21 July, 12:37

# Three crates with various contents are pulled by a force F pull = 3615 N across a horizontal, frictionless roller‑conveyor system. The group of boxes accelerates at 1.516 m / s 2 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 m 1 and the box of mass m 2, the force meter reads F 12 = 1387 N. Between the box of mass m 2 and the box of mass m 3, the force meter reads F 23 = 2304 N. Assume that the ropes and force meters are massless.

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1. 21 July, 13:26
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m1=914.9kg

m2=604.9kg

m3=864.75kg

Explanation

I think we are suppose to find the mass of the crate.

The effective force that moves the body in positive x direction is 3615N

ΣFx = Σma

Then Fx=3615N

Then the masses be m1, m2 and m3

Then,

ΣF = Σ (ma)

3615 = (m1+m2+m3) a

Given that a=1.516

The masses are

m1+m2+m3=, 2384.56. Equation 1

Between mass 1 and mass 2 is, F12=1387.

The effective force that pull mass 1 is 1387.

F12=m1 * a

Therefore,

m1=F12/a

m1=1387/1.516

m1=914.9kg.

The effective force that pulls crate 1 and crate 2 is F23

F23 = (m1+m2) a

Therefore

2304 = (m1+m2) a

Therefore, since a=1.516

m1+m2=2304/1.516

m1+m2=1519.8kg

Since m1=914.9kg

So, m2=1519.8-m1

m2=1519.8-914.9

m2=604.9kg

Also from equation 1

m1+m2+m3=2384.56

Since m1=914.9kg and m2=604.9kg

Then, m3=2384.56-604.9-914.9

m3=864.75kg