I have two doubts in this equilibrant problem (resultant force)

[Benjamin_harsh]

Homework Statement

How to guess directions like "North East" in equilibrant problems and where does "51" angle fits in the triangle?

Relevant Equations

How to guess directions like "North East" in equilibrant problems and where does "51" angle fits in the triangle?

Two forces are pushing an object along the ground. One force is 10 N [W] and the other is 8.0 N. Sketch a diagram showing the equilibrant of these two forces, and determine the equilibrant.

Ans) Calculate the equilibrant...
C² = a² + b² = 102 + 82
c = 13 NDiagram:

tanΘ = opp/adj = 10 / 8
Θ = 51°

The equilibrant is 13 N [N51°E]. How to guess directions like "North East" in equilibrant problems and where does "51" angle fits in the triangle?

 

[berkeman]

By "equilibrant" I think you mean "resultant force" (just different terms for the same thing, I think).

You shouldn't have to guess anything. If you are given the magnitude and direction of the two vectors (in either polar or rectangular coordinates), you should be able to draw them nose-to-tail to see the resultant (or add them using algebra in rectangular coordinates).

One force is 10 N [W] and the other is 8.0 N.

Is the "W" in your problem statement supposed to mean "West"? If so that is usually drawn as a horizontal vector pointing to the left (just like on a map with "North" at the top of the map). And why is there no direction given for the 2nd vector?

 

[jbriggs444]

By "equilibrant" I think you mean "resultant force" (just different terms for the same thing, I think).

As I read the drawing, the "equilibriant" is that force which when added to the existing forces results in a total of zero. So it is the opposite of the "resultant". [Google agrees]

 

[CWatters]

where does "51" angle fits in the triangle?

You did this calculation...

Tan(angle)= Opposite/Adjacent = 10/8

So 51 degrees is the angle opposite the side of length 10.