Three crates with various contents are being pulled by a 3695-N force across a h

Question

Three crates with various contents are being pulled by a 3695-N force across a horizontal, frictionless roller- conveyor system. The group of boxes is accelerating on the conveyor at 1.408 m/s2 to the right. Between each adjacent pair of boxes is a force meter that measures the magnitude of the tension in the connecting rope. Assume that the ropes and force meters are massless. I13 3695 N 2365 N 1401 N What is the total mass of the three boxes? Number Kg 11t otal nm3 What is the mass of each box? nt2 iit Number Number Number kg kg kg

Explanation / Answer

according to newton's second law of motion

Net force is Fnet = mtotal*a

3695 = mtotal*1.408

mtotal = 3695/1.408= 2624.3 Kg

for mass m3

Fnet = 3695 - 2365

m3*a = 1330

m3 = 1330/1.408 = 944.6 Kg

for mass m2

Fnet = 2365 - 1402

m2*a = 963

m2 = 963/1.408 = 684 Kg

for mass m1

Fnet = 1401

m1*a = 1401

m1= 1401/1.408 = 995 Kg

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

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