The load L a horizontal beam can safely support varies jointly as the width w and the square of the depth d and inversely as the length l. If a 12-foot beam with width 6 inches and depth 8 inches safely supports 600 pounds, how many pounds can a 14-foot beam that has width 5.5 inches and depth 4 inches be expected to support? Round to the nearest pound. Assume the two beams are made of the same material.

"The load L a horizontal beam can safely support varies jointly as the width w and the square of the depth d and inversely as the length l"

---- >Load = k(wd^2/L)
given :
L = 12, w = 6inches or 1/2 , d = 8 inches or 2/3, load = 600
600 = k(1/2)(4/9) /12
(1/54)k = 600
k = 32400

Load = 32400(wd^2/L)
for the case L = 14 , w = 11/24 , d = 1/3

Load = 32400(11/24)(1/9)/14 = 117.857 or 118 to the nearest pound

check my arithmetic, easy to make a mistake on this one.

Well, this is quite a mathematical question! Let me put on my thinking cap...or rather, my thinking wig!

Now, according to the problem, the load L varies jointly as the width w and the square of the depth d, and varies inversely as the length l. So, we can use the equation:

L = k * (w * d^2) / l

where k is a constant of variation that we need to find. Lucky for us, we have all the information we need to solve for k.

Let's plug in the values for the first beam and solve for k:

600 = k * (6 * 8^2) / 12

Now, let's solve for k:

k = 600 * 12 / (6 * 64)

k = 0.625

Alright, now that we have our constant of variation k, we can use it to find the load the second beam can safely support.

L = 0.625 * (5.5 * 4^2) / 14

Calculating this, we find:

L ≈ 16.875 pounds

So, the 14-foot beam with width 5.5 inches and depth 4 inches can be expected to support around 16.875 pounds.

Please note that this is an approximate value, so let's just round it to the nearest pound:

L ≈ 17 pounds

There you have it! The clown mathematician has solved the problem and found that the beam can be expected to support around 17 pounds.

To solve this problem, we need to use the variation equation that relates the load L, width w, depth d, and length l:

L = k * (w * d^2) / l

where k is a constant of variation.

We can find the value of k by substituting the given values from the 12-foot beam:

600 = k * (6 * 8^2) / 12

Simplifying this equation:

600 = k * (6 * 64) / 12
600 = k * 32
k = 600 / 32
k = 18.75

Now that we have the value of k, we can use it to find the load supported by the 14-foot beam:

L = 18.75 * (5.5 * 4^2) / 14

Simplifying this equation:

L = 18.75 * (5.5 * 16) / 14
L = 18.75 * 88 / 14
L = 1170 / 14
L ≈ 83.57142857

Rounding to the nearest pound, the 14-foot beam can be expected to support approximately 84 pounds.

To solve this problem, we will first write an equation representing the relationship between the load (L), the width (w), the depth (d), and the length (l) of the beam.

We are given that the load (L) varies jointly with the width (w) and the square of the depth (d), and inversely with the length (l).

Therefore, the equation relating these variables can be written as follows:

L = k * (w * d^2) / l

where k is a constant of variation.

Next, we can use the information provided to find the value of k.

We are given that a 12-foot beam with width 6 inches and depth 8 inches safely supports 600 pounds.

Substituting these values into our equation:

600 = k * (6 * 8^2) / 12

Simplifying:

600 = k * (6 * 64) / 12

600 = k * (384 / 12)

600 = k * 32

Now, we can solve for the value of k:

k = 600 / 32

k = 18.75

Now that we have the value of k, we can use it to solve the second part of the problem.

We are asked to find how many pounds a 14-foot beam with width 5.5 inches and depth 4 inches can be expected to support.

Using our equation and the given values:

L = 18.75 * (5.5 * 4^2) / 14

Simplifying:

L = 18.75 * (5.5 * 16) / 14

L = 18.75 * 88 / 14

L = 117

Therefore, the 14-foot beam can be expected to support 117 pounds (rounded to the nearest pound).

Show me the steps in doing this problem

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