Magnitude of the Tension in a Connection String

[Weasler]

1. Two masses are connected by a light string. The horizontal mass of 2 kilograms is being pulled to the left with a force of 30 Newtons along a frictionless surface. The vertical mass is 1 kilogram. What is the magnitude of the tension in the connection string?



2. Fnet=ma



3. My attempt was simply adding the two forces acting upon the string together; 30 Newtons and 10 Newtons respectively.

 

[PhanthomJay]

Hi there Weasler, welcome to PF! When you say vertical and horizontal mass, you mean that one mass is on a horizontal frictionless table and the other is hanging down off the table via a string a pulley? Try using free body diagrams of each mass, identifying all forces acting on each, and applying Newton's laws to solve 2 equations with 2 unknowns.

 

[Weasler]

Ahhh I had seen where my logic was flawed; one would use the entirety of the system to calculate the acceleration (30-10=3a), and then take that and find the Tension between the second block (but not the first as the force is being applied to it directly and there is no friction) and the string (T-10=1*(20/3)). Thank you kindly!

 

[PhanthomJay]

Ahhh I had seen where my logic was flawed; one would use the entirety of the system to calculate the acceleration (30-10=3a), and then take that and find the Tension between the second block (but not the first as the force is being applied to it directly and there is no friction) and the string (T-10=1*(20/3)). Thank you kindly!

Yes, your answer is correct, but sometimes you will get into trouble when first trying to isolate the system. It is best in these problems to isolate each mass separately. for the mass on the table,
(Eq. 1) 30 -T = 2a, and for the hanging mass,
(Eq. 2) T - 10 = a, then multiplying this 2nd equation by 2,
(Eq. 3) 2T - 20 = 2a. Now looking at (Eq 1)and (Eq.3), then
(Eq. 4) 30 - T = 2T - 20 , from which
(Eq. 5) 50 = 3T
which yields your same result. It may seem longer, and I can't argue with your method since you have the correct answer, but nevertheless you must proceed systematically.

 

[rwisz]

This would give a total force of 40 Newtons acting downwards on the string.

The magnitude of the tension in the connection string can be calculated using the equation Fnet=ma, where Fnet is the net force acting on the string and m is the mass of the object attached to the string. In this case, the net force acting on the string is the sum of the horizontal force (30 Newtons) and the weight of the vertical mass (10 Newtons, calculated as 1kg x 10m/s^2). Therefore, the magnitude of the tension in the connection string would be 40 Newtons, as correctly calculated in the attempt provided. It is important to note that the tension in the string is equal throughout its length and acts in both directions, balancing the forces acting on the masses. This tension allows the masses to move together without the string breaking.