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A wooden crate weighs 780lb.What force is needed to start the crate sliding on a wooden floor when the coefficient of starting...Asked by J
A wooden crate weighs 780 lb. What force is needed to start the crate sliding on a wooden floor when the coefficient of starting friction is 0.40?
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Answered by
Henry
Mass = 780Lbs * 0.454kg/Lb = 354.1 kg.
Wc = 354.1kg * 9.8N/kg = 3470. N.
Fp = 3470*sin0 = 0. = Force parallel to floor.
Fv = 3470*cos0 = 3470 N. = Force perpendicular to floor.
Fs = u*Fv = 0.4 * 3470 = 1388 N. = Force
of starting friction.
Fap-Fp-Fs = m*a.
Fap-0-1388 = m*0 = 0
Fap = 1388 N. = Force applied.
Wc = 354.1kg * 9.8N/kg = 3470. N.
Fp = 3470*sin0 = 0. = Force parallel to floor.
Fv = 3470*cos0 = 3470 N. = Force perpendicular to floor.
Fs = u*Fv = 0.4 * 3470 = 1388 N. = Force
of starting friction.
Fap-Fp-Fs = m*a.
Fap-0-1388 = m*0 = 0
Fap = 1388 N. = Force applied.
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