An object with a mass of #5 kg# is on a plane with an incline of # - pi/8 #. If it takes #18 N# to start pushing the object down the plane and #3 N# to keep pushing it, what are the coefficients of static and kinetic friction?

1 Answer
Feb 8, 2016

Answer:

Coefficient of static friction : #\mu_s = 0.8188#
Coefficient of kinetic friction : #\mu_k = 0.4805#

Explanation:

Step1 : Draw a free body diagram showing all these forces.
(1) Gravitational force (weight) : #\vec{w_{}}#, #\qquad# (2) Normal force : #\vec{N_{}}#,
(3) Frictional force : #\vec{F_f}#, #\qquad# (4) Applied Force : #\vec{F_{}}_{app}#.

Step 2 : Pick a 2D cartesian system with its X axis parallel to the incline and directed downward and its Y axis perpendicular to the plane.

Step 3: Resolve these forces into their X and Y components carefully assigning the signs:
#\vec{w_{}}=\vec{w_x}+\vec{w_y}#;
#\vec{w_{x}}=(+mgsin\theta,0)#; #\quad# #\vec{w_{y}}=(0, -mgcos\theta)#;
#\vec{N_{}}=(0, N)#; #\quad# #\vec{F_{f}}=(-\muN, 0)#; #\quad# #\vec{F_{}}_{app}=(F, 0)#

Step 4: Equilibrium condition is that the net force acting on the object must be zero. Apply this condition by adding the vectors component-wise and set each component separately to zero.

Net Force : #\vec{F_{N}} = \vec{w_{}}+\vec{N_{}}+\vec{F_{f}}+\vec{F_{}}_{app}=(0, 0)#

#\vec{F_{N}}=(mgsin\theta,-mgcos\theta)+(0, N)+(-\mu N, 0)+(F, 0)#
#\qquad##=(mgsin\theta-\muN+\F, N-mgcos\theta)=(0, 0)#

Y Component: #N-mgcos\theta=0 \rightarrow N = mgcos\theta##\quad# (Eqn 1)
X Component: #mgsin\theta-\muN+\F=0 \rightarrow \mu N = mgsin\theta-F#
Therfore, #\mu = \frac{mgsin\theta+F}{N}=\frac{mgsin\theta+F}{mgcos\theta}=tan\theta+\frac{F}{mgcos\theta}#

Static Equilibrium: When the vector sum of these forces is zero and the object is at rest, the object is in static-equilibrium and the friction in that cases is the static friction. The coefficient of friction is the static friction coefficient (#\mu\rightarrow\mu_s#)

Dynamics Equilibrium: When the vector sum of these forces is zero and the object slides with a constant velocity, the object is in dynamic equilibrium and the friction in that case is kinetic friction.The coefficient of friction is the coefficient of kinetic friction,
(#\mu\rightarrow\mu_k#)
#\mu_s = tan\theta+\frac{F}{mgcos\theta}=0.8188#
#\mu_k = tan\theta+\frac{F}{mgcos\theta}=0.4805#

#\theta=\pi/8 rad#; #\qquad# #m=5kg#; #\qquad# #g=9.8ms^{-2}#

#F=18N# (static) #\qquad# #F=3N# (kinetic)