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?

Answer 1

#mu_(s*"min")=0.2041

Explanation:

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`