Q35E - University Physics with Modern Physics

Question

Three identical 8.50-kg masses are hung by three identical springs (Fig. E6.35). Each spring has a force constant of 7.80 kN/m and was 12.0 cm long before any masses were attached to it. 

(a) Draw a free-body diagram of each mass. 

(b) How long is each spring when hanging as shown? (Hint: First isolate only the bottom mass. Then treat the bottom two masses as a system. Finally, treat all three masses as a system.)

Step-by-Step Solution

Answer

a) The free body diagram of three masses is shown as follows,

b) The lengths of the springs from first spring to third spring after stretching are 15.2 cm , 14.1 cm and 13.1 cm .

Step 1: Identification of given data

The given data can be listed below,

• The mass of three identical masses is, $m=8.50 kg$ .
• The force constant of spring is, $k=7.80 kN/m$ .
• The length of the spring is, $l=12 cm$ .

Step 2: Concept/Significance of tension in the string

The force that a string produces as it is pulled. Tension is the force felt by a person holding a block of weight W that is connected to the end of a string.

Step 3: (a) Determination of the free-body diagram of each mass

Gravity and the tension from the third string are the forces that are acting on the final body. There are three forces at work on the second body: the force of gravity, the tension from the third string, and the tension from the second string.

 

Since the first string must support three bodies, the second two, and the third one, the first string is stretched the greatest, but the third is the least, as seen in the illustration.

Step 4: (b) Determination of the long is each spring when hanging, as shown in the diagram

The balancing force on the strings is given by,

 

$$\begin{gathered} F=kx \\ mg=kx \end{gathered}$$

 

Here, k is the force constant of the spring, and x is the stretched length of the spring.

The stretching in third spring is given by,

 

$$\begin{gathered} mg=k{x}_3 \\ x_3=\frac{mg}{k} \end{gathered}$$ 

 

Here, m is the mass of the spring, and g is the acceleration due to gravity.

Substitute all the values in the above,

 

 $$\begin{gathered} x_3=\frac{\left(8.50kg\right)\left(9.80 m/s^2\right)}{7.80\times {10}^{3}N/m} \\ =1.068 cm \end{gathered}$$

 

The total length of the third spring after stretching is given by,

 

$$\begin{gathered} d=x_3+l \\ =\left(1.068+12\right)cm \\ =13.1 cm \end{gathered}$$

 

The stretched length of the second spring is given by,

 

$$\begin{gathered} k{x}_2=mg+k{x}_3 \\ =mg+mg \\ x_2=\frac{2mg}{k} \end{gathered}$$

 

Substitute all the values in the above,

 

$$\begin{gathered} x_2=\frac{2\times \left(8.50kg\right)\left(9.80 m/s^2\right)}{7.80\times {10}^{3}N/m} \\ =2.136 cm \end{gathered}$$ 

 

The total length of the second spring after stretching is given by,

 

$$\begin{gathered} d=x_2+l \\ =\left(2.136+12\right) cm \\ =14.1 cm \end{gathered}$$

 

The stretched length of the first spring is given by,

 

$$\begin{gathered} k{x}_2=mg+k{x}_2 \\ =mg+2mg \\ x_2=\frac{3 mg}{K} \end{gathered}$$

 

Substitute all the values in the above,

$$\begin{gathered} x_2=\frac{3\times \left(8.50kg\right)\left(9.80m/s^2\right)}{7.80\times {10}^{3}N/m} \\ =3.204 cm \end{gathered}$$ 

 

The total length of the first spring after stretching is given by,

$$\begin{gathered} d=x_1+l \\ =\left(3.204+12\right)cm \\ =15.2 cm \end{gathered}$$ 

Thus, the lengths of the springs from first spring to third spring after stretching are 15.2 cm , 141.1 cm and 13.1 cm .