A 14.0-m uniform ladder weighing 520 N rests against a frictionless wall. The la
ID: 1417027 • Letter: A
Question
A 14.0-m uniform ladder weighing 520 N rests against a frictionless wall. The ladder makes a 57.0° angle with the horizontal.
(a) Find the horizontal forces the ground exerts on the base of the ladder when an 830-N firefighter has climbed 4.20 m along the ladder from the bottom.
magnitude of horizontal: ___________
(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?
coeffecient of static friction: _________
Explanation / Answer
Using force balance
in x-direction
total Fx = 0, since it's not moving
0 = Ffriction - Nwall
Nwall = Ffriction
in y-direction
total force = 0
Nground - Wfire - Wladder = 0
Nground = 830 + 520 = 1150 N
using torque balance
torue in CCW = torque in CW
torque)wall = torque)ladder + torque)fire
Nw*cos 33 deg*14 = 830*4.2*sin 33 deg + 520*7.5*sin 33 deg
Nw = (830*4.2*sin 33 deg + 520*7*sin 33 deg)/(14*cos 33 deg)
Nw = 330.54 N to the left
Ff = 330.54 N to the right
Ng = 1350 N
B.
again same torque balance
torue in CCW = torque in CW
torque)wall = torque)ladder + torque)fire
Nw*cos 33 deg*14 = 830*9.2*sin 33 deg + 520*7.5*sin 33 deg
Nw = (830*9.2*sin 33 deg + 520*7*sin 33 deg)/(14*cos 33 deg)
Nw = 523.05
Nw = Ff
Ff = uk*Ng
uk*Ng = Nw
uk = 523.05/1350
uk = 0.387
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