INFORME DE LABORATORIO N 02/FISICA II
03 de agosto de 2013
UNIVERSIDAD NACIONAL “SANTIAGO ANTÚNEZ DE MAYOLO”
FACULTAD DE INGENIERÍA CIVIL
CURSO
:
FISICA I
TEMA
:
“FUERZAS-ESTATICA”
FECHA
:
26 de septiembre de 2011
DOCENTE :
Reyes Pareja Carlos
ALUMNO :
Garay Sánchez Ríder Jherson
CODIGO
:
122.0904.388
HUARAZ-PERU 2013
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INTRODUCCIÓN:
Cuando experimentamos en la la práctica las leyes físicas (ley de Hooke, primera y llegamos a la la conclusión que solo se cumplen en la segunda condición de equilibrio) llegamos teoría y utilizando materiales ideales, y que no se cumple en el experimento, debido a diversos factores, como errores que se cometen a lo largo de la experimentación. Pero, los resultados son próximos a las leyes físicas. En esta práctica analizaremos las leyes de Hooke, la primera y segunda condición de equilibrio demostrando que no se cumple con total exactitud en los experimentos, debido a varios factores como errores en los cálculos de las magnitudes, etc.
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I.
OBJETIVOS:
1.1.
Verificar experimentalmente la ley de Hooke.
1.2.
Representar gráficamente los esfuerzos aplicados a un resorte en función de las deformaciones que le producen y a partir de la gráfica, determinar la constante elástica de los resortes.
1.3.
Verificar la primera condición de equilibrio.
1.4.
Verificar la igualdad de momentos respecto a un punto en un cuerpo de equilibrio.
II.
MATERIALES:
2.1.
Tres resortes helicoidales.
2.2.
Un soporte universal con dos varillas de hierro y una nuez.
2.3.
Una regla regla graduada en milímetros.
2.4.
un juego de pesas calibradas con porta pesas.
2.5.
Una argolla.
2.6.
Un soporte de madera.
2.7.
Una prensa.
2.8.
Una barra barra metálica metálica con orificios. orificios.
2.9.
3 ganchos, tres clavos.
2.10.
Un tablero.
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III. 3.1.
MARCO TEORICO:
Ley de Hooke: Consideremos un resorte hecho de alambre de sección circular enrollado en forma de hélice cilíndrica fijo por uno de sus extremos y el otro libre, tal como se muestra en la fig. 01. al aplicar ala extremo libre una fuerza una fuerza externa como por ejemplo colocando una pesa experimenta una deformación
m, el resorte
, se demuestra que la fuerza aplicada es
directamente proporcional al desplazamiento
o al cambio de longitud del
resorte. Es decir, en forma de ecuación se escribe:
…………….. …………….. (1)
Donde, k es una constante de proporcionalidad comúnmente llamada “constante elástica”. elástica”. Mientras mayor sea
k , más rígido o fuerte será el resorte. Las
unidades de k en el sistema internacional es el newton por metro. N/m. La relación mostrada en la ecua. (1) se mantiene solo para resortes ideales. Los resortes verdaderos se aproximan a esta relación lineal entre fuerza y deformación, siempre que no se sobrepase el límite elástico, limite a partir del cual el resorte se deformara permanentemente. Por otro lado debe observarse que el resorte ejerce una fuerza igual y opuesta
cuando su longitud cambia en una magnitud ElEl signo
menos indica
que la fuerza del resorte resorte esta esta en dirección opuesta al
desplazamiento si el resorte se estira o se comprime. Esta ecuación es una forma de lo que se se conoce como “LEY DE HOOKE”. HOOKE”.
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Δx
F i g. 1 Res Resorte or te some sometitido do a carga externa. extern a.
3.2.
Equilibrio estático de un cuerpo rígido. Si un objeto esta estacionario
y permanece estacionario, se dice que se
encuentra en equilibrio estático. La determinación de las fuerzas que actúan sobre un cuerpo estático estático tiene múltiples aplicaciones aplicaciones de interés, sobre todo en ingeniería. Ha sido establecido plenamente que la condición necesaria para el equilibrio es que la fuerza neta sobre un objeto sea cero. Si el objeto se grata como una partícula, esta es la única que se debe cumplir para asegurar que la partícula está en equilibrio. Esto es si la fuerza neta sobre la partícula es cero, esta permanecerá en reposo (si inicialmente inicialmente se encontraba en reposo) o se moverá en línea recta con velocidad constante (si originalmente estaba en movimiento). La situación con objetos reales es un poco más compleja ya que los objetos no se pueden tratar como partículas. Para que un objeto se encuentre en equilibrio estático, la fuerza neta sobre él debe ser cero, y el objeto no debe tener una tendencia a girar. Esta segunda condición de equilibrio requiere que el momento de una fuerza neta alrededor de cualquier origen Matemático, lo expresado anteriormente se escribe:
∑ = 0 ∑ = 0
(2) (3)
sea cero
en lenguaje.
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IV.
METODOLOGÍA
4.1.
Para verificar experimentalmente la ley de Hooke. a. Utilizando los resortes helicoidales realizamos el montaje del equipo como se muestra a continuación, el resorte fue ajustado firmemente del anillo de su extremo.
F ig. 2. I nstalación nstalación del del equipo parar ver ver if icar l a ley de H ooke y calcul ar l a constante elástica ti ca k. b.
Con la regla medimos tres veces la longitud del resorte sin carga externa, llamando a esta longitud Lo.
c. En el extremo libre del resorte se cuelgue el porta pesas.
d. Colocamos una pesa m1en el porta pesa, el resorte se estirada y esperamos que se alcance su equilibrio estático. Con la regla medimos la longitud del resorte, L1. La diferencia de L1 – L0 = Δx, Δx, es el alargamiento producido por el peso m1.registramos los valores en la tabla I. e. Agregamos al porta pesas sucesivamente, sin quitar los anteriores, pesas m2, m3, etc., y calculamos los alargamientos producidos en todos los casos con respecto a L . Registramos sus valores en tabla I.
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sucesivamente cargas). Para cada valor de peso agregado, se tomará como lectura x el promedio de las lecturas ascendentes correspondientes a un mismo peso. g. Repitamos los los pasos de “a”. hasta f. con los otros resortes. Registramos resortes. Registramos los valores en la Tabla I.
Tabl a I. I . Datos y cá cál cul os para ve verr if icar l a L ey de H ooke.
RESORTE I N°
Masa (gr)
1 2 3 4 5 6 7 8
69.6 89.6 109.6 125,6 149,6 169,6 189,6 209,6
Longitud inicial (cm) RESORTE II L=6,8 Longitud final Lf (cm) N° Masa (gr) Carga Carga ascendente descendente 7.2 7.2 1 69.6 7.4 7.4 2 89.6 7.9 7.6 3 109.6 8.5 8.4 4 129.6 9.2 9.2 5 149.6 9.7 9.8 6 169.6 10.3 10.6 7 189.6 11.4 11.4 8 209.6
RESORTE III N°
Masa (gr)
Longitud inicial (cm) L0=6,6 Longitud final Lf (cm) (cm) Carga Carga ascendente descendente 9.8 12.1 14.5 16.8 18.1 21.5 23.9 25.2
Longitud inicial (cm) L0 =6.3 Longitud final Lf (cm) (cm) carga ascendente
carga descendente
9.9 12.1 14.5 16.9 18.2 21.6 23.9 25.2
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b.
Fije uno de los extremos de cada resorte a la argolla y el otro extremo a la base del soporte, tal como se muestra en la Fig. 3. los marcamos con una cinta adhesiva para identificarlos.
Fig. 3. Instalación de los resortes para verificar la Primera condición de equilibrio c.
d. e.
Al realizar el paso “b” los resortes se deben estirar. Mida con la regla la longitud final del resorte y a partir de ella determine la deformación L o. Con el valor de Δxy Δx = L f – – L Δxy el valor de k obtenido en el procedimiento (4.1.). Determine la fuerza en el resorte. En una hoja de papel milimetrado colocada debajo de los resortes, trace un sistema de referencia OXY y en él grafique las direcciones de las fuerzas. Proceda a verificar la valides de las condiciones de equilibrio.
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a.
Fije el soporte soporte de madera en la mesa y asegúrelo mediante una prensa.
b. Suspenda la varilla en la cuchilla y por su orificio central (centro de gravedad), tal como se muestra la Fig. 4.
Fig.4 Barra suspendida en un punto. c. Utilizando ganchos, cuelgue de la palanca, a izquierda y a derecha del eje, porta pesas y pesas hasta que la barra quede en equilibrio, en posición horizontal. d. Con la regla mida las distancias de las cargas al eje de rotación. Registre su lectura en la tabla III. e.
Con la balanza mida la masa total de la pesas
m1, m2, m3, m4 conjuntamente
con los ganchos. Registre sus lecturas en la t abla III
Tabla III. Datos para verificar la segunda condición de equilibrio.
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V.
5.1.
CUESTIONARIO:
Verificación de la ley de Hooke
a. En papel milimetrado trace una gráfica Fuerza vs. Desplazamiento, para cada uno de los resortes R 1, R 2 Y R 3 y a partir de ella determine la constante elástica de los resortes. Utilice mínimos cuadrados.
Los mínimos cuadrados son los siguientes: Resorte I m= (36.62±0.085) N b= (0.538±3.025) N
Resorte II m= (1.618 ± 1.151) N b= (0.064 ±9.5) N
Resorte III m= (38.79± 0.158) N b= (0.175 ± 4.58) N
A) Cálculos para verificación de la ley de Hooke Sabemos: F=mg; donde: m=masa; g= gravedad
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RESORTE I N
Masa (kg)
1 2 3 4 5 6 7 8
0.0696 0.0896 0.1096 0.1256 0.1496 0.1696 0.1896 0.2096 1.1128
∑
Los datos del resorte I en metros y kilogramos
Longitud inicial (m) L0=0.068 Longitud final Lf (m) (m) Carga Carga ascendente descendente 0.072 0.074 0.079 0.085 0.092 0.097 0.103 0.114 0.716
0.072 0.074 0.076 0.084 0.092 0.098 0.106 0.114 0.716
promedio Xi(m)
F
K
0.072 0.074 0.0775 0.0845 0.092 0.0975 0.1045 0.114 0.716
0.682 0.878 1.074 1.230 1.466 1.662 1.858 2.054 10.604
0.004 0.006 0.0095 0.0165 0.024 0.0295 0.0365 0.046 0.172
170.5 146.33 113.05 74.54 61.08 56.33 50.90 44.65 717.38
Hallamos la pendiente y el intercepto del resorte I:
n 1 2 3 4 5 6 7 8 ∑
=X
Xi
xiyi
0.004 0.006 0.0095 0.0165 0.024 0.0295 0.0365 0.046
F=yi 0.682 0.878 1.074 1.230 1.466 1.662 1.858 2.054
0.000016 0.000036 0.00009 0.00027 0.00057 0.00087 0.0013 0.0021
0.0027 0.0052 0.010 0.0202 0.035 0.049 0.067 0.094
0.172
10.604
0.0052
0.283
i
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∑ ∑ ∑ ∑ ∑ ∑ =0.538N/m Hallamos los errores de m y b, para ello se usara procesos estadísticos Datos:
=b+mx
i
N
xi (m)
1 2 3 4 5 6 7
0.004 0.006 0.0095 0.0165 0.024 0.0295 0.0365
yi(N) 0.682 0.878 1.074 1.230 1.466 1.662 1.858
(N) 0.684 0.757 0.885 1.142 1.416 1.618 1.874
)N
(yi-
-0.002 0.121 0.189 0.088 0.05 0.044 -0.016
(yi-
)
2
0.000004 0.0146 0.0357 0.007 0.0025 0.0019 0.00025
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Grafico (fuerza vs desplazamiento) para el resorte Nº 1
Resorte I 3 2 XF
1 0 0
0.01
RESORTE II N°
Masa (Kg)
1
0.069
0.02
0 .0 3
0.04
0.05
Los datos del resorte II en metros y en kilogramos kilogramos.
Longitud inicial (m) L0=0.068 Longitud final L0 = (m) Carga Carga ascendente descendente 0.098
0.099
F
̅
k
0.0305
0.6762
0.0985
22.170
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∑
0.8755
10.898
0.1169
1.377
∑∑ ∑∑ ∑ =1.618 N ∑ ∑ ∑ ∑ ∑ ∑ =0.064 N/m Hallamos los errores de m y b, para ello se usara procesos estadísticos Datos:
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=√ ∑∑ ∑ ∑
=9.5N/m
=√ m= (1.618 ± 1.151) N b= (0.064 ±9.5) N
Grafico (fuerza vs desplazamiento) para el resorte nº2 Resorte II 2 1.5 1 0.5
F
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Hallamos el pendiente y el intercepto del resorte III n 1 2 3 4 5 6 7 8 ∑
=X
F=yi
Xi
xiyi
0.008 0.013 0.018 0.026 0.033 0.043 0.048 0.056
0.682 0.878 1.074 1.23 1.466 1.662 1.858 2.054
0.000064 0.000169 0.000324 0.000676 0.001089 0.001849 0.002304 0.003136
0.005 0.011 0.019 0.031 0.048 0.071 0.089 0.115
0.245
10.904
0.009611
0.392
i
∑ ∑ ∑
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Ahora hallamos los errores de m y b, para ello se usara proceso estadísticos Datos-:
=b+mx (N) i
N
xi (m) 1 2 3 4 5 6 7 8
0.008 0.013 0.018 0.026 0.033 0.043 0.048
yi(N)
0.682 0.878 1.074 1.23 1.466 1.662 1.858
0.485 0.679 0.873 1.183 1.455 1.842 2.036 2.347
)N
(yi-
0.179 0.199 0.201 0.047 0.011 -0.18 -0.178
(yi-
)
2
0.032041 0.039601 0.040401 0.002209 0.000121 0.0324 0.031684
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b. ¿Se cumple la ley de Hooke? Explique La ley sí se cumple pero materiales ideales, pero pero también depende de distintos factores que lo determinan como la precisión a la hora de realizar las mediciones, los resortes mismos no tienen que estar vencidos, y entre otros. En este experimento observamos que no se cumple la ley de Hooke debido a las causas antes mencionadas, razón por la cual se recomienda sumo cuidado a la hora de realizar las mediciones.
c. Utilizando la gráfica, cómo determinaría el peso de un cuerpo si se conoce la
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errores sistemáticos.- son aquellas que se repiten constantemente en el transcurso del tiempo, o bien durante una serie particular de medidas; puede ser:; a. Debido a la mala calibración de los instrumentos. e. Debido a las condiciones experimentales no adecuadas. f. Debido al uso de técnicas imperfectas. g. Debido al uso de fórmulas incorrectas. h. Debido al uso de teorías incorrectas. -
Al verificar verificar la segunda condición de equilibrio, no se garantizó garantizó que la barra barra esté
totalmente horizontal porque la vista humana no es tan precisa para ello se necesitaba un nivel, para dejarlo horizontalmente en equilibrio.
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c.
Con los datos de la tabla II descomponga las fuerzas en componentes X e Y, verifique la condición de equilibrio. R x = Σxi = 0 R y = Σyi = 0
Calcule la desviación relativa en las direcciones ortogonales. ¿A qué atribuye Ud. las desviaciones observadas? Físicamente, ¿cuál es la principal causa de la desviación? Cálculos para la verificación de la primera condición de equilibrio
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De la tabla sabemos que: F1 = 1.325 F2 = 1.241 F3 = 1.363 Sumatoria Sumatoria de fuerzas en el eje “x” ∑Fx=0 1.241cos38-1.325cos31=0 -0.0157≡0 Sumatoria Sumatoria de fuerzas en el eje “y” ∑Fy=0 1.241sen38+1.325sen31-1.363=0
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De la tabla sabemos que: F1 = 1.325 F2 = 1.241
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De la tabla sabemos que: F1 = 1.325 F2 = 1.241 F3 = 1.363
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5.3.
verificación de la segunda condición de equilibrio a)
Dibuje un diagrama de la fuerzas que actúan sobre la barra (incluidos las pesas y los ganchos).
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Hallando la fuerza: Dato: g = 9.8N
F=mg (N)
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b) Calcule la reacción en el eje. masa de la barra
m1 (kg)
m2 (kg)
m3 (kg)
m4 (kg)
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