Ask Question
24 October, 14:01

A 40 N crate rests on a rough horizontal floor. A 12 N horizontal force is then applied to it. If the coecients of friction are μs=0.5 and μk=0.4, the magnitude of the frictional force on the crate is:

+1
Answers (1)
  1. 24 October, 17:32
    0
    Fr = 20 (N)

    Explanation: See atached file (free body diagram)

    As for Newton's low

    ∑ Fy = 0

    -mg + N = 0 ⇒ - 40 + N = 0 ⇒ N = 40 [Newtons]

    by definition : Fr = μs * N ⇒ Fr = 0,5 * 40 ⇒ Fr = 20 (N)

    ∑ Fx = 0 body is at rest

    Fe - Fr = 0

    Fr > Fe

    Fr > 12 (N) the body is at rest
Know the Answer?
Not Sure About the Answer?
Find an answer to your question ✅ “A 40 N crate rests on a rough horizontal floor. A 12 N horizontal force is then applied to it. If the coecients of friction are μs=0.5 and ...” in 📘 Physics if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers