Asked by Sarah
A taut string connects a 5-kg crate to a 12-kg crate (see figure below). The coefficient of static friction between the smaller crate and the floor is 0.573; the coefficient of static friction between the larger crate and the floor is 0.443. Find the minimum horizontal force required to start the crates in motion.
Answers
Answered by
Henry
Crate 1:
Wc = m*g = 5kg * 9.8N/kg = 49 N. = Wt.
of crate.
Fs = u * Wc = 0.573 * 49 = 28.08 N. =
Force of static friction.
Crate 2:
Wc = m*g = 12kg * 9.8N/kg = 117.6 N.
Fs = 0.443 * 117.6 = 52.1 N.
Fap-Fs1-Fs2 = m*a.
Fap-28.08-52.1 = m*0 = 0
Fap-80.2 = 0
Fap = 80.2 N. = Force applied.
Wc = m*g = 5kg * 9.8N/kg = 49 N. = Wt.
of crate.
Fs = u * Wc = 0.573 * 49 = 28.08 N. =
Force of static friction.
Crate 2:
Wc = m*g = 12kg * 9.8N/kg = 117.6 N.
Fs = 0.443 * 117.6 = 52.1 N.
Fap-Fs1-Fs2 = m*a.
Fap-28.08-52.1 = m*0 = 0
Fap-80.2 = 0
Fap = 80.2 N. = Force applied.