How Does Tension Affect Acceleration in a Two-Mass System?

[John O' Meara]

A body of mass .5kg, rests on a smooth horizontal table, and is connected by a string, which passes over the edge of the table, with a mass of .02kg hanging freely. find the acceleration of the system.
My attempt: let the tension of the piece of string between the table edge and the .5kg body be T directed towards the table edge. This tension is the same as the tension between the table edge and the suspended .02kg body, yes,no?
Since the string between the table edge and the .5kg body in not horizontal, the Tension T has components, Tcos(theta) parallel to the table and Tsin(theta) perpendicular to the table, but the angle theta is unknown.
Thanks.

 

[Hootenanny]

As the table is frictionless I would assume that the only force acting is that of gravity. Hence the force acting on the system is $0.02g$. Can you go from here? Don't forget the total mass of the system is not 0.02kg.

 

[kyle8921]

Yes, you are correct that the tension in the string is the same for both the .5kg body and the .02kg body. However, the angle theta is not unknown. We can use the fact that the system is in equilibrium to find the value of theta. Since the system is not moving, the acceleration of the system must be zero. This means that the net force on the system is also zero.

We can set up an equation for the forces in the horizontal direction:

Tcos(theta) = mg

Where T is the tension, m is the mass of the .02kg body, and g is the acceleration due to gravity. We can also set up an equation for the forces in the vertical direction:

Tsin(theta) = mg

Now, we can solve for theta by dividing the second equation by the first:

tan(theta) = Tsin(theta)/Tcos(theta) = mg/mg = 1

This means that theta must be equal to 45 degrees. Now that we know the angle, we can find the value of T using either of the original equations. Let's use the first equation:

Tcos(45) = mg

T = (0.02kg)(9.8m/s^2)/cos(45) = 0.028N

Now, we can use this tension value in the second equation to find the acceleration:

Tsin(45) = mg

a = Tsin(45)/m = (0.028N)(sin(45))/0.5kg = 0.0196m/s^2

Therefore, the acceleration of the system is 0.0196 m/s^2.