a body is at equilibrium under the action of three forces.one force is 10N acting due east and one is 5N in the direction 60¡ã northeast . what is the magnitude and direction of the third force
Northeast is E45°N or N45°E.
Assuming you mean E60°N, that means
<10,0> + <5/2,5√3/2> + <x,y> = <0,0>
To find the magnitude and direction of the third force, we can use the method of vector addition.
First, let's break down the given forces into their horizontal and vertical components.
Given:
Force 1 (10N) acting due east: This force has no vertical component, and its horizontal component is 10N.
Force 2 (5N) at an angle of 60° northeast: To find the horizontal and vertical components, we need to use trigonometry.
The horizontal component is given by cosθ * F, where θ is the angle (60°) and F is the magnitude of the force (5N). So, the horizontal component is (cos 60°) * 5N = (1/2) * 5N = 2.5N.
The vertical component is given by sinθ * F, so the vertical component is (sin 60°) * 5N = (√3/2) * 5N = 2.5√3N.
Now, let's add up the horizontal and vertical components separately to find the resultant force:
Horizontal component: 10N + 2.5N = 12.5N (east direction)
Vertical component: 0N + 2.5√3N = 2.5√3N (north direction)
Using the Pythagorean theorem, we can find the magnitude of the resultant force:
Magnitude = √(horizontal component^2 + vertical component^2) = √((12.5N)^2 + (2.5√3N)^2) = 13.96N (approximately).
To find the direction of the resultant force, we can use trigonometry again:
Direction = tan^(-1)(vertical component / horizontal component)
Direction = tan^(-1)(2.5√3N / 12.5N)
Direction = tan^(-1)(√3 / 5)
Direction ≈ 30.96°
Therefore, the magnitude of the third force is approximately 13.96N, and its direction is approximately 30.96° north of east.