How Does a Spring Balance Measure Tension in a Pulley System?

[SiYuan]

Homework Statement

A light spring balance is attached to two strings that passes over smooth pulleys. Loads of m1 and m2 are attached to the ends of the strings. If m1=m2=4.0kg, or m1=6kg m2=4kg, what is the reading of the spring balance.

Ans: 40N and 48N

I just need to know how do i do it. And a general theory behind it all. Sorry for this very simple question here.

It's basically like this

o------Spring-----o
|```````````````|
m1``````````````m2


Cheers for the upcoming help.

 

[SiYuan]

Alright got it anyway.

Must've been too easy for youse lot here.

34 views with no reply.

 

[baba89]

I am happy to help you with this problem. The first step in solving this problem is to draw a free body diagram, which is a visual representation of all the forces acting on the system. In this case, the system includes the spring balance, the two masses (m1 and m2), and the pulleys.

Next, we need to understand the concept of tension in a string. When a string is attached to two masses, it exerts a force on each mass in opposite directions. This force is known as tension, and it is equal in magnitude for both masses.

Now, let's consider the two scenarios given in the problem. In the first scenario, both masses (m1 and m2) are equal to 4.0kg. This means that the tension in the strings will also be equal, and we can represent it as T. Using Newton's second law, we can write the equation:

T = m1a1 = m2a2

where a1 and a2 are the accelerations of masses m1 and m2, respectively. Since the pulleys are smooth, we can assume that there is no friction, and the acceleration of both masses is the same. So, we can rewrite the equation as:

T = m1a = m2a

Now, we can use the equation for the magnitude of tension in a string, which is given by:

T = mg

where g is the acceleration due to gravity (9.8 m/s^2). Substituting this in our equation, we get:

m1g = m2g

Solving for g, we get:

g = 0 m/s^2

This means that the system is in equilibrium, and the reading on the spring balance will be zero.

In the second scenario, m1 = 6kg and m2 = 4kg. Following the same steps as above, we can write the equation:

T = m1a = m2a

Substituting the values, we get:

T = (6 kg)(9.8 m/s^2) = 58.8 N

Therefore, the reading on the spring balance will be 58.8 N.

In general, the theory behind this problem is based on Newton's second law, which states that the net force on an object is equal to its mass times its acceleration. In this case, the net force is the sum of the