1. 57. Analyzing forces at the knot (particularly helpful is a graphical view of the vector right-triangle with
horizontal “side” equal to the static friction force fs and vertical “side” equal to the weight W5 of the
5.0-kg mass), we find fs = W5 tan θ where θ = 30◦ . For fs to be at its maximum value, then it must
equal µs W1 0 where the weight of the 10 kg object is W1 0 = (10 kg)(9.8 m/s2 ). Therefore,
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µs W1 0 = W5 tan θ =⇒ µs = tan 30◦ = 0.29 .
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