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nolangriffin
14.12.2019 •
Physics
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?
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Ответ:
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
Ответ:
sorry I don't really understand you ^^