Solving Spring Stiffness and Un-deformed Length for a Weighing Machine

This problem involves calculating the un-deformed length and stiffness of a spring used in a weighing machine. We are given the initial state with just the pan and a second state with an additional mass placed on the pan. The spring has linear characteristics, meaning it obeys Hooke's Law.

Understanding the Problem Setup

• Mass of the pan, $m_{pan} = 2$ kg.
• Initial length of the spring with the pan, $l_1 = 200$ mm.
• Mass added to the pan, $m_{added} = 20$ kg.
• Final length of the spring with the pan and added mass, $l_2 = 100$ mm.
• We need to find the un-deformed length, $L$ (in mm), and the spring stiffness, $k$ (in N/m).
• We will use the standard value for acceleration due to gravity, $g = 9.8$ m/s².

Applying Hooke's Law

Hooke's Law states that the force ($F$) exerted by a spring is directly proportional to its extension or compression ($x$) from its un-deformed position. The constant of proportionality is the spring stiffness ($k$). The formula is:

$$ F = kx $$

In this problem, the force acting on the spring is the weight due to the mass placed on it.

Calculating Forces and Compressions

State 1: Pan on the spring

• Total mass: $m_1 = m_{pan} = 2$ kg.
• Force (weight): $F_1 = m_1 \times g = 2 \times 9.8 = 19.6$ N.
• The spring is compressed from its un-deformed length $L$ to $l_1 = 200$ mm.
• Compression: $x_1 = L - l_1 = L - 200$ mm.
• Converting compression to meters: $x_1 = \frac{L - 200}{1000}$ m.

State 2: Pan with added mass on the spring

• Total mass: $m_2 = m_{pan} + m_{added} = 2 + 20 = 22$ kg.
• Force (weight): $F_2 = m_2 \times g = 22 \times 9.8 = 215.6$ N.
• The spring is compressed from its un-deformed length $L$ to $l_2 = 100$ mm.
• Compression: $x_2 = L - l_2 = L - 100$ mm.
• Converting compression to meters: $x_2 = \frac{L - 100}{1000}$ m.

Setting up the Equations

Using Hooke's Law ($F = kx$) for both states:

• For State 1: $19.6 = k \times \frac{L - 200}{1000}$
• For State 2: $215.6 = k \times \frac{L - 100}{1000}$

We can rewrite these equations as:

1. $19600 = k(L - 200)$
2. $215600 = k(L - 100)$

Calculating the Un-deformed Length (L)

To find $L$, we can divide the second equation by the first equation:

$$ \frac{215600}{19600} = \frac{k(L - 100)}{k(L - 200)} $$

Simplifying the fraction:

$$ 11 = \frac{L - 100}{L - 200} $$

Now, solve for $L$:

$$ 11(L - 200) = L - 100 $$

$$ 11L - 2200 = L - 100 $$

$$ 11L - L = 2200 - 100 $$

$$ 10L = 2100 $$

$$ L = \frac{2100}{10} = 210 \text{ mm} $$

Calculating the Spring Stiffness (k)

Now substitute the value of $L = 210$ mm back into the first equation ($19600 = k(L - 200)$):

$$ 19600 = k(210 - 200) $$

$$ 19600 = k(10) $$

Solve for $k$:

$$ k = \frac{19600}{10} = 1960 \text{ N/m} $$

Conclusion

The calculated values are:

• Un-deformed length, $L = 210$ mm.
• Spring stiffness, $k = 1960$ N/m.

These values match option 3.