Questions: Consider two pieces of tubing 1/2 in and 3/4 in, two washers 3/16 in each, and a nut 1/16 in thick. The smallest length of a bolt that will pass through all these is in. (Type a whole number, proper fraction, or mixed number.)

$Consider two pieces of tubing 1/2 in and 3/4 in, two washers 3/16 in each, and a nut 1/16 in thick. The smallest length of a bolt that will pass through all these is in. (Type a whole number, proper fraction, or mixed number.)$

Transcript text: Consider two pieces of tubing $\frac{1}{2}$ in $\& \frac{3}{4}$ in, two washers $\frac{3}{16}$ in each, and a nut $\frac{1}{16}$ in thick. The smallest length of a bolt that will pass through all these is $\square$ in. (Type a whole number, proper fraction, or mixed number.)

Solution

Solution Steps

To find the smallest length of a bolt that will pass through all the given pieces, we need to sum the thicknesses of all the components: two pieces of tubing, two washers, and one nut.

Solution Approach

Sum the thicknesses of the two pieces of tubing.
Sum the thicknesses of the two washers.
Add the thickness of the nut.
Combine all these sums to get the total length required.

Step 1: Sum the Thicknesses of the Tubing

We have two pieces of tubing with thicknesses $ \frac{1}{2} $ in and $ \frac{3}{4} $ in. We sum these thicknesses: \[ \frac{1}{2} + \frac{3}{4} = 0.5 + 0.75 = 1.25 \text{ in} \]

Step 2: Sum the Thicknesses of the Washers

We have two washers, each with a thickness of $ \frac{3}{16} $ in. We sum these thicknesses: \[ 2 \times \frac{3}{16} = 2 \times 0.1875 = 0.375 \text{ in} \]

Step 3: Add the Thickness of the Nut

The nut has a thickness of $ \frac{1}{16} $ in: \[ \frac{1}{16} = 0.0625 \text{ in} \]

Step 4: Combine All the Sums

We combine the sums from the previous steps to get the total length required: \[ 1.25 + 0.375 + 0.0625 = 1.6875 \text{ in} \]