A 9.1-kg monkey climbs a uniform ladder with weight w = 1.22 102Nand length L =

Question

A 9.1-kg monkey climbs a uniform ladder with weight w = 1.22 102Nand length L = 3.50 m as shown in the figure below. Theladderrests against the wall at an angle of = 53°. Theupperand lower ends of the ladder rest on frictionless surfaces,withthe lower end fastened to the wall by a horizontal rope thatisfrayed and that can support a maximum tension of only 80.0N.
Find the normal force exerted by the bottom of the ladder.
Find the tension in the rope when the monkey is two-thirds oftheway up the ladder.
Find the maximum distance d that the monkey can climb up theladderbefore the rope breaks.

Explanation / Answer

Fn = Wm +WL normal force on floor = weight ofmonkey + weight ofladder Fn = 9.10 * 9.8 + 122 =211.18 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 = 91 N for tension inrope 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 now80 N d = [Fw L sin 53 - WL cos 53 (L / 2) ] / (Wm cos 53) d = [80 * 3.5* .80 - 122 * .60 * 1.75] / (89.18 * .60) = 1.7922m

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