physics problem- force and motion?

Question

three ballot boxes are connected by cords, one of which wraps over a pulley having negligible friction on its axle and negligible mass. the three masses are mA= 30kg, mB=40 kg and mC=10kg. when the assembly is released from rest:
a. what is the tension in the cord connecting B and C
b. how far does A move in the first 0.250 s
could you help me figure out the right equations to use?
thanks.

Configuration:
Box A is on a flat surface and Boxes B and C are hanging off of the side. all three boxes are connected by the rope.

Answer 1

Start out with a free body diagram in each box, remember that for A and B that are on the surface, their sum of all forces in the y- direction will be equal to zero but in the x-direction they'll be equal to ma respectively

Also, for the box C in this case the sum of all forces in the y direction will be equal to ma and 0 in the x direction, since they're all connected to the same cords... the tension between them all is the same and so is the acceleration of the system.

It's pretty much solvable from there. :)

oh and for part b, since you've found the acceleration you can use this formula x = xi + vit + 1/2 a t^2 but xi = O and so does vit so you can figure out how far A moves by substituting here:

x = 1/2 a t^2

hope it helped =)

Answer 2

Which box hangs over the pulley?

I will assume that box c hangs over the pulley in frictionless air, and that a and b are on a horizontal, frictionless surface

O---T1--[B]--T2--[A]
|
T1
|
[C]

Since they are all connected with no friction, the accelerations will be the same.

motion down for c and left for a and b are positive.

using F=m*a, look at a FBD for each box

10*g-T1=10*a
Isolate T1
98.1-T1=10*a
T1=98.1-10*a


T1-T2=40*a

T2=30*a

now solve for all of the tensions using g=9.81

98.1-10*a-30*a=40*a
98.1/80=a
a=1.23

T2=1.23*30
T1=98.1-10*1.23

b) use s(t)=.5*a*t^2
or
s(0.250)=.5*1.23*0.250^2

j