Suma De Vectores Laboratorio De Fisica

Páginas: 5 (1206 palabras) Publicado: 28 de mayo de 2012
Métodos gráficos y tablas.

Método nº 1 Polígono.

Ejercicio nº 1


A= 96.8[gr] ± 0.03[gr] Escala = 0.095 [N] = 1[cm]
0.0968[kg] ± 0.00003 [kg]
0.94864[N] ± 0.000294[N] 0.95[N]/0.095[N] = 10[cm]
≈ 0.95 [N] ±0.000294[N]



B= 99.5[gr] ± 0.03[gr] Escala= 0.095 [N] = 1[cm]
0.995[kg] ± 0.00003[kg]
0.975[N] ± 0.000294[N] 0.98[N]/0.095[N] = 10.3[cm]
≈ 0.98[N] ± 0.000294[N]



C= 100[gr] ±0.03[gr] Escala = 0.095[N] = 1[cm]
0.095[kg] ±0.00003[kg]
0.99[N] ±0.000294[N] 0.99[N]/0.095[N]= 10.4[cm]Ejercicio nº2



A= 91.8[gr] ± 0.03[gr] Escala = 0.095[N] = 1[cm]
0.0918[kg] ± 0.00003[kg]
0.89964[N] ±0.000294 [N] 0.90[N]/0.095[N] = 9.4 [cm]
≈ 0.90[N] ± 0.000294[N]



B= 170[gr] ± 0.03 [gr] Escala = 0.095 [N] = 1 [cm]
0.17[kg] ± 0.00003[kg]1.666[N] ± 0.000294[N] 0.90[N]/0.095[N] = 9.4[cm]
≈ 1.7[N] ± 0.000294[N]



C= 99.5 [gr] ± 0.03 [gr] Escala 0.095[N] = 1 [cm]
0.0995 [kg] ± 0.00003[kg]
0.9751[N] ± 0.000294 [N] 0.98[N]/0.095[N] = 10.3[cm]
≈ 0.98[N] ± 0.000294[N]







Método nº 2trigonometría elemental.

fx= Fcos ө y fy= Fsen ө , fx y fy componentes cartesianos

Ejercicio nº1

→ → →
A = 0° ; B =120 ° ; C=240°

Ax = 0.95 cos0°= 0.95[N]
Ay = 0.95 sen0°= 0

Bx= 0.98 cos120° = -0.49[N]
By= 0.98 sen120°= 0.85[N]

Cx= 0.99 cos240°= -0.49[N]
Cy= 0.99 sen 240°= -0.85 [N]

Rx= Ax+ Bx+ Cx= ( 0.95 - 0.49 -0.49 ) = -0.03 [N]
Ry= Ay+ By+Cy= (0 + 0.85 – 0.85) = 0 [N]

→ → → →
A=(0.95 ; 0) B=(-0.49 ; 0.85) C=(-0.49; -085) R=(-0.03 ; 0)

tgө= Ry/Rx = 0/-0.03 = 0 donde ө = tg¹ 0 = 0°


R= √(-0.03)² + (0)²= 0.03[N]

R=0.03[N]Ejercicio nº2

→ → →
A = 0° ; B =150 ° ; C=210°

Ax = 0.90cos0° = 0.90[N]
Ay = 0.90sen0° =0 [N]

Bx= 1.7cos150° = -1.47 [N]
By=1.7sen150° =-0.85 [N]

Cx= 0.98cos210° = -0.84 [N]
Cy= 0.98cos210° = -0.49 [N]

Rx= Ax+ Bx+ Cx=(0.90 – 1.47 – 0.84) = -1.41[N]
Ry= Ay+ By+ Cy=(0 + 0.85 – 0.49) = 0.36[N]

tgө= Ry/Rx =0.36/-1.41= -0.25 donde ө = tg¹-0.25 = -14°
→ →
R=√(-1.41)² + (0.36)²= R= 1.45[N]

Metodo nº3 del coseno.

D = A + B

D²= A² + B² - 2ABcosө


A= 0.95 [N] D²=0.95² + 0.98² - 2(0.95)(0.98)cos120°

B= 0.98 [N] =1.862-1.862
Cos= 60° D²= 0

D= ¿? → D= 0[N]


La dirección delvector resultante en C.

Ejercicio nº1

Ө= ¿? B²= A² + D² - 2ADcosө

A= 0.95 [N] ө= cos¹(A²+D²-B²) ——→ cos¹ ( (0.95)²+(0)²-(0.98)²)
→ --------------- -----------------------------------
B= 0.98 [N] 2AD( 2-(0.95)(0) )

D= 0 [N] cos¹ (0.8637)
----------- -------→ ө= 90°
(0)

Ejercicio nº2

D= A + B


A= 0.90 [N] D²=A² + B² - 2ABcosө

B= 1.7 [N] D²= 0.90² + 1.7² - 2(0.90)(1.7)cos150°
Cos=150°...
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