Two forces 4n and 5n act on a body in direction due north and due east respectively calculate their equilibria ( physics)
I sent the questions
please answer the question
the resultant force is r = 5i+4j
so the equilibrant is -r
H CV UK
Yes
To calculate the equilibrium resulting from two forces acting on a body, we need to find the net force and the direction of this force. The equilibrium occurs when the net force is zero.
In this case, we have two forces acting on the body: 4N due north and 5N due east. Let's break down these forces into their components.
The force due north has a magnitude of 4N and acts purely in the y-direction (upwards). So its components are:
F_north_x = 0N (no force in the x-direction)
F_north_y = 4N (force in the y-direction)
Similarly, the force due east has a magnitude of 5N and acts purely in the x-direction (to the right), so its components are:
F_east_x = 5N (force in the x-direction)
F_east_y = 0N (no force in the y-direction)
To find the net force in the x-direction, we sum the x-components of the individual forces:
Net force in the x-direction = F_north_x + F_east_x = 0N + 5N = 5N
To find the net force in the y-direction, we sum the y-components of the individual forces:
Net force in the y-direction = F_north_y + F_east_y = 4N + 0N = 4N
Now, we have the net forces in both x and y-directions:
Net force in the x-direction = 5N
Net force in the y-direction = 4N
The equilibrium occurs when the net force is zero in both x and y-directions.
So, for the given forces, there is no equilibrium, as the net force is nonzero in both directions.