Beer Jhonson Solucionario
Páginas: 2 (356 palabras)
Publicado: 27 de mayo de 2012
1.70:
[pic]
El tercer tramo debe haber tomado el este marinero a distancia.
[pic]
y una distancia al norte.
[pic]
La magnitud del desplazamiento es:
[pic]
y la dirección es arctan[pic]= [pic] al norte de este, que es [pic] al este del norte.
Una respuesta más precisa será necesario retener adicional cifras significativas en los cálculos intermedios.
4.31: Take the[pic]-direction to be along[pic] and the [pic]-direction to be along [pic]. Then [pic] and [pic], so [pic], at an angle of [pic] from [pic].
4.32: Get g on X:
[pic]
[pic][pic]
[pic]
4.33: a) The resultant must have no y-component, and so the child must push with a force with y-component [pic] For the child to exertthe smallest possible force, that force will have no x-component, so the smallest possible force has magnitude 16.6 N and is at an angle of [pic], or [pic] clockwise from the
[pic]-direction.b)[pic].
5.3: a) The two sides of the rope each exert a force with vertical component T [pic], and the sum of these components is the hero’s weight. Solving for the tension T,
[pic]
b) Whenthe tension is at its maximum value, solving the above equation for the angle [pic] gives
[pic]
5.6: [pic]
5.7: a) [pic]
b) [pic]
5.11: a) [pic] so[pic] or [pic]
b) [pic]5.46: a) The analysis of Example 5.22 may be used to obtain [pic] but the subsequent algebra expressing R in terms of L is not valid. Denoting the length of the horizontal arm as r and the length ofthe cable as [pic] The relation [pic] is still valid, so [pic] Solving for the period T,
[pic]
Note that in the analysis of Example 5.22,[pic] is the angle that the support (string or cable) makeswith the vertical (see Figure 5.30(b)). b) To the extent that the cable can be considered massless, the angle will be independent of the rider’s weight. The tension in the cable will depend on the...
[pic]
El tercer tramo debe haber tomado el este marinero a distancia.
[pic]
y una distancia al norte.
[pic]
La magnitud del desplazamiento es:
[pic]
y la dirección es arctan[pic]= [pic] al norte de este, que es [pic] al este del norte.
Una respuesta más precisa será necesario retener adicional cifras significativas en los cálculos intermedios.
4.31: Take the[pic]-direction to be along[pic] and the [pic]-direction to be along [pic]. Then [pic] and [pic], so [pic], at an angle of [pic] from [pic].
4.32: Get g on X:
[pic]
[pic][pic]
[pic]
4.33: a) The resultant must have no y-component, and so the child must push with a force with y-component [pic] For the child to exertthe smallest possible force, that force will have no x-component, so the smallest possible force has magnitude 16.6 N and is at an angle of [pic], or [pic] clockwise from the
[pic]-direction.b)[pic].
5.3: a) The two sides of the rope each exert a force with vertical component T [pic], and the sum of these components is the hero’s weight. Solving for the tension T,
[pic]
b) Whenthe tension is at its maximum value, solving the above equation for the angle [pic] gives
[pic]
5.6: [pic]
5.7: a) [pic]
b) [pic]
5.11: a) [pic] so[pic] or [pic]
b) [pic]5.46: a) The analysis of Example 5.22 may be used to obtain [pic] but the subsequent algebra expressing R in terms of L is not valid. Denoting the length of the horizontal arm as r and the length ofthe cable as [pic] The relation [pic] is still valid, so [pic] Solving for the period T,
[pic]
Note that in the analysis of Example 5.22,[pic] is the angle that the support (string or cable) makeswith the vertical (see Figure 5.30(b)). b) To the extent that the cable can be considered massless, the angle will be independent of the rider’s weight. The tension in the cable will depend on the...
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