To solve this problem, we can use the concept of joint variation. The relationship can be expressed as:

$w = k \cdot l \cdot d^2$

where $k$ is the constant of variation.

Given:

• $w = 140$
• $d = 4$
• $l = 54$

First, we can substitute these values into the equation to find $k$:

$140 = k \cdot 54 \cdot 4^2$

Calculating $4^2$:

$4^2 = 16$

Now substitute that back into the equation:

$140 = k \cdot 54 \cdot 16$

Calculating $54 \cdot 16$:

$54 \cdot 16 = 864$

So, the equation becomes:

$140 = k \cdot 864$

Now, solving for $k$:

$k = \frac{140}{864}$

Simplifying $\frac{140}{864}$:

$k = \frac{35}{216}$

Now that we have $k$, we can express $d$ in terms of $w$ and $l$:

Starting from the original equation:

$w = k \cdot l \cdot d^2$

Rearranging it to solve for $d^2$:

$d^2 = \frac{w}{k \cdot l}$

Taking the square root to find $d$:

$d = \sqrt{\frac{w}{k \cdot l}}$

Substituting $k = \frac{35}{216}$:

$d = \sqrt{\frac{w}{\left(\frac{35}{216}\right) \cdot l}} = \sqrt{\frac{216w}{35l}}$

Thus, the final expression for $d$ in terms of $w$ and $l$ is:

$d = \sqrt{\frac{216w}{35l}}$

Summary

• $d$ is expressed as $d = \sqrt{\frac{216w}{35l}}$.

Would you like any further details or have any questions? Here are some related questions you might consider:

1. What is the significance of joint variation in real-world applications?
2. How does changing the length or diameter affect the weight of the bar?
3. Can you provide an example with different values for $w$ and $l$?
4. How would this relationship change if the weight depended only on diameter and not on length?
5. What are some other mathematical relationships that involve variation?

Tip: When solving variation problems, always identify the relationship first and ensure you understand the role of each variable.