What Is the Minimum and Maximum Acceleration for a Block on an Inclined Wedge?

americanforest
Messages
220
Reaction score
0

Homework Statement



A block rests on a wedge inclined at an angle \theta. The coefficient of friction between the block and the plane is \mu.Let the wedge move with horizontal acceleration a to the right. Assuming
that tan tan(\theta) < \mu, find the minimum value of a for the block to remain on the wedge without sliding.

Also find the maximum value of a.

Homework Equations



The Attempt at a Solution



mgcos(\theta)+masin(\theta)=F_{normal};<br /> \tex{mgcos(\theta)+\mu (F_{normal})=macos(\theta);<br /> <br /> a=\frac{g(sin+\mu cos}{cos - \mu sin}

which gives wacky limiting values and isn't a range at all but just one value.
 
Last edited:
Physics news on Phys.org
Have a bit correction with this one: F_{static friction} \leq \mu F_{normal}
Also, there must be a condition for a to be positive, as a acts to the right.
 
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...
Back
Top