Respuesta :
Answer:
Magnitude (in kilogram) of the resultant force, [tex]|M| = 1958.53 kg[/tex]
Direction = 164.37°
Explanation:
The two forces exerted by the two tugboats are non zero vectors, therefore they can be expressed as:
For the 1620 kg tugboat in the direction N35degreesW
[tex]M_{1} = 1620 cos \theta_1 i + 1620 sin \theta_1 j\\[/tex]
The direction of the force from the positive x - axis, [tex]\theta_1 = 90 - 35 = 55^{0}[/tex]
The equation above then becomes:
[tex]M_{1} = 1620 cos 55 i + 1620 sin 55 j\\ M_{1} = (1620 * 0.57) i + (1620*0.82) j\\M_{1} = 923.4 i + 1328.4j[/tex]
For the 1250 kg tugboat in the direction S50degreesW
[tex]M_{2} = 1250 cos \theta_2 i + 1250 sin \theta_2 j\\[/tex]
The direction of the force from the positive x - axis,
[tex]\theta_2 = - (90 - 50) = - 40^{0}[/tex]
The equation above then becomes:
[tex]M_{2} = 1250 cos(-40) i + 1250 sin (-40) j\\ M_{2} = (1250 * 0.77) i - (1250*0.64) j\\M_{2} = 962.5 i - 800j[/tex]
The resultant vector will be given by:
[tex]M = M_{1} + M_{2} \\ M = 923.4 i + 1328.4j + 962.5 i - 800j\\M = 1885.9i + 528.4j[/tex]
The magnitude of the resultant is given by:
[tex]|M| = \sqrt{1885.9 ^{2} + 528.4^2 } \\|M| = \sqrt{3556618.81 + 279206.56}\\|M| = \sqrt{3835825.37} \\|M| = 1958.53 kg[/tex]
The direction will be given by :
[tex]cos \theta = \frac{a}{|M|} \\cos \theta = \frac{1885.9}{1958.53}\\cos \theta = 0.963\\\theta = cos^{-1} 0.963\\\theta = 15.63^0[/tex]
Direction = 180 - 15.63 = 164.37°
