A rock climber, 70.0 kg, climbs along a fissure with is hands pushed against one

Question

A rock climber, 70.0 kg, climbs along a fissure with is hands pushed against one side of the fissure and his feet against the other as shown in the figure. The width w of the fissure is 1.08 m and the man's center of mass is at distance x-0.648 m from the right wall of the fissure. The coefficient of static friction between his hands and rock is ?1-0.750 and between his feet and rock is u21.17. What is the least horizontal push by the hands and feet that will keep the man stable? Submit Answer Tries 0/10 Using the horizontal push by the hands, what must be the vertical distance h between the feet and the hands for stability? Submit Answer Tries 0/10 Assuming the man climbs on a rainy day when 1- 0.300 and ?2 = 0.468, what is the new least horizontal push by the hands and feet that will keep the man stable? Submit Answer Tries 0/10 Using the new horizontal push by the hands, what must be the vertical distance h between the feet and the hands for stability? Submit Answer Tries 0/10

Explanation / Answer

(a) 700/(1.17+0.75) = 364N

(b) moment about either hands of feet should be equal to zero

we take about feet, uN*(1.08m) +Nh= Mg*(0.648m)

thus, 0.75*364*1.08 +364*h = 70*10*0.648

thus h = 0.436m

(c)

similarly

700/(0.3+0.468) = 911N

(d)

moment about either hands of feet should be equal to zero

we take about feet, uN*(1.08m) +Nh= Mg*(0.648m)

thus, 0.75*911*1.08 +911*h = 70*10*0.648

thus h = -0.312m (i.e. below his feet)

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