Insane Problem I think :( A ladder (IL = 8.40 m) of weight WL = 335 N leans agai
ID: 2215452 • Letter: I
Question
Insane Problem I think :(
A ladder (IL = 8.40 m) of weight WL = 335 N leans against a smooth vertical wall. The term "smooth" means that the wall can exert only a normal force directed perpendicular to the wall and cannot exert a frictional force parallel to it. A firefighter, whose weight is 866 N, stands 6.20 m up from the bottom of the ladder (this distance goes along the ladder, it is not the vertical height). Assume that the ladder's weight acts at the ladder's center, and neglect the hose's weight. What is the minimum value for the coefficient of static friction between the ladder and the ground, so that the ladder (with the fireman on it) does not slip? (Assume 6 = 48.0 degree.)Explanation / Answer
distance of center of ladder from ground point horizontally = (8.40/2)(cos 48) = 2.81m similarly distance where fireman weight acts = 6.2* cos48 = 4.15m let normal force by wall be F making moment at the poin of contact of ladder and ground 0 we get 2.81*335+866*4.15 = 8.4 cos48 *F = 5.62F so F = 941.35+3593.9 /5.62 = 806.98N normal force at point of contact with ground = sum of vertical forces = 335+866 = 1201N so uN = 806.98 so u*1201 = 806.98 so u = 0.67 (the procedure is absolutely correct, it there is any discrepency with answer check calculations once, and please comment if u have any doubts i can clear them and rate lifesaver, because if u rate some other u cannot change it)