Apply the second condition for equilibrium to the ladder, evaluating torques about an axis perpendicular to the page thr

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Apply The Second Condition For Equilibrium To The Ladder Evaluating Torques About An Axis Perpendicular To The Page Thr 1 (43.89 KiB) Viewed 9 times

Apply the second condition for equilibrium to the ladder, evaluating torques about an axis perpendicular to the page through 0: Solve for tan 0: Under the conditions that the ladder is just ready to slip, 8 becomes 8min and Pmax is given by Equation (5). Substitute: MASTER IT To = Pe sin e-mg Enter a number. ***** sin 8 cos 8 Additional 0 min -mg-cos 8 = 0 = tan 8 = mg8= tan 2P Finalize Notice that the angle depends only on the coefficient of friction, not on the mass or length of the ladder. = tan = 42.79 1-¹(2mg) = tan-¹(211) = tan-¹ [2(0.54)] 2P max. If the coefficient of static friction is 0.55, and the same ladder makes a 47° angle with respect to the horizontal, how far along the length of the ladder can a 56.0-kg person climb before the ladder begins to slip2.The.ladder.js 7.5 m in length and has a mass of 21 kg.