Question

wall length of Ladder=7.om man on middle of ladder mass of man man Ladder's center of ma 2.5mm floor 3. 2 m Static Equilibrium - Ladder Comprehensive As shown in the figure, the top of a 7.0-m long ladder (mass 2 = 20 kg) leans against a frictionless wall. The bottom of the ladder rests on a rough floor, with 3.2 m from the wall. The distance from the ladder's center of mass to its bottom end is 2.5 m (note: this ladder is not uniform, therefore its center of mass is not at the midpoint). A man of mass 1 = 81 kg stands on the middle of the ladder. The diagram also shows all the relevant forces.

Answer 1

As per problem the man tot Person (mp=81K1) and Vet ladder cm = 2009). B AB= 7mn BC = 3.5m, AD = 2.5m OA= 3.2m In AOBA. (afhen Sino = 0 AB. __Sing = 3.2=0.4571_no = 27. 202° NAN SN hormal floor - Sing body in in equilibrium For vertical equilibrium. N = mpgtmag reaction from - I SP - Normal reaction from wall f friction fore from floor). N = (81+20)x9.& 3) N=989.8 N.! Since axis of rotation and Normal force poosing through pivot poin't A So torque produce by Nis TNZO. 9 (N=ON-m ? b) As the friction force of and the axis of rotation is - passing through point A, so Corque due to friction of

c) The torque produced by man is an Cm= mpg CA SINQ._ - Om =_BLXI.BX 3:5X 3-2 1Ş CA = 3.5m} Em = 1270.08 N-m, D) Torqe due to ladder about Ais , Et E lez meydasince = 20x9-2X 2-5X 3.2 k le = 224 N-mi LE) Sing the torque im 8 Ce are anti clock wise about L A 4and body is in rotational equilibrium, so forcel will provide the clock wire Coropue to keep the body in Tequilibrium. I Porque due to pis. Cp = PXOB. OBE OA tend = 3.2 fon (27202) 6.226 m. as let Im = 1270.06+ 224. = 1494.08 6.226.p. 1494.06. p=239.97 N _ A Nam Torque due to Pin ?p= 239, 97 x 6 226 = 1494.08 Sino body in in equilibrium in a direction -- Ref #= 23.1.97N