This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Massachusetts Institute of Technology  Physics Department Physics8.01 Fall 1999 Solutions for Assignment # 10 by Dru Renner dru@mit.edu December 8, 1999 9:54 am Problem 10.1 (a) We pull with a force F (both ropes combined). At the point O the sidewalk exerts an unknown force F on the cylinder. The condition for just lifting the cylinder off the street is N = 0 where N is the usual normal force the street exerts on the cylinder. We can eliminate F by measuring all torques about O . (A torque is defined as positive if it would cause the cylinder to roll up the sidewalk.) The condition for just rotating about the corner is τ = 0 where τ is the total torque about O . 0 = FR sin( π − ( θ + α )) − MgR sin µ π − µ π 2 − θ ¶ = FR sin( θ + α ) − MgR cos θ where we use the trigonometric identities sin( π − x ) = sin x and sin ³ π 2 + x ´ = cos x . Solving the above equation for F gives F Mg = cos θ sin( θ + α ) (b) A local minimum or maximum occurs when 0 = dF dα = − Mg · cos θ cos( θ + α ) sin 2 ( θ + α ) The only solution to the above equation is cos( θ + α ) = 0 = ⇒ α = 90 ◦ − θ For θ = 30 ◦ then α = 60 ◦ . By plotting F Mg we see that α = 60 ◦ is a local minimum and the global minimum and α = 0 is the global maximum (not local). It’s not hard to show that if θ < 45 ◦ then the maximum is at α = 0 ◦ , and if θ > 45 ◦ then the maximum is at α = 45 ◦ . 1 Problem 10.2 Let L = 5 m, r = 1 2 · . 01 m, A = πr 2 , M = 400 kg, and Y = 0 . 36 × 10 10 N / m 2 (please see Table 14.1 on page 366 for nylon). Equation (27) on page 366 gives ∆ L L = 1 Y F APP A = ⇒ F APP = Y A L · ∆ L where F APP is the total applied force and ∆ L is the corresponding increase in length. It is convenient to define a constant k as k = Y A L ≈ 5 . 7 × 10 4 N / m = ⇒ F APP = k ∆ L First we need to calculate the amount the rope is stretched, ∆, due to the weight of 400 kg alone. ∆ = Mg k ≈ . 070 m (a) We need to find the additional force required to stretch the rope another 0.03 m. F + Mg = k (∆ + 0 . 03) = ⇒ F = k · . 03 ≈ 1 . 78 × 10 3 N (b) Let x denote the displacement down from equilibrium. (Equilibrium corresponds to the rope length L + ∆.) The equation above gives the necessary applied force F APP to stretch the rope by ∆ L ; therefore the rope must exert a restoring force of equal and opposite magnitude. Therefore for a displacement of x down from equilibrium, now with the applied force not present, the total force is F TOT = − k (∆ + x ) + Mg = − kx where in the above, F is the total force down corresponding to the displacement x down from equilibrium....
View
Full
Document
This note was uploaded on 11/23/2009 for the course 640 373 taught by Professor Staff during the Spring '08 term at Rutgers.
 Spring '08
 Staff

Click to edit the document details