A 16.0-m uniform ladder weighing 520 N rests against a frictionless wall. The la
ID: 1793459 • Letter: A
Question
A 16.0-m uniform ladder weighing 520 N rests against a frictionless wall. The ladder makes a 65.0° angle with the horizontal.
(a) Find the horizontal and vertical forces the ground exerts on the base of the ladder when an 830-N firefighter has climbed 3.90 m along the ladder from the bottom.
Horizontal Force
Vertical Force
(b) If the ladder is just on the verge of slipping when the firefighter is 8.80 m from the bottom, what is the coefficient of static friction between ladder and ground?
Explanation / Answer
(a)
In equilibrium net torque about the base = 0
Fh*L*sintheta - Wladder*L/2*costheta - Wman*l*costheta = 0
Fh*16*sin65 - 520*16/2*cos65 - 830*3.9*cos65 = 0
Fh = 215.6 N
along vertical net force = 0
Fnet,y = 0
Fv - Wladder - Wman = 0
Fv = 520 + 830 = 1350 N
========================
part (b)
In equilibrium net torque about the base = 0
Fh*L*sintheta - Wladder*L/2*costheta - Wman*l*costheta = 0
Fh*16*sin65 - 520*16/2*cos65 - 830*8.8*cos65 = 0
Fh = 334.1 N
along horizontal Fx net = 0
fs - Fh = 0
fs = us*Fv
us*Fv - Fh = 0
us = Fh/Fv = 334.1/1350
coefficient of static friction us = 0.248 <<<------ANSWER
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