An object with a mass of #4 kg# is lying still on a surface and is compressing a horizontal spring by #1/4 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
Jan 7, 2017

Answer:

#mu_s~~0.04#

Explanation:

We will solve this first by setting up a force diagram and writing out sum of forces statements.

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Where #vecn# is the normal force, #vecF_g# is the force of gravity, #vecF_(sp)# is the force of the spring, and #vecf_s# is the force of static friction. I will define rightward as positive.

Statements of the net force:

#sumF_x=vecF_(sp)-vecf_s=mveca_x#

#sumF_y=vecn-vecF_g=mveca_y#

The object is not moving, and experiences no net acceleration in either #x#- or #y#-direction.

#sumF_x=vecF_(sp)-vecf_s=cancel(mveca_x)=vec0#

#sumF_y=vecn-vecF_g=cancel(mveca_y)=vec0#

We can rearrange these statements to find:

#vecF_(sp)=vecf_s#

#vecn=vecF_g#

Where #vecF_g=mg#

The equation for the frictional force is #vecf=muvecn#. Therefore, we have:

#vecF_(sp)=mumg#

The equation for the spring force is given by Hooke's law as #vecF_(sp)=k(Deltas)#, where #k# is the spring constant and #Deltas# is the displacement from equilibrium (how far the spring is stretched or compressed).

#k(Deltas)=mumg#

Solving for #mu#, the coefficient of friction:

#mu=(k(Deltas))/(mg)#

Using our known values, we can calculuate #mu#:

#mu=((6N/m)(1/4m))/(4kg*9.8m/s^2)#

#mu=0.038...~~0.04#

Therefore, the minimum coefficient of static friction to keep the object still is #~~0.04#.