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sergey [27]
3 years ago
8

A hydraulic 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?

Physics
2 answers:
ExtremeBDS [4]3 years ago
4 0
Pressure is defined as the force applied per unit area of the press. In this case, we are given with a force equivalent to 350 N and an area equal to 4 cm2. 4 cm2 is equal to 0.0004 m2. We divide 350Newton by 0.0004 square meters equal to 875,000 Pascals or 875 kiloPascals.
Andre45 [30]3 years ago
3 0

The pressure on the sample is 4.375 × 10⁷ Pa

\texttt{ }

<h3>Further explanation</h3>

Let's recall Hydrostatic Pressure formula as follows:

\boxed{ P = \rho g h}

\boxed{ P = F \div A}

<em>where:</em>

<em>P = hydrosatic pressure ( Pa )</em>

<em>ρ = density of  fluid ( kg/m³ )</em>

<em>g = gravitational acceleration ( m/s² )</em>

<em>h = height of a column of liquid ( m )</em>

<em>F = force ( N )</em>

<em>A = cross-sectional area ( m² )</em>

Let us now tackle the problem!

\texttt{ }

<u>Given:</u>

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

<u>Asked:</u>

pressure on the sample = P = ?

<u>Solution:</u>

<em>Firstly , we will find the force acting on the small cylinder:</em>

\Sigma \tau = 0

F(2L) - F_2(L) = 0

2FL = F_2L

F_2 = 2F

F_2 = 2(350)

\boxed{F_2 = 700 \texttt{ N}}

\texttt{ }

<em>Next, we could find the force acting on the large cylinder:</em>

P_1 = P_2

F_1 \div A_1 = F_2 \div A_2

F_1 = \frac{A_2}{A_1} F_2

F_1 = \frac{\frac{1}{4} \pi (D_2)^2}{\frac{1}{4} \pi (D_1)^2} F_2

F_1 = (\frac{D_2}{D_1})^2 F_2

F_1 = (\frac{10.0}{2.0})^2 \times 700

\boxed{F_1 = 17500 \texttt{ N}}

\texttt{ }

<em>Finally, we could calculate the pressure on the sample:</em>

P = F_1 \div A

P = 17500 \div (4 \times 10^{-4})

\boxed{P = 4.375 \times 10^7 \texttt{ Pa}}

\texttt{ }

<h3>Learn more</h3>
  • Buoyant Force : brainly.com/question/13922022
  • Kinetic Energy : brainly.com/question/692781
  • Volume of Gas : brainly.com/question/12893622
  • Impulse : brainly.com/question/12855855
  • Gravity : brainly.com/question/1724648

\texttt{ }

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Pressure

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