An object with a mass of #12 kg# is lying still on a surface and is compressing a horizontal spring by #1/3 m#. If the spring's constant is #6 (kg)/s^2#, what is the minimum value of the surface's coefficient of static friction?

1 Answer
Aug 16, 2017

#mu_s<=0.017#

Explanation:

We can use Newton's second law to solve.

Diagram:

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We have the following information:

  • #|->m=12" kg"#
  • #|->Deltas=-1//3" m"#
  • #|->k=6" kg"//"s"^2#
  • #|->g=9.81"m"//"s"^2#

#=>#Note that the displacement is negative for a compressed spring as the final position is more negative than the initial position.

We can sum the forces parallel and perpendicular:

#sumF_x=F_s-f_s=0#

#sumF_y=n-F_g=0#

  • Note that the net force in both directions is 0 as the object remains at rest. The system is in a state of static equilibrium.

We also know:

  • #f_(s"max")=mu_sn=>f_s<=mu_sn#

  • #F_s=-kDeltas#

  • #F_g=mg#

Substituting the above equations into the statements we formed above:

  • #n=mg#

  • #-kDeltas-mu_(s"max ")mg=0#

We can solve for #mu_(s"max")#:

#color(darkblue)(mu_(s"max ")=(-kDeltas)/(mg))#

Submitting in our known values:

#mu_(s"max ")=-((6" kg"//"s"^2)(-1//3" m"))/((12" kg")(9.81"m"//"s"^2)#

#=>~~0.017#

Therefore, #color(darkblue)(mu_s<=0.017)#.