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The line of action of the force \vec{F}=5\vec{i}-10\vec{j} F =5 i −10 j ​ N passes through the point with coordinates (in meters) (2, 2). What is the moment of this force about the coordinate origin?

Respuesta :

Answer:

[tex]\tau = - 30 \hat k[/tex]

Explanation:

Position vector of the point of application of point of application of force is given as

[tex]\vec r = 2\hat i + 2\hat j[/tex]

now we have have force

[tex]\vec F = 5 \hat i - 10\hat j[/tex]

now the moment of force is given as

[tex]\tau = \vec r \times \vec F[/tex]

[tex]\tau = (2\hat i + 2\hat j) \times (5\hat i - 10\hat j)[/tex]

[tex]\tau = -20\hat k - 10 \hat k[/tex]

[tex]\tau = - 30 \hat k[/tex]

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