An 81 kg person stands on a uniform 9.0 kg ladder that is 4.0 m long, as shown i

Question

An 81 kg person stands on a uniform 9.0 kg ladder that is 4.0 m long, as shown in the figure(Figure 1) . The floor is rough; hence, it exerts both a normal force, f1, and a frictional force, f2, on the ladder. The wall, on the other hand, is frictionless; it exerts only a normal force, f3.

A) Using the dimensions given in the figure, find the magnitudes of f1

B) Using the dimensions given in the figure, find the magnitudes of f2

C) Using the dimensions given in the figure, find the magnitudes of f3

Explanation / Answer

here,

Height of ladder = a = 3.8m
mass of person = mp = 81 Kg
mass of Ladder = ml = 9.0 kg
length of ladder = 4.0 m

the angle that the ladder makes with the floor :

a = Lsin
= arcSin(a/l)
= arcSin(3.8/4)
= 71.80 degrees

To find the length l that the person is up the ladder :

b = Lp*Cos
Lp = b/Cos
Lp = 0.70/cos71.80
Lp = 0.70/0.312
Lp = 2.24 m

Talking Torque abot Point O i.e point of contant of ladder and wall :

tf3 = L*f3*Sin71.80

also for person,
tmpg = -Lp*mp*g*Cos71.80

tmlg = -L/2 *ml*g*Cos71.80

As Tnet = tf3 + tmpg + tmlg = 0

L*f3*Sin71.80 -Lp*mp*g*Cos71.80 -L/2 *ml*g*Cos71.80 =0

f3 = ( (lp*mp + (L/2)*ml)*gCos71.80 ) / (L*sin71.8)

f3 = ( (2.24*81 + (4/2)*9)*9.8*0.312 ) / (4*0.94)

f3 = 162.18 N

From Newton Secon law :
Sum(Fx) = 0
f2 - f3 = 0
f2 = f3 = 162.18 N

also,for y direction
Sun(Fy) = 0
f1 - mp*g - ml*g
f1 = g(mp+ ml)
f1 = 9.8(81+9)
f1 = 882 N

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