sin\theta = ma_2\cos\theta - m\left(\frac{\left[\frac{mg - ma_2\sin\theta}{\cos\theta}\right]\sin \theta - \mu \left(Mg + \left[\frac{mg - ma_2\sin\theta}{\cos\theta}\right]\cos\theta\right)}{M}\right)


(mg−ma2sin⁡θcos⁡θ)(Msin⁡θ+msin⁡θ−μmcos⁡θ)=Mma2cos⁡θ+μmMg\displaystyle \left(\frac{mg - ma_2\sin\theta}{\cos\theta}\right)(M\sin\theta + m\sin \theta - \mu m\cos\theta) = Mma_2\cos\theta + \mu mMg


2gM+2mg−mgμ−2a2Msin⁡θ−2a2msin⁡θ+μa2msin⁡θ=Ma2cos⁡θ+μMg\displaystyle 2gM + 2mg - mg\mu - 2a_2M\sin\theta - 2a_2m\sin \theta + \mu a_2m\sin\theta = Ma_2\cos\theta + \mu Mg


−4a2M−4a2m+2μa2m−Ma2=μMg−2gM−2mg+mgμcos⁡θ\displaystyle - 4a_2M - 4a_2m + 2\mu a_2m - Ma_2 = \frac{\mu Mg - 2gM - 2mg + mg\mu}{\cos\theta}


a2=μMg−2gM−2mg+mgμcos⁡θ(−4M−4m+2μm−M)\displaystyle a_2 = \frac{\mu Mg - 2gM - 2mg + mg\mu}{\cos\theta (-4M - 4m + 2\mu m - M)}


Np(sin⁡θ−μcos⁡θ)=Ma1+μMg\displaystyle N_p(\sin \theta - \mu \cos\theta) = Ma_1 + \mu Mg


mg−ma2sin⁡θ=3Ma1+3μMg5\displaystyle mg - ma_2\sin\theta = \frac{3Ma_1 + 3\mu Mg}{5}


mg−3Ma1+3μMg5=ma2sin⁡θ\displaystyle mg - \frac{3Ma_1 + 3\mu Mg}{5} = ma_2\sin\theta


mg−3Ma1+3μMg5=2mμMg−2gM−2mg+mgμ(−4M−4m+2μm−M)\displaystyle mg - \frac{3Ma_1 + 3\mu Mg}{5} = 2m\frac{\mu Mg - 2gM - 2mg + mg\mu}{(-4M - 4m + 2\mu m - M)}


4mg−3ma1+3μmg5=8mμmg−2gm−8mg+4mgμ(−4m−16m+8μm−m)\displaystyle 4mg - \frac{3ma_1 + 3\mu mg}{5} = 8m\frac{\mu mg - 2gm - 8mg + 4mg\mu}{(-4m - 16m + 8\mu m - m)}


4g−3a1+3μg5=8μg−2g−8g+4gμ(−4−16+8μ−1)\displaystyle 4g - \frac{3a_1 + 3\mu g}{5} = 8\frac{\mu g - 2g - 8g + 4g\mu}{(-4 - 16 + 8\mu - 1)}


20g−3a1−3μg=40μg−2g−8g+4gμ(−4−16+8μ−1)\displaystyle 20g - 3a_1 - 3\mu g = 40\frac{\mu g - 2g - 8g + 4g\mu}{(-4 - 16 + 8\mu - 1)}


20g−3a1−g=40μg−2g−8g+4gμ(−4−16+8μ−1)\displaystyle 20g - 3a_1 - g = 40\frac{\mu g - 2g - 8g + 4g\mu}{(-4 - 16 + 8\mu - 1)}


19g−3a1=120μg−2g−8g+4gμ(−55)\displaystyle 19g - 3a_1 = 120\frac{\mu g - 2g - 8g + 4g\mu}{(-55)}


−209g+33a1=24(μg−2g−8g+4gμ)\displaystyle -209g + 33a_1 = 24(\mu g - 2g - 8g + 4g\mu)


−209g+33a1=24(5μg−10g)\displaystyle -209g + 33a_1 = 24(5\mu g - 10g )


33a1=120μg−240g+209g\displaystyle 33a_1 = 120\mu g - 240g +209g


33a1=9g\displaystyle 33a_1 = 9g


a1=9g33=3g11\displaystyle a_1 = \frac{9g}{33} = \frac{3g}{11}
 
Jaysus !!!

I did manage to get 3g/11, all I needed to do was include the normal force in calculating the Friction ...
Ill post my solution later ...
 
pazzy78 said:
Jaysus !!!

I did manage to get 3g/11, all I needed to do was include the normal force in calculating the Friction ...
Ill post my solution later ...
Click to expand...
Then, you are smarter than me. But my method of calculation has a benefit that yours might not have. I will tell you about it later. For now, let us see how you did it!
 
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