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natulia [17]
2 years ago
12

En la figura los émbolos son de masa despreciable y están en reposo, como se ve en la figura, siendo M=300 kg. Determine: a) la

presión generada por el bloque de masa M sobre el embolo. b) La presión manométrica en un punto a 50 cm de profundidad en el embolo más grande. c) La fuerza F para que el sistema esté en equilibrio. Area del emb. grande= 1,2 m2; A del emb.pequeño=20 cm2; P liquido=600kg/m2

Physics
1 answer:
Leni [432]2 years ago
4 0

The image mentioned is in the attachment

Answer: a) P = 2450 Pa;

b) P = 2940 Pa;

c) F = 4.9 N

Explanation:

a) Pressure is a force applied to a surface of an object or fluid per unit area.

The image shows a block applying pressure on the large side of the piston. The force applied is due to gravitation, so:

P = \frac{F}{A}

P = \frac{m.g}{A}

P = \frac{300.9.8}{1.2}

P = 2450 Pa

The pressure generated by the block is P = 2450 Pa.

b) A static liquid can also exert pressure and can be calculated as:

P_{staticfluid} =ρ.g.h

where

ρ is the density of the fluid

h is the depth of the fluid

g is acceleration of gravity

P_{staticfluid} = 600.9.8.0.5

P_{staticfluid} = 2940 Pa

The pressure in the fluid at 50 cm deep is P_{staticfluid} = 2940 Pa.

c) For the system to be in equilibrium both pressures, pressure on the left side and pressure on the right side, have to be the same:

P_{s} = P_{b}

\frac{F}{A_s} = \frac{F_b}{A_b}

F = \frac{F_b}{A_b}.A_s

Adjusting the units, A_{s} = 0.002 m².

F = \frac{300.9.8.0.002}{1.2}

F = 4.9 N

The force necessary to be equilibrium is F = 4.9 N.

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2 years ago
The uniform slender bar AB has a mass of 6.4 kg and swings in a vertical plane about the pivot at A. If θ˙ = 2.7 rad/s when θ =
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Answer:

F=√[(1.5(14.58L+11.96))² + (3.2(2.97L - 157.03) + 62.72)²]

Explanation:

Given data,

The mass of the bar AB, m = 6.4 kg

The angular velocity of the bar,  θ˙ = 2.7 rad/s

The angle of the bar at A, θ = 24°

Let the length of the bar be, L = l

The angular moment at point A is,

                        ∑ Mₐ = Iα

Where,     Mₐ - the moment about A

                 α  - angular acceleration

                 I - moment of inertia of the rod AB

                       -mg(\frac{lcos\theta}{2})=\frac{1}{3}(ml^{2})\alpha

                        \alpha=\frac{-3gcos\theta}{2l}

Let G be the center of gravity of the bar AB

The position vector at A with respect to the origin at G is,

                          \vec{r_{G}}=[\frac{lcos\theta}{2}\hat{i}-\frac{lcos\theta}{2}\hat{j}]

The acceleration at the center of the bar

                          \vec{a_{G}}=\vec{a_{a}}+\vec{\alpha}X\vec{r_{G}}-\omega^{2}\vec{r_{G}}

Since the point A is fixed, acceleration is 0

The acceleration with respect to the coordinate axes is,

                         (\vec{a_{G}})_{x}\hat{i}+(\vec{a_{G}})_{y}\hat{j}=0+(\frac{-3gcos\theta}{2l})\hat{k}\times[\frac{lcos\theta}{2}\hat{i}-\frac{lcos\theta}{2}\hat{j}]-\omega^{2}[\frac{lcos\theta}{2}\hat{i}-\frac{lcos\theta}{2}\hat{j}]

(\vec{a_{G}})_{x}\hat{i}+(\vec{a_{G}})_{y}\hat{j}=[-\frac{cos\theta(2l\omega^{2}+3gsin\theta)}{4}\hat{i}+(\frac{2l\omega^{2}sin\theta-3gcos^{2}\theta}{4})\hat{j}]

Comparing the coefficients of i

=-\frac{cos\theta(2l\omega^{2}+3gsin\theta)}{4}

Comparing coefficients of j

(\vec{a_{G}})_{y}=\frac{2l\omega^{2}sin\theta-3gcos^{2}\theta}{4}

Net force on x direction

F_{x}=(\vec{a_{G}})_{x}

substituting the values

F_{x}=1.5(14.58L+11.96)

Similarly net force on y direction

F_{y}=(\vec{a_{G}})_{y}+mg

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Where L is the length of the bar AB

Therefore the net force,

F=\sqrt{F_{x}^{2}+F_{y}^{2}}

F=√[(1.5(14.58L+11.96))² + (3.2(2.97L - 157.03) + 62.72)²]

Substituting the value of L gives the force at pin A

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