A1/2 inch diameter piston moves 10 inches when 30 Lbs of force is applied, how far will it move a 6 inch Diamter piston, and how much weight will this piston lift?
pressure is the same
30/PI*(.5)^2=forcemoved/PI(3)^2
forcemoved=30 (6)^2
So the mechanical advantage is 36
which means the large piston moves 10/36
o;o
To find out how far the 6-inch diameter piston will move, we can use the principle of cross-sectional area equivalence. The cross-sectional area of a circle is calculated using the formula A = π * r^2, where A represents the area and r represents the radius.
In this case, we have the following information:
Piston 1:
Diameter = 1/2 inch
Radius = 1/4 inch (1/2 inch ÷ 2)
Movement = 10 inches
Piston 2:
Diameter = 6 inches
Radius = 3 inches (6 inches ÷ 2)
Movement = ?
Now, let's calculate the cross-sectional areas for both pistons:
Piston 1:
A1 = π * (1/4 inch)^2
Piston 2:
A2 = π * (3 inches)^2
Since the volume of a piston is constant, we can equate the two cross-sectional areas:
A1 * Movement1 = A2 * Movement2
Solving for Movement2, we get:
Movement2 = (A1 * Movement1) / A2
Now, let's substitute the values:
Movement2 = (π * (1/4 inch)^2 * 10 inches) / (π * (3 inches)^2)
We can simplify the equation:
Movement2 = (1/16 * 10 inches) / (9 inches)
Movement2 = 10/144 inches
To find out the weight the 6-inch diameter piston will lift, we need to consider the relationship between the area and force applied. The force exerted on a piston is directly proportional to the cross-sectional area.
Let's calculate the cross-sectional area for the 6-inch diameter piston:
A2 = π * (3 inches)^2
Now, let's find the force for the 6-inch diameter piston using the formula:
Force2 = (A2 * Force1) / A1
Substituting the values:
Force2 = (π * (3 inches)^2 * 30 lbs) / (π * (1/4 inch)^2)
Simplifying the equation:
Force2 = (9 * 30 lbs) / (1/16 lbs)
Force2 = 540 lbs / 0.0625
Force2 = 8640 lbs
Therefore, the 6-inch diameter piston will move 10/144 inches and will be able to lift 8640 lbs of weight.