Saturday
March 25, 2017

Post a New Question

Posted by on .

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.

  • Physics 1 - ,

    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.

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question