How Do Newton's Laws Apply to a System of Hanging Masses?

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Homework Statement

Three ballot boxes are connected by cords, one of which wraps over a pulley having negligible friction on its axle and negligible mass. The masses are mA = 32.0 kg, mB = 40.0 kg, mC = 18.0 kg. (box a is on the horizontal suface and b and c hang down.

(a) When the assembly is released from rest, what is the tension in the cord that connects boxes B and C?

(b) How far does box A move in the first 0.250 s (assuming it does not reach the pulley)?[/B]

Homework Equations

F=ma

The Attempt at a Solution

First I found acceleration.
T=Mass of box a * acceleration
F gravity = (mass of b + mass of c) * g
T-F gravity = (Mass of b + mass of c) * -accelleration
acceleration=((mass of b + mass of c) *g)/(mass of a +mass of b + mass of c)
a=(58 * 9.8)/(32 + 40 + 18)=6.68
T=ma
T=32 * 6.68=187 N
and that's as far as I got

 

[Hootenanny]

Welcome to Physics Forums,

A hint for part (b): you know the acceleration of box A as well as the initial velocity and the time period.

 

[thomate1]

I would like to clarify that this problem involves the application of Newton's Laws of Motion, specifically the second law, which states that the net force acting on an object is equal to its mass multiplied by its acceleration (F=ma). In this case, we have three masses connected by cords, and we need to determine the tension in one of the cords and the acceleration of the system.

To solve for the tension in the cord connecting boxes B and C, we first need to find the acceleration of the system. This can be done by considering the forces acting on the masses. The only external force acting on the system is gravity, which pulls down on boxes B and C. The tension in the cord connecting boxes B and C is equal in magnitude to the force of gravity on boxes B and C (since the system is at rest). Therefore, we can set up an equation using Newton's second law:

T = (mB+mC)g

where T is the tension in the cord, mB and mC are the masses of boxes B and C, and g is the acceleration due to gravity. Substituting the given values, we get T = (40+18) * 9.8 = 548 N.

To solve for the acceleration of the system, we can use the same equation, but this time, the net force acting on the system is the difference between the tension in the cord and the force of gravity on box A. Therefore, our equation becomes:

(mB+mC)g - mA*g = (mB+mC+mA)a

Substituting the given values and solving for a, we get a = ((40+18+32) * 9.8)/(40+18+32) = 6.68 m/s^2.

To solve for the distance that box A moves in the first 0.250 s, we can use the equation of motion:

x = x0 + v0t + 0.5at^2

where x0 is the initial position, v0 is the initial velocity (which is zero in this case), a is the acceleration we just calculated, and t is the time. Substituting the given values, we get x = 0 + 0 + 0.5*6.68*(0.250)^2 = 0.208 m.

In conclusion, the tension in the cord connecting boxes B and C is