All categories

3 years ago

La tension que se transmite en la cuerda BD es de 75 lb. Calcula el momento de fuerza generada por la cuerda respecto al punto C

Physics

1 answer:

Bas_tet [7]3 years ago

7 0

Answer:

Mc = 1920[lb*in]

Explanation:

Para poder solucionar este problema debemos realizar un análisis estático, por tal motivo lo primero es realizar un diagrama de cuerpo libre con las respectivas fuerzas actuando sobre la barra ABC. DE igual manera calcular la geometría de la configuración mostrada.

El diagrama de cuerpo libre se puede ver en la imagen adjunta, con la solución de este problema.

Lo primero es determinar el angulo t, el cual por medio de las propiedades del triangulo rectángulo se puede determinar.

Con este angulo (t) ya determinado, fijamos la atención en el triangulo BCD, este triangulo no es rectángulo, pero por medio de la ley de senos podemos determinar el angulo omega.

Después de determinar el angulo omega, restamos el angulo (t) para poder determinar el angulo (a).

Seguidamente realizamos una sumatoria de momentos alrededor del punto C, utilizado las respectivas fuerzas con los ángulos descompuestos.

El momento en el punto C es de 1920 [Lb*in].

Nota: ya que no se menciona la fuerza en el punto A, esta se desprecia y no se tiene en cuenta en los calculos. En la imagen adjunta se puede ver el procedimiento desarrollado.

You might be interested in

Does Anybody Know The Answers?

ch4aika [34]

Answer:

I was going to give you the paper where I saw it but since you are not giving enough points I can not give you so I am only going to give you some of these that are here sorry

Explanation:

$9^{2} + 12^2 = x^2\\81 + 144= x^2\\\sqrt{225} = \sqrt{x} \\ 15=x\\\\ 2.\\x^2+12^2+=13^2\\x^2+144 =169\\x^2 = 25\\\sqrt{x^2 =\sqrt{25\\\\$

x=5

$3.\\12^2+32^2 = x^2\\34.176= x$

5,12,13

$\frac{x}{4} ,\frac{12}{4} ,\frac{20}{4}\\\\\frac{x}{4},3,5 \\\\x=16\\\\12. \\x^2 + 48^2=50^2\\\\x^2=196\\x=14$

Download docx

5 0

3 years ago

A tube is sealed at both ends and contains a 0.0100-m long portion of liquid. The length of the tube is large compared to 0.0100

Ahat [919]

Answer:

31.321 rad/s

Explanation:

L = Tube length

A = Area of tube

$\rho$ = Density of fluid

v = Fluid velocity

m = Mass = $\rho Al$

Centripetal force is given by

$F=\dfrac{mv^2}{L}\\ F=\dfrac{m(\omega L)^2}{L}\\ F=m\omega^2\\ F= 0.01A\rho\omega^2L$

Pressure is given by

$P=\dfrac{F}{A}=\rho gL\\\Rightarrow \dfrac{0.01A\rho\omega^2L}{A}=\rho gL\\\Rightarrow 0.01\omega^2=g\\\Rightarrow \omega^2=\dfrac{g}{0.01}\\\Rightarrow \omega=\sqrt{\dfrac{g}{0.01}}\\\Rightarrow \omega=\sqrt{\dfrac{9.81}{0.01}}\\\Rightarrow \omega=31.321\ rad/s$

The angular speed of the tube is 31.321 rad/s

5 0

3 years ago

A sample of gas in a syringe releases heat to its surroundings while the

blsea [12.9K]

Answer: b

Explanation:

When heat is released by the system i.e. system loses heat. So, we take it as negative -Q

When the work is done on the system then it is considered as negative work on the system i.e. -W

In this case, the plunger is pulled out, and work is done on the system. So, we take work as negative work -W

Correct option is b

3 0

3 years ago

Read 2 more answers

An electrician finds that a 1 m length of a certain type of wire has a resistance of 0.24 Ω . What is the total resistance of th

zlopas [31]

The resistance of a given conductor depends on its electrical resistivity ( $\rho$ ), its length(L) and its cross-sectional area (A), as follows:

$R=\frac{\rho L}{A}$

In this case, we have $L'=138L$ , $\rho'=\rho$ and $A'=A$ . So, the total resistance of the wire with length of 138m is:

$R'=\frac{\rho' L'}{A'}\\R'=\frac{\rho 138L}{A}\\R'=138\frac{\rho L}{A}\\R'=138R\\R'=138(0.24\Omega)\\R'=33.12\Omega$

5 0

3 years ago

A pitcher can throw a fastball that reaches home plate at 95 mph. What is this speed in m/s

Taya2010 [7]

42.47 meters per second

5 0

3 years ago