Mechanics problem - Forces

[jonnburton]

I was wondering if anyone could tell me where I'm going wrong with this question. I've done it several times and have not come to the answer specified in the book.


A mass 10kg is at rest on an inclined plane, where the coefficient of friction is . Find the Magnitude P, of the applied force, if the particle is on the point of sliding down the plane.

I have copied the diagram from the book as follows:




This is my working:


\(\displaystyle F+ P cos 20 = 98sin 20 = 33.52\)

\(\displaystyle R = 98cos 20 + P sin 20 = 92.09+ P sin 20\)



\(\displaystyle F = \mu R\)

\(\displaystyle 33.52 - P cos20 =\mu (92.09+Psin20)\)

\(\displaystyle Pcos20= 33.52 - \mu(92.09+Psin20)\)

\(\displaystyle Pcos20 = 33.52 - 92.09\mu +\mu Psin20\)

\(\displaystyle P cos 20 - \mu Psin20 = 33.52-92.09\mu\)

\(\displaystyle P (cos20 - \mu sin20) = 33.52 -92.09\mu\)

\(\displaystyle P = \frac{33.52-92.09\mu}{cos20-\mu sin 20} = \frac{33.52-9.209}{cos20-0.1sin20} = 26.84N\)

The book says the solution is 25N, however.

 

[Deleted member 4993]

I was wondering if anyone could tell me where I'm going wrong with this question. I've done it several times and have not come to the answer specified in the book.


A mass 10kg is at rest on an inclined plane, where the coefficient of friction is . Find the Magnitude P, of the applied force, if the particle is on the point of sliding down the plane.

I have copied the diagram from the book as follows:

View attachment 3075


This is my working:


\(\displaystyle F+ P cos 20 = 98sin 20 = 33.52\)

\(\displaystyle R = 98cos 20 + P sin 20 = 92.09+ P sin 20\)



\(\displaystyle F = \mu R\)

\(\displaystyle 33.52 - P cos20 =\mu (92.09+Psin20)\)

\(\displaystyle Pcos20= 33.52 - \mu(92.09+Psin20)\)

\(\displaystyle Pcos20 = 33.52 - 92.09\mu +\mu Psin20\)...... Incorrect

should be

\(\displaystyle Pcos20 = 33.52 - 92.09\mu -\mu Psin20\)

and so on

.

 

[DrPhil]

I was wondering if anyone could tell me where I'm going wrong with this question. I've done it several times and have not come to the answer specified in the book.


A mass 10kg is at rest on an inclined plane, where the coefficient of friction is . Find the Magnitude P, of the applied force, if the particle is on the point of sliding down the plane.

I have copied the diagram from the book as follows:

View attachment 3075


This is my working:


\(\displaystyle F+ P cos 20 = 98sin 20 = 33.52\)

\(\displaystyle R = 98cos 20 + P sin 20 = 92.09+ P sin 20\)

I'll work it through with a slightly different viewpoint, to see which answer I agree with. I like to decompose vectors \(\displaystyle \vec{W}, \vec{P}\) into components parallel (\(\displaystyle \hat{f}\)) and perpendicular (\(\displaystyle \hat{r}\)) to the inclined plane. In this case

\(\displaystyle \displaystyle \vec{W} = -(98\mathrm{N})\cos\theta\ \hat{r} - (98\mathrm{N})\sin\theta\ \hat{f} \)

\(\displaystyle \displaystyle \vec{P} = -P\ \sin\theta\ \hat{r} + P\ \cos\theta\ \hat{f}\)

The Force F is the sum of three balanced contributions:

\(\displaystyle \displaystyle F = - (98\mathrm{N})\sin\theta\ + P\ \cos\theta\ + \mu\ R = 0\),

where the normal force \(\displaystyle \displaystyle R = (98\mathrm{N})\cos\theta\ + P\ \sin\theta \)

Note that I have made the sign of \(\displaystyle R\) positive and that the friction force is up the plane, opposing downward motion.

\(\displaystyle \displaystyle P \left[ \cos\theta + \mu\ \sin\theta \right] = (98\mathrm{N}) \left[ \sin\theta - \mu\ \cos\theta \right]\)

............\(\displaystyle \displaystyle P = (98\mathrm{N})\ \dfrac{\sin\theta - \mu\ \cos\theta}{\cos\theta + \mu\ \sin\theta}= \cdot\cdot\cdot\)

This differs from yours in the sign of the sine in the denominator, and gives the result \(\displaystyle P = 25.0\mathrm{N}\).

 

[jonnburton]

Thanks a lot Subhotosh for pointing out my mistake, and DrPhil for showing an alternative way to approach these problems!