Please Answer A 16.0-m uniform ladder weighing 480 N rests against a frictionles
ID: 2098523 • Letter: P
Question
Please Answer
A 16.0-m uniform ladder weighing 480 N rests against a frictionless wall. The ladder makes a 61.0 degree angle with the horizontal. Find the horizontal and vertical forces the ground exerts on the base of the ladder when an 830-N firefighter has climbed 4.10 m along the ladder from the bottom. Horizontal Force magnitude N direction y Vertical Force magnitude N direction If the ladder is just on the verge of slipping when the firefighter is 9.40 m from the bottom, what is the coefficient of static friction between ladder and ground?Explanation / Answer
a)
Horizontal reaction force = normal reaction from wall
Vertical reaction force = weight of ladder + weight of man
= 480 + 830 = 1310 N
moment about bottom is zero
N * 16 * sin(61) = 480 * 8 * cos(61) + 830 * 4.1 * cos(61)
N = 480 * 8 * cot(61) + 830 * 4.1 * cot(61))/16
= 250.93 N
horizontal reaction = 250.93N
----------------------------------------------------
(b)
vertical reaction is same = 1310 N
horizontal reaction = normal at wall
= (480 * 8 * cot(61) + 830 * 9.4 * cot(61))/16
= 403.33 N
coefficient of friction = horizontal reaction / vertical reaction
= 403.3 / 1310
= 0.308