Tension in Strings

When particles are released from rest in the diagram below, what is the tension in the string that holds them?

Tension in string example

Answer:

In a situation like this, the particle with the highest mass will be the one to drop, and the particle with the lowest mass will rise. Let's take the particle with 2kg mass as particle a and the one with 5kg mass as particle b.

To clarify the weight of each particle, we have to multiply its mass with gravity.

Weight of a = 2g

Weight of b = 5g

Now you can model an equation for each particle's acceleration and tension.

T -2g = 2a [Particle a] [Equation 1]

5g -T = 5a [Particle b] [Equation 2]

You now solve this simultaneously. Add both equations to eliminate the T variable.

3g = 7a

If you take 9.8 ms^-2 gas

\(a = 4.2 ms^{-2}\)

You can substitute acceleration into any of the equations to give you tension.

Substitute acceleration into equation 1.

\(T = -2g = 2 \cdot 4.2 \rightarrow T -19.6 = 8.4 \rightarrow T = 28 N\)

There are two particles, one with a 2kg mass sitting on a smooth table and the other with a 20kg mass hanging on the side of the table over a pulley connecting both particles – demonstrated below. These particles have been held in place all this time, and they are now released. What will happen next? What is the acceleration and tension in the string?

Tension in a string with one particle on a smooth table

Answer: Let us add to the diagram to see what we are working with.

Tension in a string with one particle on a smooth table

Take particle with 2kg mass to be particle A.

And particle with 20kg mass to be particle B.

Now let's resolve particle A horizontally.

T = ma [equation 1]

Resolving particle B vertically

mg -T = ma [Equation 2]

We substitute the figures in them:

T = 2a [Equation 1]

20g - T = 20a [Equation 2]

We can now add both equations to cancel tensions.

20g = 22a

\(a = \frac{98}{11} = 8.9 ms^{-2}\)

Now factorise acceleration into either of the equations. We would do the first.

\(T = 2 \cdot \frac{98}{11} = 17.8 N\)

Find the tension in each part of the string in the diagram below.

Tension at an angle

Answer: what we will need to do is to make two equations out of the entire diagram – one for the vertical forces and another for the horizontal. So what we are going to do is resolve tension for both strings into their respective vertical and horizontal components.

Tension at an angle

\(T_1 \sin 20 = T_2 \sin 30 \space [Equation \space 2] [Horizontal]\)

Since we have two equations and two unknowns here, we are going to use the simultaneous equation procedure to do this by substitution.

Now we will rearrange the second equation and substitute it into the first equation.

\(T_1 = \frac{T_2 \sin 30}{\sin 20}\)

\((\frac{0.5T_2}{0.342}) = \cos 20 + T_2 \cos 30 = 50\)

\((\frac{0.5T_2}{0.342})0.94 + 0.866 \space T_2 = 50\)

\(1.374 \space T_2 + 0.866 \space T_2 = 50\)

\(2.24 T_2 = 50\)

\(T_2 = 22.32 N\)

Now that we have a value for T₂, we can go ahead to substitute that into any of the equations. Let's use the second.

\(T_1 \sin 20 = 22.32 \space \sin 30\)

\(T_1 = \frac{11.16}{0.342} = 32.63\)