Auniform ladder 7.0 m long weighing 450 n rests with one end on the ground and the other end against a perfectly smooth vertical wall. the ladder rises at a 60.0 degrees above the horizontal floor. a 750 n painter finds that she can climb 2.75 m up the ladder measured along its length, before it begins to slip. what is the friction force and normal force that the floor exerts on the ladder?

fs = 300.03 N

FN = 1200N

Explanation

The equilibrium equation are:

∑Fx=0

∑Fy=0

∑M = 0  

M = F*d  

Where:  

∑M : Algebraic sum of moments

M : moment  ( N*m)  

F : Force ( N)  

d :Perpendicular distance of the force to the point  ( N*m )

Forces acting on the ladder

Look at the attached graph of the free body diagram of the ladder

W₁ =450 N : Weight of the ladder (vertical downward)  

W₂ =750 N : Weight of the painter (vertical downward)  

FN :Normal force that the floor exerts on the ladder (vertical upward)  

fs : friction force that the floor exerts on the ladder (vertical downward)  

N :  Normal Force that the wall exerts on the ladder (point A)

Calculation of the distances of the forces at the point B (contact point of the ladder on the floor)

d₁ = 3.5*cos60° (m):  Distance from W₁ to the point B

d₂ = 2.75*cos60° (m)  : Distance from W₂ to the point B

d₃ = 7*sin60° (m) : Distance from N to the point B

The equilibrium equation of the moments at the point B  (contact point of the ladder with the floor)  

∑MB = 0  

N( d₃) - W₁( d₁) - W₂(d₂) = 0

N(d₃) = W₁(d₁) + W₂(d₂)

N(7*sin60°) = (450)(3.5*cos60°) + (750)( 2.75*cos60°)

N(6.062) = (450)(1.75) + (750)( 1.375)

N(6.062) = 787.5 + 1031.25)

N = (1818.75) / (6.062)  

N = 300.03 N

∑Fx=0

fs - N = 0

fs = N

fs = 300.03 N

∑Fy=0

FN -W₁ -W₂= 0

FN = W₁ + W₂

FN = 450N+ 750N

FN = 1200N

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