Expressing moment as a vector in i,j,

[chandran]

there is a beam. A moment is applied with a value of 3NM at one end of the beam(right end). the direction of moment is clockwise in the z direction. z direction is away from the screen and x-axis is along the beam. the origin of coordinate system is is at the other end(left end).
ho do i express the moment in terms of vectors like 3i+4j etc.

 

[supersix2]

If the moment is in the Z direction then the problem is already solved. Remember that in i,j,k coordinates x=i, y=j, z=k. Therefore if the moment is acting in the z direction what would your vector be? Use the right hand rule to figure out the direction of the moment. Hint (curl your fingers of your right hand so it resembles the letter "C" then stick your thumb out. Now curl your hand in the clockwise direction. Which way does your thumb point? That is the direction of the moment along that particular axis.)

 

[chandran]

supersix,
that is ok. what i need is the moment vector in the form of ai+bj+ck. In this problem the moment vector is 0i+0j+3k. Is it correct?

But on seeing a moment vector like this ai+bj+ck how can one know at which point the moment vector is acting? In this problem

the moment vector is acting at a point at the right end of the beam which is SAY a distance L from the left end of the beam(i.e the orgin of
the vector's coordinate system). But this vector ai+bj+ck alone is not able to give the position. Why

 

[supersix2]

You are getting confused I think between what a vector is and what a point is. A vector only describes the direction something something is acting in. Not at which point it is acting it.
Your answer is still incorrect because of the direction in the k direction. Use your right hand rule again. Your thumb should be pointing into the paper. If that is the case then the moment is acting along the negative Z axis or negative k direction.

The moment has no components in the x or y directions therefore the magnitude in the i and j directions are 0.

Think of it this way. The magnitude of a moment is equal to
$$rFsin\theta$$ Where $$\theta$$ is the angle between the force creating the moment and the axis in which the moment is rotating around. Therefore the maximum moment is when the angle is 90 degrees a thus the sine of it equals 1. This also means that there is no component of the force or the moment in any other directions aside from the line of action. As a result your moment is only in one direction the negative z direction.