Question

(I) Calculate the force needed to move the wire in Fig. 10–34 if it holds a soapy solution (Table 10–4) and the wire is \(21.5\;cm\) long.

Short answer

The force needed to move the wire is \(0.01075\;{\rm{N}}\).

Step-by-step solution

Concept

Assuming a U-shaped apparatus is enclosing a soapy solution. The free-body diagram of the apparatus is given below:

Here, \(F\) is the surface tension force applied on the wire, \(l\) is the length of the wire and \(\gamma \) is the surface tension of the enclosed soap solution.

The soap solution enclosed in the apparatus is also described as the ‘thin film of liquid.’ This thin film consists of two surfaces, a bottom surface and a top surface. So, the total length of the surface under tension is twice the initial length.

The expression for the surface tension of the enclosed soap solution is given by,

\(\gamma = \frac{F}{{2l}}\)

Given data

The length of the wire is, \(l = 21.5\;{\rm{cm}} \times \frac{{1\;{\rm{m}}}}{{100\;{\rm{cm}}}} = 0.215\;{\rm{m}}\).

Calculation

Using the Table 10-4, the surface tension of a soap solution at \(20^\circ {\rm{C}}\) is given by,

\(\gamma = 0.025\;{\rm{N/m}}\)

The expression for the surface tension of the enclosed soap solution is given by,

\(\begin{array}{c}\gamma = \frac{F}{{2l}}\\F = \gamma \times 2l\end{array}\)

Substitute the values in the above expression,

\(\begin{array}{l}F = 0.025\;{\rm{N/m}} \times 2 \times 0.215\;{\rm{m}}\\F = 0.01075\;{\rm{N}}\end{array}\)

Therefore, the force needed to move the wire is \(0.01075\;{\rm{N}}\).