Question

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?

Answer 1

Answer:

$\tau = - 30 \hat k$

Explanation:

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

$\vec r = 2\hat i + 2\hat j$

now we have have force

$\vec F = 5 \hat i - 10\hat j$

now the moment of force is given as

$\tau = \vec r \times \vec F$

$\tau = (2\hat i + 2\hat j) \times (5\hat i - 10\hat j)$

$\tau = -20\hat k - 10 \hat k$

$\tau = - 30 \hat k$