A 14.0-m uniform ladder weighing 520 N rests against a frictionless wall. The la

Question

A 14.0-m uniform ladder weighing 520 N rests against a frictionless wall. The ladder makes a 61.0° 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 4.20 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 9.10 m from the bottom, what is the coefficient of static friction between ladder and ground?

Explanation / Answer

(a)
...the wall is frictionless so the force it exerts is only a push on the ladder which is a horizontal force parallel to the ground
...the floor has friction so the forces experienced by the lower part of the ladder is divided into x and y components
...equilibrium concepts apply since the ladder is not moving

for the magnitude of the force of the wall, we will use summation of moment equals zero... this is a tricky one because it involves trigonometry the equation goes like this (assume positive as clockwise)

let R be the magnitude of the force of the wall
M(t) = 430(4.2*cos51) + 520(7*cos61) - R(14sin61) = 0

then,

1136.55 + 1764.7 - R*12.245 = 0

R = 236.93 N

direction: to the left

for the horizontal component on ground, the forces acting on the x axis (horizontal) are the force of the wall which is R and Gx (ground horizontal) therefore, the force Gx = 236.93 N directed to the right or towards the wall

for the vertical component of the force, the weight of ladder and the man acted in the opposite direction of the force Gy (ground vertical) therefore, the force Gy = 520 + 830 = 1350 N directed upwards

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