Ahydraulic press for compacting powdered samples has a large cylinder which is 10.0 cm in diameter, and a small cylinder with a diameter of 2.0 cm. a lever is attached to the small cylinder as shown in (figure 1) . the sample, which is placed on the large cylinder, has an area of 4.0 cm2.

what is the pressure on the sample if f = 350n is applied to the lever?

The pressure on the sample is 4.375 × 10⁷ Pa

Let's recall Hydrostatic Pressure formula as follows:

where:

P = hydrosatic pressure ( Pa )

ρ = density of  fluid ( kg/m³ )

g = gravitational acceleration ( m/s² )

h = height of a column of liquid ( m )

F = force ( N )

A = cross-sectional area ( m² )

Let us now tackle the problem!

Given:

diameter of large cylinder = D₁ = 10.0 cm

diameter of small cylinder = D₂ = 2.0 cm

area of the sample = A = 4.0 cm² = 4.0 × 10⁻⁴ m²

magnitude of force = F = 350 N

Asked:

pressure on the sample = P = ?

Solution:

Firstly , we will find the force acting on the small cylinder:

Next, we could find the force acting on the large cylinder:

Finally, we could calculate the pressure on the sample:

Grade: High School

Subject: Physics

Chapter: Pressure

potential energy is directly proportional to height.

more height = more potential energy.

greatest increase in height = greatest increase in potential energy

where are the points with the greatest increase in height ?

has to be c-e or d-e .