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3 years ago

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Problem 4: Hydraulic systems utilize Pascal's principle by transmitting pressure from one cylinder (called the primary) to anoth

er (called the secondary). Since the pressures will be equal, if the surface areas are different then the forces applied to the cylinders' pistons will be different. Suppose in a hydraulic lift, the piston of the primary cylinder has a 1.95-cm diameter and the piston of the secondary cylinder has a 24.5-cm diameter.
What force, in newtons, must be exerted on the primary cylinder of this lift to support the weight of a 2250 kg car (a large car) resting on the secondary cylinder?

1 answer:

artcher [175]3 years ago

8 0

Answer: F = 139.65 N

Explanation:

Using Pascal principle:

F = P * A, where

F = Force applied to the piston

P = Pressure

A = Piston area

also, we know generally,

A = πd²/4, in this case,

d= piston diameter

Calculation of the force Fp necessary to support the weight of the car.

Now, we use the Pascal principle: to solve,

Fp = P * Ap, where

Fp = Force on the primary piston

Ap = Primary piston area

we know our Ap but not P, so to find P

We use Newton's first law for a balanced Secondary piston -automobile system

Newton's first law: ∑F = 0

W - Fs = 0, where

W = car weight

Fs = Force on the secondary piston

Thus, W = Fs, if we put this in the equation, it becomes

W = P * As, where

As= Secondary piston area

making P subject of formula,

P = W / As

Substituting this in the initial equation, we have

Fp = W/As * Ap

From this, we know that

As = πds²/4

Ap = πdp²/4, so that

Fp = W/(πds²/4) * πdp²/4

Fp = 4W/πds² * πdp²/4

Fp = W/ds² * dp²

Fp = 2250/24.5² * 1.95²

Fp = 2250/600.25 * 3.8025

Fp = 14.25 kg

Converting to N, we have

14.25 * 9.8 = 139.65 N

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From the given information:

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$\rho (r) \ \alpha \ \delta (r -R)$

$\rho (r) = k \ \delta (r -R) \ \ at \ \ (r = R)$

$\rho (r) = 0\ \ since \ r< R \ \ or \ \ r>R---- (1)$

To find the constant k, we examine the total charge Q which is:

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$(2 \pi)(2) * \int ^{R}_{0} k \delta (r -R) * r^2dr = \sigma \times 4 \pi R^2$

Thus;

$k * 4 \pi \int ^{R}_{0} \delta (r -R) * r^2dr = \sigma \times R^2$

$k * \int ^{R}_{0} \delta (r -R) r^2dr = \sigma \times R^2$

$k * R^2= \sigma \times R^2$

$k = R^2 --- (2)$

Hence, from equation (1), if k = $\sigma$

$\mathbf{\rho (r) = \delta* \delta (r -R) \ \ at \ \ (r=R)}$

$\mathbf{\rho (r) =0 \ \ at \ \ rR}$

To verify the units:

$\mathbf{\rho (r) =\sigma \ * \ \delta (r-R)}$

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$Q = (2 \pi) (2) \int ^R_0 \sigma * \delta (r-R) r^2 \ dr$

$Q = 4 \pi \sigma \int ^R_0 * \delta (r-R) r^2 \ dr$

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