• anonymous
let v1 = (1,2), v2 = (2,3) s = {v1, v2} is a basis for R2 if w = (-2, -2) find (w)s
  • Stacey Warren - Expert
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  • chestercat
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  • anonymous
hint: ________1_________ As you will see, the following is the only value we need to know... ω₀ = 6 rev/s I am not certain I am correct, but I will take the phrase "shortened by 47%" to mean that the final length is 53% the original. Since the system is closed, angular momentum is conserved. Initial angular momentum is... L₀ = I₀ ω₀ I₀ = m d² + m d² + m D² + m D² = 2 m (d² + D²) L₀ = 2 m (d² + D²) ω₀ Final angular momentum is... L = I ω = 2 m ( (0.53 d)² + (0.53 D)² ) ω = 2 m (d² + D²) 0.53² ω Setting L₀ = L and solving for ω gives... 2 m (d² + D²) ω₀ = 2 m (d² + D²) 0.53² ω ω = 0.53ˉ² ω₀ ≈ 21.36 rev/s ________1 edit_________ I will now take the phrase "shortened by 47%" to mean that the final length is 47% the original instead of 53%. Everything will be the same except we will get... ω = 0.47ˉ² ω₀ ≈ 27.16 rev/s Unless I am completely mistaken, either this or the answer I gave previously must be the correct answer. If the answer I just gave is the correct answer, the question should have said "shortened TO 47%" instead of "shortened BY 47%". Please let me know if this new answer is correct! ________2_________ The values given are... m = 69 kg M = 370 kg L = 7.3 m I will define... D = L - d I will assume that the beam has constant mass per length. When amount d is hanging off the cliff and when the student is standing on the edge of beam, there will be no normal force anywhere but at edge of cliff, and this point will be a great choice for the pivot point because then we can ignore this force's contribution to torque. We now just need to insure that the two torques cancel in order for the beam to be at rest... τ₁ = τ₂ Note that I am taking τ₁ and τ₂ to be magnitudes and are therefore positive. The weight of the beam of length D creates τ₁. Keep in mind that weight always acts at the center of gravity, which would be at D/2. Also, the mass of this beam section is D/L M. The force of gravity is... F₁ = D/L M g which means that torque is... τ₁ = D/2 D/L M g = D² M g / (2 L) The weight of the beam of length d and the weight of the student create τ₂, which is... τ₂ = d² M g / (2 L) + d m g Using τ₁ = τ₂ and D = L - d to solve for d gives... (L - d)² M g / (2 L) = d² M g / (2 L) + d m g Which, after some work, simplifies to... d M + d m = L M / 2 Which gives... d = L M / (2 (M + m) ) ≈ 3.1 m ________3_________ We are given... w = 807 N Since the force at A is 0, there are three forces on the beam: w, W, and the force at B. Let's choose B to be the pivot so that we can ignore its contribution to torque. We now only have w and W creating torque. τ₁ = τ₂ gives... L/4 w = L/2 W Solving for W gives... W = w / 2 ≈ 404 N

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