Equilibrium Problem 4: Shown in the figure below is a 5 meter, 10 kg ladder lean
ID: 1461256 • Letter: E
Question
Equilibrium
Problem 4: Shown in the figure below is a 5 meter, 10 kg ladder leaning on a frictionless wall at an angle of 60 degrees to the floor. The COM for the ladder is located half-way up the ladder. The floor has a static coefficient of friction equal to 0.42. A person climbs the ladder and at some point the ladder slips out and falls.
a) Calculate the normal force of the floor on the ladder (with the person on the ladder).
b) Calculate the normal force of the wall on the ladder (with the person on the ladder).
c) Calculate the greatest distance up the ladder the person can climb without the ladder slipping.
Explanation / Answer
ans a.
normal force of the wall on the ladder (with the person on the ladder)
N = m*g + M*g =10* 9.8 + 85* 9.8
N= 98 +833 =931 Newtons.....................ans a.
torque caused by the wall on the ladder (call it ). The other torques are caused by the weight of the ladder () and the weight of the man ().
frictional force = Fs = µs * FN = 0.42 * 931 =391.02N
T1 = rwsin = 5 * 391.02 sin 60 =1693.17Nm.......................ans b.
T2 = rwsin ( The COM for the ladder is located half-way up the ladder so r1=2.5m ) and
T3 =rwsin ( let person is at x meter on the ladder so r2= x m.)
T2= (2.5m)(98N)sin60° and T3 = (x m)(833N)sin60°
we can sum the torques and set them equal to zero since it is in equilibrium. that is
1693.17+ 212.18 N+ 721.4 x N=0
x=1905.35 / 721.4 =2.64m ...................ans c.
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