Equilibrium Problem 1-3: Chapter 12 Problems 10, 14, 28 Problem 4: Shown in the

Question

Equilibrium Problem 1-3: Chapter 12 Problems 10, 14, 28 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

M= mass of ladder , m= mass of person, L= length of ladder, x = distance of person from the bottom of the ladder

Applying Newton’s second law horizontally

Fw-Fx= 0

Fw=Fx-----------------(1)

Applying Newton’s second law vertically

Fy-Mg-mg=0

Fy= Mg+mg ----------------(2)

a) Fy= 10*9.8+85*9.8 = 931 N

b)Fx=us*Fy = 0.42*931 = 391 N

Fw= 391 N

c) Applying law of conservation of torque at pint of F

-Fw*h +Mg*b +mg*a +Fy*0+Fx*0 =0

-Fw*h +Mg*b +mg*a=0

-Fw*Lsinq +Mg*L/2*cosq+mg*x*cosq = 0

Plugging values,

-391*5*sin60+10*9.8*5/2*cos60+85*9.8*x*cos60 =0     => x= 3.8 m

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