Physics

Physics · mechanics

Problem

A square based pyramid (that is symmetrical) is standing on top of a cube with side length ℓ=10 cm\ell=10 \mathrm{~cm} such that their square faces perfectly line up. The cube is initially standing still on flat ground and both objects have the same uniform density. The coefficient of friction between every surface is the same value of μ=0.3\mu=0.3. The cube is then given an initial speed vv in some direction parallel to the floor. What is the maximum possible value of vv such that the base of the pyramid will always remain parallel to the top of the cube? Answer in meters per second. <image_1> Given constants: - Proton mass, mp=1.67⋅10−27 kgm_{p}=1.67 \cdot 10^{-27} \mathrm{~kg} - Neutron mass, mn=1.67⋅10−27 kgm_{n}=1.67 \cdot 10^{-27} \mathrm{~kg} - Electron mass, me=9.11⋅10−31 kgm_{e}=9.11 \cdot 10^{-31} \mathrm{~kg} - Avogadro's constant, N0=6.02⋅1023 mol−1N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1} - Universal gas constant, R=8.31 J/(mol⋅K)R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) - Boltzmann's constant, kB=1.38⋅10−23 J/Kk_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K} - Electron charge magnitude, e=1.60⋅10−19Ce=1.60 \cdot 10^{-19} \mathrm{C} - 1 electron volt, 1eV=1.60⋅10−19 J1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J} - Speed of light, c=3.00⋅108 m/sc=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s} - Universal Gravitational constant, G=6.67⋅10−11( N⋅m2)/kg2G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} - Solar Mass M⊙=1.988⋅1030 kgM_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} - Acceleration due to gravity, g=9.8 m/s2g=9.8 \mathrm{~m} / \mathrm{s}^{2} - 1 unified atomic mass unit, 1u=1.66⋅10−27 kg=931MeV/c21 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} - Planck's constant, h=6.63⋅10−34 J⋅s=4.41⋅10−15eV⋅sh=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} - Permittivity of free space, ϵ0=8.85⋅10−12C2/(N⋅m2)\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) - Coulomb's law constant, k=14πϵ0=8.99⋅109( N⋅m2)/C2k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} - Permeability of free space, μ0=4π⋅10−7 T⋅m/A\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} - Magnetic constant, μ04π=1⋅10−7( T⋅m)/A\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} - 1 atmospheric pressure, 1 atm=1.01⋅105 N/m2=1.01⋅105 Pa1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} - Wien's displacement constant, b=2.9b=2.9. 10−3 m⋅K10^{-3} \mathrm{~m} \cdot \mathrm{K} - Stefan-Boltzmann constant, σ=5.67⋅10−8 W/m2/K4\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
Intuitive explanation

The key is recognizing that when the cube accelerates, friction acts on the pyramid, creating a torque about the pyramid's center of mass. The pyramid stays level as long as this torque doesn't cause it to tip—which happens when the friction force tries to shift the pyramid's effective contact point beyond the edge of the cube's top face. Find the maximum acceleration the cube can have before the pyramid's center of mass would need to be supported outside the square base, then convert that to the speed limit using kinematics and friction constraints.

Intuitive explanation

The key is recognizing that when the cube accelerates, friction acts on the pyramid, creating a torque about the pyramid's center of mass. The pyramid stays level as long as this torque doesn't cause it to tip—which happens when the friction force tries to shift the pyramid's effective contact point beyond the edge of the cube's top face. Find the maximum acceleration the cube can have before the pyramid's center of mass would need to be supported outside the square base, then convert that to the speed limit using kinematics and friction constraints.

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