Subfield: Mechanics

$g=9.8 \mathrm{~N} / \mathrm{kg}$



$k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}$



$c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s}$



$N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1}$



$\sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right)$



$1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J}$



$m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2}$



$m_{p}=1.673 \times 10^{-27} \mathrm{~kg}=938 \mathrm{MeV} / \mathrm{c}^{2}$



$\sin \theta \approx \theta-\frac{1}{6} \theta^{3}$ for $|\theta| \ll 1$



$$

\begin{aligned}

& G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\

& k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\

& k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\

& R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\

& e=1.602 \times 10^{-19} \mathrm{C} \\

& h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times 10^{-15} \mathrm{eV} \cdot \mathrm{s} \\

& (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\

& \ln (1+x) \approx x \text { for }|x| \ll 1 \\

& \cos \theta \approx 1-\frac{1}{2} \theta^{2} \text { for }|\theta| \ll 1

\end{aligned}

$$

$$

\begin{array}{lrl}

\text { velocity parameter } & \beta & =\frac{v}{c} \\

\text { Lorentz factor } & \gamma & =\frac{1}{\sqrt{1-\beta^{2}}} \\

\text { relativistic momentum } & p & =\gamma \beta m c \\

\text { relativistic energy } & E & =\gamma m c^{2} \\

\text { relativistic doppler shift } & \frac{f}{f_{0}} & =\sqrt{\frac{1-\beta}{1+\beta}}

\end{array}

$$

Extra Supplementary Reading Materials:

A pair of wedges are located on a horizontal surface. The coefficient of friction (both sliding and static) between the wedges is $\mu$, the coefficient of friction between the bottom wedge B and the horizontal surface is $\mu$, and the angle of the wedge is $\theta$. The mass of the top wedge A is $m$, and the mass of the bottom wedge B is $M=2 \mathrm{~m}$. A horizontal force $F$ directed to the left is applied to the bottom wedge as shown in the figure.



<img_4579>

Question: Determine the range of values for $F$ so that the the top wedge does not slip on the bottom wedge. Express your answer(s) in terms of any or all of $m, g, \theta$, and $\mu$.

Solution:
Assume the block does not slip. Considering the horizontal forces on the entire system gives



$$

F-3 \mu m g=3 m a \quad \Rightarrow \quad a=\frac{F}{3 m}-\mu g

$$



When $F$ is small, the top wedge wants to slide downward, so static friction points up the ramp. Considering the horizontal and vertical forces on the block gives



$$

N \cos \theta+f \sin \theta=m g, \quad N \sin \theta-f \cos \theta=m a

$$



When the minimal force is applied, the friction is maximal, $f=\mu N$. Eliminating $N$ gives



$$

m a=m g \frac{\sin \theta-\mu \cos \theta}{\cos \theta+\mu \sin \theta}

$$



and plugging in our first equation gives



$$

F_{\min }=3 m g\left(\mu+\frac{\sin \theta-\mu \cos \theta}{\cos \theta+\mu \sin \theta}\right)=3 m g \frac{\left(1+\mu^{2}\right) \tan \theta}{1+\mu \tan \theta}

$$



When $F$ is large, the top wedge wants to slide upward, so static friction points down the ramp, and



$$

N \cos \theta-f \sin \theta=m g, \quad N \sin \theta+f \cos \theta=m a

$$



Now setting $f=\mu N$ gives



$$

F_{\max }=3 m g\left(\mu+\frac{\sin \theta+\mu \cos \theta}{\cos \theta-\mu \sin \theta}\right)=3 m g \frac{2 \mu+\left(1-\mu^{2}\right) \tan \theta}{1-\mu \tan \theta}

$$



Therefore, naively the range of forces so that the block will not slip is



$$

F \in\left[F_{\min }, F_{\max }\right]

$$







However, to get full credit, students must account for two edge cases. First, when $\mu>\tan \theta$, no force is required at all to keep the block in place, so the minimum force is zero. Second, when $\mu>\cot \theta$, the block will not slip up under any circumstances, so there is no maximal force.
Assume the block does not slip. Considering the horizontal forces on the entire system gives



$$

F-3 \mu m g=3 m a \quad \Rightarrow \quad a=\frac{F}{3 m}-\mu g

$$

The problem can also be solved geometrically. In general, the no slip condition is



$$

\mu>\tan \phi

$$



where $\phi$ is the angle between the vertical and the normal to the plane. Working in the noninertial reference frame of the plane, the fictitious force due to the acceleration is equivalent to a tilting of the gravity vector by an angle



$$

\tan \beta=\frac{a}{g}

$$



where, as in above,



$$

a=\frac{F}{3 m}-\mu g

$$



Then the top block will not slip as long as $|\theta-\beta| \leq \phi$. At the minimum acceleration $a_{\min }, \beta=\theta-\phi$, and taking the tangent of both sides gives



$$

\frac{a_{\min }}{g}=\frac{\tan \theta-\mu}{1+\mu \tan \theta}

$$



This is only meaningful for $\tan \theta>\mu$, otherwise the answer is simply $a_{\min }=0$. At the maximum acceleration $a_{\max }, \beta=\theta+\phi$, which gives



$$

\frac{a_{\max }}{g}=\frac{\tan \theta+\mu}{1-\mu \tan \theta}

$$



This is only meaningful for $\cot \theta>\mu$, otherwise the answer is simply $a_{\max }=\infty$.

Final answer: $[3 m g \frac{\left(1+\mu^{2}\right) \tan \theta}{1+\mu \tan \theta}, 3 m g \frac{2 \mu+\left(1-\mu^{2}\right) \tan \theta}{1-\mu \tan \theta}]$