An object with a mass of #9 kg# is on a plane with an incline of # - pi/4 #. If it takes #15 N# to start pushing the object down the plane and #5 N# to keep pushing it, what are the coefficients of static and kinetic friction?

Question

An object with a mass of #9  kg# is on a plane with an incline of # - pi/4  #.  If it takes #15   N# to start pushing the object down the plane and #5  N# to keep pushing it, what are the coefficients of static and kinetic friction?

Benjamin Asiedu · Accepted Answer

#\mu_s = (F_{app}^s/(mg\cos heta) + an heta) = 1.24#
#\mu_k = (F_{app}^k/(mg\cos heta) + an heta) = 1.08# Consider a body of mass #m# on an incline with an inclination angle of # heta#. To this a force is applied without disturbing the equilibrium. The forces acting on the object are #vec W # : Weight of the object acting vertically downward; #vec F_{app} # : Force applied parallel to the inclined plane, #vecF_f# : Frictional force acting parallel to the plane of the incline, #vec N # : Normal force acting perpendicular to the plane of the incline, Consider a coordinate system with its X-axis parallel to the plane downward and Y-axis perpendicular to the plane upward. Resolve the weight into components that are parallel and perpendicular to the inclined plane #vec W_{||} = -mg\sin heta; \qquad vec W_{\_|_} = -mg\cos heta# Equilibrium: #\qquad vec W + vec F_{app} + vec N + vec F_f = vec 0# (A) Perpendicular Component: #\qquad vec N + vec W_{\_|_} = vec 0# #N - W_{\_|_} = 0; \qquad N = mg\cos heta;# (B) Parallel Component: #\qquad vec W_{||} + F_{app} + vec F_f = vec 0# #W_{||} + F_{app} - F_f = 0; \qquad F_f = F_{app} + W_{||} # ...... (2) Static Equilibrium: In this case the frictional force is due to static friction. #F_f = \mu_sN = \mu_s.mg\cos heta; \qquad F_{app} = F_{app}^s = 15# #N;# Expanding (2), #\mu_s.mg\cos heta = F_{app}^s + mg\sin heta# #\mu_s = (F_{app}^s/(mg\cos heta) + an heta)# ...... (3) Dynamic Equilibrium: In this case the frictional force is due to kinetic friction. #F_f = \mu_kN = \mu_k.mg\cos heta; \qquad F_{app} = F_{app}^k = 5# #N;# Expanding (2), #\mu_k.mg\cos heta = F_{app}^k + mg\sin heta# #\mu_k = (F_{app}^k/(mg\cos heta) + an heta)# ...... (4) Calculate: #\sin heta = \cos heta = 1/\sqrt{2}; \qquad an heta = 1; \qquad mg\cos heta = 62.37# #N# #mu_s = 1.24; \qquad \mu_k = 1.08#

Parker Abbott · Answer

undefined The coefficient of static friction is μs = 0.50 and the coefficient of kinetic friction is μk = 0.17.