A 11.6-kg monkey climbs a uniform ladderwith weight w = 1.06 x 10^2 N and length

Question

A 11.6-kg monkey climbs a uniform ladderwith weight w = 1.06 x 10^2 N and length L = 2.80 m as shown in the figure below. The ladder restsagainst the wall at an angle of = 53 degree. Theupper and lower ends of the ladder rest on frictionless surfaces,with the lower end fastened to the wall by a horizontal rope thatis frayed and that can support a maximum tension of only 80.0 N. (a) Draw a free-body diagram for the ladder. (Do this on paper.Your instructor may ask you to turn in this work.) (b) Find the normal force exerted by the bottom of the ladder. 1 N (c) Find the tension in the rope when the monkey is two-thirds ofthe way up the ladder. 2 N (d) Find the maximum distance d that the monkey can climbup the ladder before the rope breaks. 3 m

Explanation / Answer

Fn = Wm +WL    normal force on floor = weight ofmonkey + weight of ladder Fn = 11.6 * 9.8 + 106 = 220 N Fw = T    normal force exerted bywall = tension in rope Fw L sin 53 = WL cos 53 (L /2) + Wm (2 L / 3) cos 53 (Taking torque about contact with floor) Fw = T = (WL / 2 + 2 Wm / 3)cos 53 / sin 53 = 97 N   for tension in rope Fw L sin 53 = WL cos 53 (L /2) + Wm d cos 53 (replacing 2 L / 3 by d the distance up theladder)   where Fw is now 80 N d = [Fw L sin 53 - WL cos 53 (L / 2) ] / (Wm cos 53) d = [80 * 2.8 * .80 - 106 * .60 * 1.4] / (114 * .60) = 1.32m

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.