Answer:
(c) F= 51.3 N : Force exerted by the wall on the ladder
Explanation:
We apply Newton's first law to balance moments (M) at the point o at the bottom of the ladder, taking positive the moments that go counterclockwise and negative the moments that go clockwise
ΣMo=0
W₁*d₁+ W₂*d₂-F*d=0 Equation (1)
W₁ = ladder Weight (N)
W₂= person Weight (N)
F= force exerted by the wall on the ladder (N)
d₁ = distance from W₁ to point o
d₂ = distance from W₂ to point a
d₃ = distance from F to point o
Data
m₁ = 25.0 kg : ladder mass
m₂ = 65.0 kg : person mass
g = 9.8 m/s² : acceleration due to gravity
We calculate W₁ and W₂ :
W₁=m₁*g= 25.0 kg*9.8 m/s = 245 N
W₂=m₂*g= 65.0 kg*9.8 m/s = 637 N
We calculate d₁, d₂ and d₃
cosβ=( 1.5) / (7.5)
d₁= 3.75*cosβ = 3.75*( 1.5 / 7.5)=0.75m
d₂= 1.5*cosβ = 1.5*( 1.5 / 7.5) = 0.3m
[tex]d_{3} =\sqrt{(7.5)^{2}-(1.5)^{2} } = 7.3m[/tex]
look at the free body diagram of the ladder in the attached graphic
In the equation (1):
W₁*d₁+ W₂*d₂-F*d₃=0
245*( 0.75)+637*(0.3)- F*(7.3) = 0
374.85 = F*(7.3)
F= (374.85) / (7.3)
F= 51.3 N
