Posted by **garth** on Thursday, June 7, 2012 at 8:27pm.

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.

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**Henry**, Friday, June 8, 2012 at 1:10am
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.

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