posted by garth on .
What is the minimum downward force on the box in the figure that will keep it from slipping? The coefficients of static and kinetic friction between the box and the floor are 0.38 and 0.28, respectively.
Given: Fs = 0.38, Fk = 0.28, m = ?.
Did you forget the mass?
Mass = 30kg(Assumed).
Wb = m*g = 30kg * 9.8N/kg = 294 N. =
Weight of box.
Fb = 294N @ 0 Deg. = Force of box.
Fp=294*sin(0)=0=Force parallel to floor
Fv = 294*cos(0) = 294 N. = Force perpendicular to floor.
Fn = Fp - u*Fv -u*Fv' = m*a. a = 0.
0 - 0.38*294 - 0.38Fv' = 0
-0.38Fv' = 0.38*294 = 111.72
Fv' = - 294 N.