Determine the range of values for so that the the top wedge does not slip on... · OlympiadBench #1433

Physics · mechanics

Problem

Determine the range of values for FF 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,θm, g, \theta, and μ\mu. Given constants: g=9.8 N/kgg=9.8 \mathrm{~N} / \mathrm{kg} k=1/4πϵ0=8.99×109 N⋅m2/C2k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2} c=3.00×108 m/sc=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s} NA=6.02×1023( mol)−1N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1} σ=5.67×10−8 J/(s⋅m2⋅K4)\sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right) 1eV=1.602×10−19 J1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J} me=9.109×10−31 kg=0.511MeV/c2m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2} mp=1.673×10−27 kg=938MeV/c2m_{p}=1.673 \times 10^{-27} \mathrm{~kg}=938 \mathrm{MeV} / \mathrm{c}^{2} sin⁡θ≈θ−16θ3\sin \theta \approx \theta-\frac{1}{6} \theta^{3} for ∣θ∣≪1|\theta| \ll 1 G=6.67×10−11 N⋅m2/kg2km=μ0/4π=10−7 T⋅m/AkB=1.38×10−23 J/KR=NAkB=8.31 J/(mol⋅K)e=1.602×10−19Ch=6.63×10−34 J⋅s=4.14×10−15eV⋅s(1+x)n≈1+nx for ∣x∣≪1ln⁡(1+x)≈x for ∣x∣≪1cos⁡θ≈1−12θ2 for ∣θ∣≪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}  velocity parameter β=vc Lorentz factor γ=11−β2 relativistic momentum p=γβmc relativistic energy E=γmc2 relativistic doppler shift ff0=1−β1+β\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 μ\mu, the coefficient of friction between the bottom wedge B and the horizontal surface is μ\mu, and the angle of the wedge is θ\theta. The mass of the top wedge A is mm, and the mass of the bottom wedge B is M=2 mM=2 \mathrm{~m}. A horizontal force FF directed to the left is applied to the bottom wedge as shown in the figure. <image_1>

Browse the problem library