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1 year ago

5

The small piston of a hydraulic press has an area of 3.2 in2. If the applied force is 2.0 lb, what must the area of the large pi

ston be to exert a pressing force of 795 lb?

1 answer:

BartSMP [9]1 year ago

7 0

Given data:

The area of the hydraulic press is,

$A_1=3.2in^2$

The applied force is,

$F_1=2\text{ lb}$

The pressing force on the large piston is

$F_2=795\text{ lb}$

Given hydraulic press works on Pascal's law principle.

From the Pascal's law, the equation can be given as,

$\frac{F_1}{A_1}=\frac{F_2}{A_2}$

Here,

$A_2$

is the area of the larger piston.

Substituting the values in the above equation, we get:

$\begin{gathered} \frac{2\text{ lb}}{3.2in^2}=\frac{795\text{ lb}}{A_2} \\ A_2=1272in^2 \end{gathered}$

Thus, the area of the large piston is

$1272in^2$

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Explanation:

A) To prove the motion of the center of mass of the cylinders is simple harmonic:

System diagram for given situation is shown in attached Fig. 1

We can prove the motion of the center of mass of the cylinders is simple harmonic if

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where aₓ is acceleration when attached cylinders move in horizontal direction:

<h3>PROOF:</h3>

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from Fig 3 it can be seen that fs is force by which the cylinders roll without slipping as they oscillate

So above equation becomes

$f_{s}=\frac{1}{2}MR\alpha$ ------ (5)

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$a= R\alpha$

Eq (5) becomes

$f_{s}=\frac{1}{2}Ma_{x}$ ---- (6)

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From (3)

∑Fₓ = Maₓ

Fₓ is sum of fs and restoring force that spring exerts:

$\sum F_{x} = f_{s} - kx$ ----(7)

Put (7) in (3)

$f_{s} - kx = Ma_{x}$ [/tex] -----(8)

Using (6) in (8)

$\frac{1}{2}Ma_{x} - kx =Ma_{x}$

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$T=\frac{2\pi }{\omega}$

$T = 2\pi \sqrt{\frac{3M}{2k}$

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b)

We can solve this part by using the magnification equation:

$M=-\frac{y'}{y}=\frac{q}{p}$

where

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y is the size of the object

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p is the distance of the object from the mirror

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q = 15 cm

p = 30 cm

y = 4 cm

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and the negative sign means that the image is inverted.

#LearnwithBrainly

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