Question

A 14.0 m uniform ladder weighing 490 N rests against a frictionless wall. The ladder makes a 63.00-angle with the horizontal. (a) Find the horizontal and vertical forces (in N) the ground exerts on the base of the ladder when an 830-N firefighter has climbed 3.80 m along the ladder from the bottom. horizontal force magnitude direction ---Select--- vertical force magnitude direction ---Select--- V (b) If the ladder is just on the verge of slipping when the firefighter is 9.20 m from the bottom, what is the coefficient of static friction between ladder and ground? (c) What If? If oil is spilled on the ground, causing the coefficient of static friction to drop to half the value found in part (b), what is the maximum distance (in m) the firefighter can climb along the ladder from the bottom before the ladder slips?

Answer 1

1 490N Ladder length 214m Weight of Ladder = 490N Angle made by ladder = 630 Firefighten weight = 830N Distance moved from Kase = 3.8 m centre of mon of ladder at rm, Net torque is zero. Etclockwise - Econliclockwise w 32 2200 Twall = tfirefighter & a) {Fu=0 ladder Ff tNwal = 0 Horizontal forces - Newall to R7 x 14 m = 8,3 0 NX 3-8 sing + Pgonx7 sinar Nwale X 12.474 # 1432 +1557:18 wall = 239.63 N to the left Fy = 239 63N to right, - vertical prus. {Fy = Nground - Where - Wradda 20 Ng = 830 +490 = 13820N b) Figure position = 9.2m co 97° Nwau X14 = 830X9.2 Sin27 +49 0x 7 sinen

18.474 Merian = 3466.67 +1557:18 Hwalu = 408.74 to left fr = 402.74 to Right For a ung M = Fr - 40274 -0.305 ing 1320 3 If u = halfed H = 0.1525 coil spiled on ground ) fr = 203.37 wall = 202.37 2027 x2022:37 X 14 = 830 x x x sinzzo +490x 2 simeo 2524:44 = 376.81 x + 1557.18 589.96 = 376 81 x [x = 1.565 m/