Auniform ladder of mass 25.0 kg and length 7.5 m is leaned against a smooth vertical wall. a person of mass 65.0 kg stands on the ladder a distance 1.5 m from the bottom; as measured along the ladder. the foot of the ladder is 1.5 m from the bottom of the wall. the exerted by the wall on the ladder is : (a) 31.5 n (b) 53.1 n (c) 51.3 n (d) 35.1 n

(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

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

12000j is work done