6.51.
Model: The box will be treated as a particle. Because the box slides down a vertical wood wall, we
will also use the model of kinetic friction.
Visualize:
Solve: The normal force due to the wall, Gwhich is perpendicular to the wall, is here to the right.
JG The
G box slides
G
down the wall at constant speed, so a = 0 and the box is in dynamic equilibrium. Thus, Fnet = 0. Newton’s
second law for this equilibrium situation is
( Fnet ) x = 0 N = n − Fpush cos 45°
( Fnet ) y = 0 N = f k + Fpush sin 45° − FG = f k + Fpush sin 45° − mg
The friction force is f k = μ k n. Using the x-equation to get an expression for n, we see that f k = μ k Fpush cos 45°.
Substituting this into the y-equation and using Table 6.1 to find μ k = 0.20 gives,
μ k Fpush cos 45° + Fpush sin 45° − mg = 0 N
⇒ Fpush =
( 2.0 kg ) ( 9.80 m/s 2 )
mg
=
= 23 N
μ k cos 45° + sin 45° 0.20cos 45° + sin 45°