Assuming ideal frictionless conditions.?

Question

You have a table with a box on top attached to two boxes, one hanging over each side, the boxes are attached, one on each side, by a rope running through a pulley. The box on the left is m1, the box on right is m2 and the box on top is m3.

What is the acceleration of the system if
m1 = 0.25 kg,
m2 = 0.50 kg,
m3 = 0.25 kg?

What is the acceleration of the system if
m1 = 0.35 kg,
m2 = 0.15 kg,
m3 = 0.50 kg?

Answer 1

its gravity pulling the blox, so the accel woul dbe g if it was pulling all of them. But it's not, so you have to get the forces. the force pulling down is gravity on box 1 and 2 is:

F = gm1 + gm2
F = 9.8 * 0.25 + 9.8 * 0.5
F = 7.35

and F=ma. the total mass is m1+m2+m3=1 so

F=ma
a=F/m
a= 7.35/1
a=7.35.

In part b, the masses are different, so the total farce is
F=9.8*.35+9.8*.15
F=4.9
so

a=F/m=4.9/1
a=4.9

Answer 2

This is a variation on a thing called 'Atwoods Machine' (look it up on teh 'net) Calculate the force exerted by gravity (F=ma) on m1 and m2 (call it F1 and F2) and take the absolute difference between them (Ft=|F1-F2|). Then the acceleration of the entire system is simply a = Ft/(m1+m2+m3) and it's in the direction of the larger mass.

HTH

Doug

Answer 3

you need to draw the diagram and analyze it bfore you can conclude abt the forces acting on the bodies....

it will be very easy if you can draw the diagram... but sorry i cant draw it here....