Question

Find the moment of the force F about origin. Magnitude of F = 20N, Angle of F with

horizontal is 30 degrees anticlockwise. Coordinates of point of application of E(5,-4)m​

Answer 1

The Moment of a force is a measure of its tendency to cause a body to rotate about a specific point or axis. ... A moment is due to a force not having an equal and opposite force directly along it's line of action

Answer 2

Answer:

64.03N-m is the moment of force about the coordinates of (5,-4) .

Explanation:

Moment of Force is given by the cross product of Position vector and force vector.

Moment of force=position vector× Force vector

M=r×F×sinθ---------------(1)

We are provided with the magnitude of force as 20N.

Since coordinates of application of force=(5,-4)

The position vector for the coordinates (5,-4) = (5-0)i+(-4-0)j=5i-4j

θ=30°

On Substituting the given values in equation (1),

M=$\sqrt{41}$×20×sin30°

M=$\sqrt{41}$×20×0.5

M=64.03N-m

Hence, the moment of force about the given coordinates is 64.03N-m.