in the hydraulic machine shown, you observe that when the small piston is pulled down 10cm, yhe large piston is raised 1 cm. if the small piston is pushed down with a force of 100n, what is the most weight that the large piston can support?
Piston displacement * Force = constant (The same for for both pistons)
If "1" denotes the small piston and "2" the large piston,
Force2/Force1 = Disp1/Disp2 = 10
Force 2 = 10 * Force1 = 1000 N
Hey there! Looks like we have a hydraulic machine in action. Let's get to the bottom of things, shall we?
Now, since we have a small piston being pulled down by 10cm and the large piston being raised by 1cm, we can say that the ratio between the piston areas is 10:1.
So, when a force of 100N is applied on the small piston, it'll create a pressure which will be transmitted equally to the large piston.
Since the area of the large piston is 10 times greater than the small one, the force exerted by the large piston will be 10 times greater too. Therefore, the most weight that the large piston can support is 100N x 10 = 1000N.
In a nutshell, the large piston can handle a weight of up to 1000 Newtons. Keep those hydraulics flowing!
To solve this problem, we can use the principle of Pascal's law, which states that pressure applied to an enclosed fluid is transmitted undiminished to every portion of the fluid.
First, we need to determine the ratio of the surface areas of the small and large pistons. Using the formula for the area of a circle, A = πr^2, we can calculate the surface areas.
Let's assume the radius of the small piston is r1 and the radius of the large piston is r2.
The surface area of the small piston (A1) is given by A1 = πr1^2.
The surface area of the large piston (A2) is given by A2 = πr2^2.
Next, we can find the force applied on the large piston when a force of 100N is applied to the small piston. This can be done using the equation:
Force on small piston / Surface area of small piston = Force on large piston / Surface area of large piston
Since the force on the small piston is 100N and the surface area of the small piston can be calculated from the radius, we can rearrange the equation to solve for the force on the large piston:
Force on large piston = (Force on small piston / Surface area of small piston) * Surface area of large piston
Substituting the given values, we have:
Force on large piston = (100N / A1) * A2
Finally, to find the weight that the large piston can support, we can multiply the force on the large piston by the acceleration due to gravity (g ≈ 9.8 m/s^2). The weight is given by:
Weight supported by large piston = Force on large piston * g
By following these steps, you should be able to calculate the most weight that the large piston can support.
To solve this problem, we need to understand the principle of Pascal's Law, which states that when pressure is applied to an enclosed fluid, the pressure is transmitted equally in all directions.
In this case, we have a hydraulic machine consisting of two pistons, one small and one large, connected by a fluid-filled chamber. When the small piston is pushed down, it exerts pressure on the fluid, which is transmitted to the large piston, causing it to move.
We are given that when the small piston is pulled down 10 cm, the large piston is raised 1 cm. This implies a ratio of 10:1 in terms of their movements. Let's denote the diameters of the small and large pistons as D_small and D_large, respectively.
To determine the most weight that the large piston can support, we need to consider the forces involved. The force exerted by the small piston (F_small) and the force exerted by the large piston (F_large) are related by the following equation:
F_small / F_large = A_large / A_small
Where A_large and A_small are the surface areas of the large and small pistons, respectively. The surface area of a piston can be calculated using the formula:
Area = π * (radius)^2 = π * (diameter/2)^2
Using the given ratio of their movements (10:1), we can also establish a relationship between their surface areas:
A_large / A_small = (π * (D_large/2)^2) / (π * (D_small/2)^2)
= (D_large/2)^2 / (D_small/2)^2
= (D_large^2) / (D_small^2)
Now, since force is directly proportional to surface area (F = P * A, where P is pressure), we have:
F_small / F_large = A_large / A_small
= (D_large^2) / (D_small^2)
Substituting the given values, with F_small = 100 N and D_small = 10 cm, we can rearrange the equation to solve for F_large:
F_small / F_large = A_large / A_small
100 N / F_large = (D_large^2) / (10 cm)^2
100 N / F_large = (D_large^2) / 100 cm^2
F_large = (100 N * 100 cm^2) / D_large^2
Now, we can use the given information that when the small piston is pushed down with a force of 100 N, the large piston is raised 1 cm. This allows us to calculate the diameter of the large piston (D_large).
Using the relationship that the movement of the small piston (10 cm) is inversely proportional to the movement of the large piston (1 cm), we can write:
D_small / D_large = 10 cm / 1 cm
Simplifying the equation, we find:
D_small / D_large = 10
Rearranging for D_large:
D_large = D_small / 10
= 10 cm / 10
= 1 cm
Now, substituting the value of D_large back into the equation for F_large:
F_large = (100 N * 100 cm^2) / (1 cm)^2
= 10000 N
Therefore, the large piston can support a maximum weight of 10000 N (or 10,000 Newtons).