PRACTICE IT Use the worked example above to help you solve this problem. A uniform ladder...

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PRACTICE IT Use the worked example above to help you solve this problem. A uniform ladder I 10.0 m long and weighing W when i

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Answer 1

Answer #1

Normal by the ground N = W+mg = 52.9+71.8*9.8 = 756.54 N

Static friction at limiting case = uN = 0.383*756.54 = 289.75 N

normal by wall N2 = friction as net force in horizontal has to be zero

= 289.75 N

Now by torque equation about bottom end,

WL/2 * cos theta + mg*d* cos theta - N2*10*sin theta = 0

52.9*10/2*cos 64 degree + 71.8*9.8*d* cos 64 degree - 289.75*10*sin 64 degree = 0

d = 8.067 answer

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Answer 2

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