(Notice I wrote in the coefficient of kinetic friction is .3 on the right) A bod

Question

(Notice I wrote in the coefficient of kinetic friction is .3 on the right)

A body of mass 30 kg rests on a horizontal surface where the coefficient of static friction is 0 4. David is pulling upwards (20degree up from the horizontal) and to the right with a force P. Lane is pushing downwards (30degree from the horizontal) and to the right with a force of 30 N. How much does David have to pull to just start the body moving? Once the body begins to move to the right. Can David reduce his pull to keep the body moving at constant velocity? To what value of P can David reduce his pull and keep the body moving at constant velocity?

Explanation / Answer

force due to static friction = us * m * g = 0.4 * 30 * 9.8 = 117.6 N
horizontal force by Lane = Fhl = 30 cos 30 = 26 N
So david horizontal force = Fhd = 117.6 - 26 = 91.6

=> Fd cos20 = 91.6
=> Fd = 97.47 N

b) yes david can reduce his pull to keep the body moving at constant velocity, because at that kinetic friction will act, whose value is less than static friction
now friction force = 0.3 * 30 * 9.8 = 88.2 N
Fd cos 20 = 88.2 - 26 (that is lane horizontal force subtracted)

Fd = 66.2 N

so David can reduce his pull by 97.47 - 66.2 = 31.27 N

Related Questions

Navigate

• Browse (All)
• Browse #
• Subjects

---

---

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