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

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

Direction = 164.37°

Step-by-step 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| = √(3556618.81 + 279206.56)\\|M| = √(3835825.37) \\|M| = 1958.53 kg`

The direction will be given by :

`cos \theta = (a)/(|M|) \\cos \theta = (1885.9)/(1958.53)\\cos \theta = 0.963\\\theta = cos^(-1) 0.963\\\theta = 15.63^0`

Direction = 180 - 15.63 = 164.37°