An object with a mass of #7 kg# is on a ramp at an incline of #pi/12 #. If the object is being pushed up the ramp with a force of # 8 N#, what is the minimum coefficient of static friction needed for the object to remain put?

Question

An object with a mass of #7 kg# is on a ramp at an incline of #pi/12  #.  If the object is being pushed up the ramp with a force of # 8 N#, what is the minimum coefficient of static friction needed for the object to remain put?

Emily Abbate · Accepted Answer

THe coefficient of static friction is #=-0.15#

Taking the direction up and parallel to the plane as positive #↗^+#

The coefficient of static friction is #mu_s=F_r/N#

The normal force is #=N=mgcostheta#

Consequently, the object's net force is

#F=F_r+Wsintheta#

#=F_r+mgsintheta#

#=mu_sN+mgsintheta#

#=mmu_sgcostheta+mgsintheta#

#=mg(mu_scostheta+sintheta)#

Thus,

#F/(mg)=(mu_scostheta+sintheta)#

#mu_scostheta=F/(mg)-sintheta#

#mu_s=1/costheta*(F/(mg)-sintheta)#

#=1/cos(pi/12)*(8/(7*9.8)-sin(pi/12))#

#=-0.15#

Benjamin Asiedu · Answer

undefined To determine the minimum coefficient of static friction needed for the object to remain put, we use the equation: $ \mu_s = \tan(\theta) $. Given that the incline angle $ \theta = \frac{\pi}{12} $ and the force pushing the object up the ramp $ F = 8 \, \text{N} $, we calculate:

$ \mu_s = \tan\left(\frac{\pi}{12}\right) = 0.267 $