A uniform crate with a mass of 16.9 kg rests on ahorizontal floor with a coeffic

Question

A uniform crate with a mass of 16.9 kg rests on ahorizontal floor with a coefficient of static friction between thecrate and the floor (ms) of0.730. The crate is a uniform cube with sides 1.50 m inlength. What horizontal force applied to the top of the crate willinitiate tipping?

Please help. I think if I relate it w/no coefficient offriction the relationship would be 1.50m/2(1.50m)*mg but Im not sure how to factor infriction A uniform crate with a mass of 16.9 kg rests on ahorizontal floor with a coefficient of static friction between thecrate and the floor (ms) of0.730. The crate is a uniform cube with sides 1.50 m inlength. What horizontal force applied to the top of the crate willinitiate tipping?

Please help. I think if I relate it w/no coefficient offriction the relationship would be 1.50m/2(1.50m)*mg but Im not sure how to factor infriction

Explanation / Answer

weight of the crate W = mg = 16.9kg * 9.8m/sec^2 Normal force on the crate         N = W static frictional force on the crate       Fs = N = coefficient of static friction hence a minimum force of Fs , isneeded to just intiate tipping.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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