Need help on Applied Mathematics

Can anyone give me correct answer to this model questions,

a) ABC is a tringle such that AB = 6m, BC = 4m AND CA = 4m. Forces of 3 ,2 and 2N act along AB,BC and CA respectively.Show that the system is equivalent to a couple and find its moment. Find the magnitude,derection and line of action of the resultant if the force in AB reversed.

b) One end of a uniform rod of mass 4kg is freely hinged to a fixed point A. The rod is held inclined at 30 digrees to the downward vetical by a force F applied at the other end. F is at right angles to the rod. Find the magnitude of F and the components of the reacion at A along and prependicular to the rod.

Pls help on this

In the 1st one if you resolve the forces along and perpendicular to AB, the 3N force will cancel by the 2N compnents along AB. All that will be left would be 2 components of the 2, 2N forces perpendicular to AB, in opposite directions. So the couple moment would be 2N*Sin(Angle); * 6m = 2N*sqrt(7)/4*6m = 3sqrt(7) Nm

For the second one resolve the weight from the centre of gravity along and perpendicular to the rod. Now the reaction along the rod will be mgcos(30) = 4g sqrt(3)/2, directed away from A
Now write the equation for all the perpendicular forces, i.e. F + Rp = mg Sin(30)

Also take moments about A, we have:
F*L = mg Sin(30) *L/2
therefore F = 4g Sin(30)/2 = g

now solve for Rp:
Rp = 4g sin(30)-F = 2g -g = g (same direction as F)

i dont think that the first question is correct:
Vect. Ab + vect BC would give vect. AC which would cancel out with vect. CA

raghav said:

i dont think that the first question is correct:
Vect. Ab + vect BC would give vect. AC which would cancel out with vect. CA

Click to expand...

If you consider the question as a static equilibrium question with no rotation dynamics, that means the rod is a point then yes the forces cancel out. But since we are considering rotation, that means we have to consider the point of application of the forces, and their moments to find out if the thing will be in rotational equilibrium or not. And it turns out it is not and has a net couple.

Nah, nah, nah!
Even though I haven't studied the 1st answer in depth, I think it was somewhat in line. The method is correct. AB+BC+CA sure is equal to zero. But remember the question says that the force in AB is reversed meaning we now have: BA+BC+CA.

This requires a bit of trigonometry; resolving along vertical and horizontal axis and all that.

To get the resultant force you simply add the corresponding components of each individual force together.

To get the direction of the resultant force you simply calculate: arctan(y component/ x component) and you should get its inclination to the horizontal. Sorry i couldn't give it to you in numerical terms, but I'm sure you'll be able to fix that.