Question

Three crates with various contents are pulled by a force Fpull=3615 N across a horizontal, frictionless roller‑conveyor system. The group of boxes accelerates at 1.516 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. Between the box of mass m1 and the box of mass m2, the force meter reads F12=1387 N. Between the box of mass m2 and the box of mass m3, the force meter reads F23=2304 N. Assume that the ropes and force meters are massless.

Answer 1

The question is incomplete. Here is the complete question.

Three crtaes with various contents are pulled by a force Fpull=3615N across a horizontal, frictionless roller-conveyor system.The group pf boxes accelerates at 1.516m/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. Between the box of mass m1 and the box of mass m2, the force meter reads F12=1387N. Between the box of mass m2 and box of mass m3, the force meter reads F23=2304N. Assume that the ropes and force meters are massless.

(a) What is the total mass of the three boxes?

(b) What is the mass of each box?

Answer: (a) Total mass = 2384.5kg;

(b) m1 = 915kg;

m2 = 605kg;

m3 = 864.5kg;

Explanation: The image of the boxes is described in the picture below.

(a) The system is moving at a constant acceleration and with a force Fpull. Using Newton's 2nd Law:

`F_(pull)=m_(T).a`

`m_(T)=(F_(pull))/(a)`

`m_(T)=(3615)/(1.516)`

`m_(T)=2384.5`

Total mass of the system of boxes is 2384.5kg.

(b) For each mass, analyse each box and make them each a free-body diagram.

For
`m_(1)`:

The only force acting On the
`m_(1)` box is force of tension between 1 and 2 and as all the system is moving at a same acceleration.

`m_(1) = (F_(12))/(a)`

`m_(1) = (1387)/(1.516)`

`m_(1)` = 915kg

For
`m_(2)`:

There are two forces acting on
`m_(2)`: tension caused by box 1 and tension caused by box 3. Positive referential is to the right (because it's the movement's direction), so force caused by 1 is opposing force caused by 3:

`m_(2) = (F_(23)-F_(12))/(a)`

`m_(2) = (2304-1387)/(1.516)`

`m_(2)` = 605kg

For
`m_(3)`:

`m_(3) = m_(T) - (m_(1)+m_(2))`

`m_(3) = 2384.5-1520.0`

`m_(3)` = 864.5kg