A body at equilibrium is under the action of three forces. One force is 10N acting due east and one is 5N in the direction 60 degree north-east. What is the magnitude and direction of the third force?
Hint:
Find the resultant of the two given forces.
The third force (which keeps the resultant in equilibrium) is a force which is equal in magnitude to the resultant, but opposite in direction.
To determine the magnitude and direction of the third force, we need to use the concept of vector addition. The three forces acting on the body can be represented as vectors.
Let's represent the first force of 10N acting due east as F₁. Since it is acting due east, its direction can be represented as 0 degrees or π/2 radians.
Next, let's represent the second force of 5N in the direction 60 degrees north-east as F₂. To convert this to a standard coordinate system, we can split this force into its x-component and y-component. The x-component can be calculated using cos(60°) = 0.5, and the y-component can be calculated using sin(60°) = 0.866. Therefore, the x-component of F₂ is 5N * 0.5 = 2.5N, and the y-component of F₂ is 5N * 0.866 = 4.33N.
Now, let's find the net force acting on the body by summing the x-components and y-components of the forces separately. The x-component can be calculated by adding the x-components of F₁ and F₂: 10N + 2.5N = 12.5N. The y-component can be calculated by adding the y-components of F₁ and F₂: 0N + 4.33N = 4.33N.
Since the body is in equilibrium, the net force acting on it should be zero. Therefore, the magnitude of the third force (F₃) would be equal to the magnitude of the net force (12.5N in the x-direction and 4.33N in the y-direction), which can be calculated using the Pythagorean theorem:
Magnitude of F₃ = √(12.5N² + 4.33N²) ≈ 13.20N
Now, to find the direction of the third force, we can use trigonometry. The direction of the net force can be found by taking the inverse tangent of the y-component divided by the x-component:
Direction of F₃ = tan^(-1)(4.33N / 12.5N) ≈ 19.7°
Therefore, the magnitude of the third force is approximately 13.20N, and its direction is approximately 19.7 degrees with respect to the positive x-axis (east).