Illustrate in diagram a force 150N is inclined a 50 to the horizontal direction find it component in the horizontal
We do not have the tools to provide diagrams. The x-component of the vector you have described would be
150 cos50 = 96.4 N
150cos50=96.4n
Fx equal 150cos50 degree equal to the 96.4 n.
To illustrate the force and find its horizontal component, we can use a diagram and apply trigonometry.
1. Draw a coordinate system with a horizontal x-axis and a vertical y-axis.
2. Label the horizontal direction as "x" and the vertical direction as "y".
3. Mark a point at the origin (0,0) to represent the starting point of the force.
4. Draw a line from the origin at an angle of 50° to the horizontal direction. This represents the force vector of 150N.
5. To find the horizontal component of the force, draw a line perpendicular to the x-axis from the end point of the force vector (call it point B) down to the x-axis (call it point C).
6. Label the length of the force vector as "a" and the length of the horizontal component (line BC) as "b".
7. The horizontal component can be calculated using the equation: b = a * cos(angle).
In this case, b = 150N * cos(50°).
The drawn diagram should now indicate the force vector (inclined at 50°) and the length of its horizontal component.