An object with a mass of # 8 kg# is on a ramp at an incline of #pi/8 #. If the object is being pushed up the ramp with a force of # 9 N#, what is the minimum coefficient of static friction needed for the object to remain put?

1 Answer
Jul 21, 2017

#mu_s >= 0.290#

Explanation:

We're asked to find the minimum value of the coefficient of static friction #mu_s# between the incline and object so that the object remains stationary.

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Since its mass is given as #8# #"kg"#,

#mgsintheta = (8color(white)(l)"kg")(9.81color(white)(l)"m/s"^2)sin(pi/8) = 30.0# #"N"#

The normal force magnitude #n# is

#n = mgcostheta = (8color(white)(l)"kg")(9.81color(white)(l)"m/s"^2)cos(pi/8) = 72.5# #"N"#

Since the object is supposed to be stationary, the body is in equilibrium (sum of all forces acting on body equals zero), so (taking positive #x# direction as up the incline)

#sumF_x = F_"applied" + f_s - mgsintheta = 0#

So

#f_s = mgsintheta - F_"applied"#

The applied force is given as #9# #"N"#, so we have

#f_s = 30.0# #"N"# #- 9# #"N"# = #21.0# #"N"#

Now that we know the static friction force #f_s#, we can figure out the coefficient of static friction using the equation

#f_s <= mu_sn#

#mu_s >= (f_s)/n = (21.0color(white)(l)"N")/(72.5color(white)(l)"N") = color(red)(0.290#