Question

The magnitude and direction exerted by two tugboats towing a ship are 1620​kilograms, N35degrees​W,and 1250​kilograms, S50degrees​W,respectively. Find the​ magnitude, in​ kilograms, and the direction​ angle, in​ degrees, of the resultant force.

Answer 1

Answer:

Magnitude (in kilogram) of the resultant force, $|M| = 1958.53 kg$

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 N35degrees​W

$M_{1} = 1620 cos \theta_1 i + 1620 sin \theta_1 j$

The direction of the force from the positive x - axis, $\theta_1 = 90 - 35 = 55^{0}$

The equation above then becomes:

$M_{1} = 1620 cos 55 i + 1620 sin 55 j\\ M_{1} = (1620 * 0.57) i + (1620*0.82) j\\M_{1} = 923.4 i + 1328.4j$

For the 1250 kg tugboat in the direction S50degrees​W

$M_{2} = 1250 cos \theta_2 i + 1250 sin \theta_2 j$

The direction of the force from the positive x - axis,

$\theta_2 = - (90 - 50) = - 40^{0}$

The equation above then becomes:

$M_{2} = 1250 cos(-40) i + 1250 sin (-40) j\\ M_{2} = (1250 * 0.77) i - (1250*0.64) j\\M_{2} = 962.5 i - 800j$

The resultant vector will be given by:

$M = M_{1} + M_{2} \\ M = 923.4 i + 1328.4j + 962.5 i - 800j\\M = 1885.9i + 528.4j$

The magnitude of the resultant is given by:

$|M| = \sqrt{1885.9 ^{2} + 528.4^2 } \\|M| = \sqrt{3556618.81 + 279206.56}\\|M| = \sqrt{3835825.37} \\|M| = 1958.53 kg$

The direction will be given by :

$cos \theta = \frac{a}{|M|} \\cos \theta = \frac{1885.9}{1958.53}\\cos \theta = 0.963\\\theta = cos^{-1} 0.963\\\theta = 15.63^0$

Direction = 180 - 15.63 = 164.37°