dockworker loading crates on a ship finds that a 22 kg crate, initially at rest

Question

dockworker loading crates on a ship finds that a 22 kg crate, initially at rest on a horizontal surface, requires a 78 N horizontal force to set it in motion. However, after the crate is in motion, a horizontal force of 53 N is required to keep it moving with a constant speed. Find the coefficients of static and kinetic friction between crate and floor.

Explanation / Answer

Since the crate weighs 22kg, you can find the normal force: Fn = ma --> Normal Force = (22)(9.8) --> Normal Force = 215.6 N The force used to set the box in motion is the applied force against the static friction holding the box in place. To find the static friction coefficient you do this: Applied Force = µs(Normal Force) 78 / 215.6 = µs(215.6) / 215.6 µs = .36 Static friction coefficient = .36 The second applied force is less because it has overcome the static friction and is now being acted on by the kinetic friction: Applied Force = µk(Normal Force) 53 = µk(215.6) µk = .274 Kinetic friction coefficient = .245

Related Questions

Navigate

• Browse (All)
• Browse D
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.