The horsepower (hp) that a shaft can safety transmit varies jointly with its speed (in revolutions per minute, or rpm) and the cube of the diameter. If a shaft of a certain material 2 in in diameter can transmit 36 hp at 75 rpm, what diameter must the shaft have in order to transmit 150 hp at 20 rpm?

just plug and chug:

hp = k*s*d^3

36 = k * 75 * 2^3

k = 36/(75 * 8) = 36/600

now, for the unknown d

150 = k * 20 * d^3
150 = 36/600 * 20 * d^3

150 * 600/36 * 1/20 = d^3

125 = d^3

To solve this problem, we'll need to use the concept of joint variation. According to the problem, the horsepower (hp) that a shaft can transmit varies jointly with its speed (in revolutions per minute, or rpm) and the cube of the diameter.

Let's represent the variables involved:
- Let x be the diameter (in inches) of the new shaft that can transmit 150 hp at 20 rpm.
- Let y be the horsepower (in hp) that the shaft can transmit.

Now, let's set up the equation using the joint variation concept:
[horsepower] = k * [speed] * (diameter)^3

The constant k represents the joint variation constant, which we need to determine.

Using the information given in the problem, we can substitute the values to find the value of k.

When the diameter is 2 inches, the horsepower is 36 hp, and the speed is 75 rpm:
36 = k * 75 * 2^3
36 = k * 75 * 8
36 = k * 600

Dividing both sides by 600:
k = 36/600
k = 0.06

Now that we have the value of k, we can solve for the diameter (x) when the horsepower (y) is 150 hp and the speed is 20 rpm:
150 = 0.06 * 20 * x^3
150 = 1.2x^3

Dividing both sides by 1.2:
125 = x^3

Taking the cube root of both sides:
x = ∛125
x = 5

Therefore, the diameter of the new shaft that will transmit 150 hp at 20 rpm is 5 inches.

To solve this problem, we can use the concept of joint variation, which states that when a quantity varies jointly with two other quantities, it can be expressed as their product raised to some power.

Given that the horsepower the shaft can transmit varies jointly with its speed and the cube of its diameter, we can write the equation as:

hp = k * speed * diameter^3

where k is the constant of variation.

From the given information, we have the first set of values:

hp1 = 36 hp (horsepower)
speed1 = 75 rpm (revolutions per minute)
diameter1 = 2 in (inches)

Let's calculate the value of k using the first set of values:

36 = k * 75 * 2^3
36 = 600k

k = 36 / 600
k = 0.06

Now, we can use this value of k to find the diameter, diameter2, for the second set of values:

hp2 = 150 hp
speed2 = 20 rpm

Using the equation:

150 = 0.06 * 20 * diameter2^3

Simplifying the equation, we get:

150 = 1.2 * diameter2^3

Dividing both sides by 1.2:

125 = diameter2^3

Taking the cube root of both sides:

diameter2 = ∛(125)

diameter2 ≈ 5.0 inches

Therefore, the diameter of the shaft must be approximately 5.0 inches in order to transmit 150 hp at 20 rpm.