Problem 7.67 Part A A68-kg person stands on a 6.0-m-long 52-kg What distance d a

Question

Problem 7.67 Part A A68-kg person stands on a 6.0-m-long 52-kg What distance d along the ladder can the person pass without the ladder slipping? Express your answer to two significant figures and include the appropriate units. ladder. The ladder is tilted 60° above the horizontal. The coefficient of friction between the floor and the ladder is 0.38. The coefficient of friction between the wall and the ladder is zero. d= 1 Value Units Submit My Answers Give Up Incorrect; Try Again; 4 attempts remaining Provide Feedback Continue

Explanation / Answer

Normal force at the floor is equal to the combined weight of the person and ladder:
Fn = (68 + 52)kg * 9.8m/s²

Fn = 1176 N

friction force
f = µ*Fn

f = 0.38 * 1176N

f = 446.88 N

Normal force at the wall must be equal to the friction force at the floor
Fw = f = 446.88 N

Sum the moments about the base of the ladder
0 = Fw*6 m*sin60º - (68kg*d + 52kg*3m)*9.8m/s²*cos60º
0 = 446.88*6 *0.866 - (68*d + 52*3)*9.8*0.5
333.2d + 764.4 = 2321.98848
d = 4.67 m

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