An object with a mass of #10 kg# is lying still on a surface and is compressing a horizontal spring by #5/6 m#. If the spring's constant is #24 (kg)/s^2#, what is the minimum value of the surface's coefficient of static friction?

Question

An object with a mass of #10   kg# is lying still on a surface and is compressing a horizontal spring by #5/6   m#. If the spring's constant is #24    (kg)/s^2#, what is the minimum value of the surface's coefficient of static friction?

Adrian Ayala · Accepted Answer

#mu_(s*"min")=0.2041

If an object is lying still then that means the acceleration in any direction is #0"m/s"^2#. Since #veca=(SigmavecF)/m# this means the net force in any direction is #0"N"#. That means the force of static friction has to be equal in magnitude and opposite in direction to the force exerted by the spring because the horizontal net force is #0"N"#. The force exerted by a spring is given by #vec(F_s)="-"kvecx# where #k# is the spring constant and #x# is the displacement from equilibrium. Plugging in the values from the problem, #vec(F_s)="-"(24"kg/s"^2)(5/6"m")="-"20"N"#, so the force of static friction #vec(F_f)=20"N"#. The magnitude of the force of static friction is defined as #F_f=mu_sF_N#, so #mu_(s*"min")=F_f/F_N#. Since the acceleration in the vertical direction is #0"m/s"^2#, the net force in the vertical direction is #0"N"#, therefore the normal force is equal in magnitude and opposite in direction to the force of gravity. Since #vec(F_g)="-"mg#, #vec(F_N)=mg#. #mu_(s*"min")=F_f/(mg)=(20"N")/((10"kg")(9.8"m/s"^2))=0.2041#

Nathan Ashcroft · Answer

undefined The minimum value of the surface's coefficient of static friction is 0.4.