A uniform ladder stands on a rough floor and rests against a frictionless wall a
ID: 1783902 • Letter: A
Question
A uniform ladder stands on a rough floor and rests against a frictionless wall as shown in the figure.
Since the floor is rough, it exerts both a normal force
N1
and a frictional force
f1
on the ladder. However, since the wall is frictionless, it exerts only a normal force
N2
on the ladder. The ladder has a length of
L = 4.6 m,
a weight of
WL = 65.0 N,
and rests against the wall a distance
d = 3.75 m
above the floor. If a person with a mass of
m = 90 kg
is standing on the ladder, determine the following.
(a) the forces exerted on the ladder when the person is halfway up the ladder (Enter the magnitude only.)
(b) the forces exerted on the ladder when the person is three-fourths of the way up the ladder (Enter the magnitude only.)
Explanation / Answer
In equilibrium along vertical
net force = 0
N1 - WL - Wman = 0
N1 = WL + Wman = WL + mg
N1 = 65 + (90*9.8)
N1 = 947 N <<<-----ANSWER
----------------------------------------
let theta be the angle made by ladder with floor
sintheta = d/L
theta = sin^-1(d/L) = 54.6 degrees
In equilibrium net torque = 0
N2*d - WL*L/2*costheta - mg*L/2*costheta = 0
N2*3.75 - (65*4.6/2*cos54.6) - (90*9.8*4.6/2*cos54.6) = 0
N2 = 336.5 N <<<-----ANSWER
-----------------------------------------
along horizontal
Fnet = 0
f1 - N2 = 0
f1 = N2 = 336.5 N <<<-----ANSWER
======================================
(b)
In equilibrium along vertical
net force = 0
N1 - WL - Wman = 0
N1 = WL + Wman = WL + mg
N1 = 65 + (90*9.8)
N1 = 947 N <<<-----ANSWER
--------------------------------------
let theta be the angle made by ladder with floor
sintheta = d/L
theta = sin^-1(d/L) = 54.6 degrees
In equilibrium net torque = 0
N2*d - WL*L/2*costheta - mg*3L/4*costheta = 0
N2*3.75 - (65*4.6/2*cos54.6) - (90*9.8*3*4.6/4*cos54.6) = 0
N2 = 493.1 N
along horizontal
Fnet = 0
f1 - N2 = 0
f1 = N2 = 493.1 N
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