A 40-N crate rests on a rough horizontal floor. A 12-N horizontal force is then applied to it. If
the coefficients of friction are μs = 0.5 and μk = 0.4, the magnitude of the frictional force on
the crate is:
1
Expert's answer
2019-10-23T09:53:28-0400
The static friction
"F_s=\\mu_sN=0.5\\times 40\\:\\rm N=20\\: N"
The kinetic friction
"F_k=\\mu_kN=0.4\\times 40\\:\\rm N=16\\: N"
Since the horizontal force applied to the crate is less then static friction "(F<F_s)," so the crate is at rest.
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