QuestionDetails: \" Force F = (-8.0n)i (hat) + (6.0N) j (hat) acts on a particle

Question

QuestionDetails: " Force F = (-8.0n)i (hat) + (6.0N) j (hat) acts on a particlewith position vector r = (3.0m) i (hat) + (4.0m)j(hat).   What is the angle between the directions of rand F??   what is the torque? I just need to know the angle between the twocomponents! Thank you! QuestionDetails: " Force F = (-8.0n)i (hat) + (6.0N) j (hat) acts on a particlewith position vector r = (3.0m) i (hat) + (4.0m)j(hat).   What is the angle between the directions of rand F??   what is the torque? I just need to know the angle between the twocomponents! Thank you! QuestionDetails: " Force F = (-8.0n)i (hat) + (6.0N) j (hat) acts on a particlewith position vector r = (3.0m) i (hat) + (4.0m)j(hat).   What is the angle between the directions of rand F??   what is the torque? I just need to know the angle between the twocomponents! Thank you!

Explanation / Answer

F = (-8.0n)i (hat) + (6.0N) j (hat), magnitude =[(-8)2 + 62] = 10 N r = (3.0m) i (hat) + (4.0m)j (hat), magnitude =(32 + 42) = 5 m angle between F and r is cos = (r dot F)/(10*5) = [3*(-8) + 4*6]/50 = 0, so =90o torque = r cross F = [3*6 - (-8)*4] N-m k(hat) = 68 N-m k(hat)

Related Questions

Navigate

• Browse (All)
• Browse Q
• Subjects

---

---

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