A 87.9 kg clock initially at rest on a horizontal floor requires a 659.2 N horizontal force to set it in motion.
After the clock is in motion, a horizontal force of 556.0 N keeps it moving with a constant speed.
Find the coefficient of static friction of the clock on the ground, µs
+33665 The resistive force is given by μs*W, where W is the weight. The resistance is overcome a force of 659.2N is applied, so
μs*87.9*9.81 = 659.2
μs = 659.2/(87.9*9.81)
$${\frac{{\mathtt{659.2}}}{\left({\mathtt{87.9}}{\mathtt{\,\times\,}}{\mathtt{9.81}}\right)}} = {\mathtt{0.764\: \!468\: \!009\: \!356\: \!383\: \!3}}$$
or μs ≈ 0.764
+33665 The resistive force is given by μs*W, where W is the weight. The resistance is overcome a force of 659.2N is applied, so
μs*87.9*9.81 = 659.2
μs = 659.2/(87.9*9.81)
$${\frac{{\mathtt{659.2}}}{\left({\mathtt{87.9}}{\mathtt{\,\times\,}}{\mathtt{9.81}}\right)}} = {\mathtt{0.764\: \!468\: \!009\: \!356\: \!383\: \!3}}$$
or μs ≈ 0.764
