I) To illustrate the forces on a vector diagram, we will start by drawing a coordinate system.
iii) Label the x-axis as horizontal and the y-axis as vertical.
iii) Draw a vector with a length of 30 units in the positive x-direction (0°) and label it as 30N.
iiii) Draw a vector with a length of 40 units in the positive y-direction (90°) and label it as 40N.
iiii) Draw a vector with a length of 50 units at an angle of 217° between the negative x-axis and the vector and label it as 50N.
Note: To draw the vector at 217°, start from the negative x-axis and rotate in the clockwise direction by 37°.
ii) The net force is the vector sum of the three forces. To find the net force, we can add the three vectors in the x and y directions separately.
In the x-direction:
- The 30N force has an x-component of 30N * cos(0°) = 30N * 1 = 30N.
- The 50N force has an x-component of 50N * cos(217°) = -50N * (-0.459) = 22.95N.
The net force in the x-direction is 30N + 22.95N = 52.95N (approximately).
In the y-direction:
- The 40N force has a y-component of 40N * cos(90°) = 0N.
- The 50N force has a y-component of 50N * sin(217°) = 50N * (-0.888) = -44.4N.
The net force in the y-direction is 0N + (-44.4N) = -44.4N.
The net force can be found using the Pythagorean theorem where the magnitude of the net force is the square root of the sum of the squares of the x and y components.
Net force = √(52.95N^2 + (-44.4N)^2)
Net force = √(2800.7025N^2 + 1971.36N^2)
Net force = √(4772.0625N^2)
Net force ≈ 69N.
So, the net force is approximately 69N.
Three forces act on a point 30N at 0°, 40N at 90° , and 50N at 217°
I) illustrate on a vector diagram
ii) what is the net force
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