Forces of 10N; 8N; 6N and 4N; act at the same point of an object on bearings 0*; 90*; 135* and 270* respectively. Find, (a) resultant force; (b) equilibrant

Question

I know the 90* and 270* have a 4N resultant, but my scale drawing is left with a quadrilateral and I'm a little stuck there, any help would be appreciated.

Answer 1

Resolve each vector into its x and y components.   For the 6N force at 135°, the components are:

 

F = 6·cos (135°)i + 6·sin(135°)j  where i and j are the unit vectors in the x and y directions respectively

 

To get the net force, add all of the x components together then all of the y-components together. The equilibrant is equal and opposite to the net force (change the signs on the x and y components of the net force).

 

To do it graphically, put the non-arrow end of each vector at the arrow end of the previous vector.  You can add them in any order.  The resultant vector goes from the non-arrow end of the first vector to the arrow end of the last one.  The equilibrant is equal in length and opposite in direction.