Question

A 14.0-kg chair is at rest on a flat floor. The coefficient of static friction between
the chair and the floor is 0.450. A person tries to move the chair by pushing on the
chair with a force directed at an angle of 30„a below the horizontal. What is the
minimum force that the person must apply in order to move the chair?

Answer 1

Wc = m*g = 14kg * 9.8N/kg = 137.2 N. =
Wt. of chair.

Fc = 137.2 N. @ 0o = Force of chair.
Fp = 137.2*sin(0) = 0 = Force Parallel
with floor.
Fv=137.2*cos(0) + Fap*sin(-30)
Fv=137.2 - 0.5Fap. =Force perpendicular to floor. = Normal.

Fn = Fap*cos(-30)-Fp-Fs = m*a.
Fap-0-0.45(137.2-0.5Fap) = m*0 = 0
Fap-61.74+0.225Fap = 0
0.775Fap-61.74 = 0
0.775Fap = 61.74
Fap = 79.7 N @ (-30o).