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ReviewSolutions

# ReviewSolutions - Physics 8A Midterm 2 Review Solutions...

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Physics 8A Midterm 2 Review Solutions Problem 1 1. θ R M mg T f s x y N Figure 1: Free Body Diagram 2. The cylinder is in static equilibrium, so the Σ ~ F = m~a = 0 and Σ τ = = 0. Choose coordinates as indicated on figure 1, with counterclockwise as positive for torque. Σ F x = f + Tcosθ - Mgsinθ = 0 (1) Σ F y = N - Tsinθ - Mgcosθ = 0 (2) Σ τ = fR - TR = 0 (3) We can use (3) and (1) to find the magnitude of the friction. From (3), we get f = T . Pluging into (1), We get Σ F x = f (1 + cosθ ) - Mgsinθ = 0 (4) which simplifies to f = Mgsinθ (1 + cosθ ) (5) 3. To find the minimum μ s that will prevent the block from slipping, use the fact that f μ s N . so the minumum will be when f = μ s N . The rest of is just a bunch of algebra, which is probably more involved than you will see on the midterm. Don’t worry too much if you couldn’t get it to simplify all the way. Plug N = f μ s and T = f into (2) N - Tsinθ - Mgcosθ = f ( 1 μ s - sinθ ) - Mgcosθ = 0 (6) Using (5), we get Mgsinθ (1 + cosθ ) ( 1 μ s - sinθ ) - Mgcosθ = 0 (7) or Mgsinθ μ s - Mgsin 2 θ - (1 + cosθ ) Mgcosθ = 0 (8) 1

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multiplying by μ s and dividing by Mg, we get sinθ - μ s sin 2 θ - μ s cosθ - μ s cos 2 θ = 0 (9) using the fact that sin 2 θ + cos 2 θ = 1, we get sinθ - μ s - μ s cosθ = 0 (10) and solving for μ s we get μ s = sinθ 1 + cosθ = 0 (11) Problem 2 1. ρ ρ l a c F mg b F ext Cube Figure 2: the forces acting on the block are from gravity and the bouyancy force. A downward external force must be applied to keep it submerged. To find the minimum force required us Σ ~ F = 0, which tells us F ext = F b - Mg = ρ l V disp g - ρ c V c g = ρ l a 3 g - ρ c a 3 g = a 3 g ( ρ l - ρ c ) (12) The external force is now coming from a volume of material being attached to it. From the free body diagram of the material, we see that T = m m g - F bm = m m g - ρ l V disp by m g (13) so using (12) with F ext = T and using the fact that m m = V m ρ m , we get m m g - ρ l V dispbym g = V m ρ m - ρ l V disp by m ga 3 g ( ρ l - ρ c ) (14) and solving for V m (assume that the material is fully submurged as well, so that V m = V disp by m ) V m = a 3 ( ρ l - ρ c ) ρ m - ρ l (15) 3. Without doing any calculations, in order for the system of the block and material to have
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ReviewSolutions - Physics 8A Midterm 2 Review Solutions...

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