Asked by mike
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?
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?
Answers
Answered by
Henry
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).
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).
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