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
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.
I) To illustrate the forces on a vector diagram, we can represent each force as an arrow with its magnitude and direction.
1) The first force, 30N at 0°, can be represented as an arrow of length 30 units pointing to the right (east).
2) The second force, 40N at 90°, can be represented as an arrow of length 40 units pointing upward (north).
3) The third force, 50N at 217°, can be represented as an arrow of length 50 units at an angle of 217° from the positive x-axis.
Now, we can draw these arrows on a graph, keeping in mind the scales for length and angles.
II) To find the net force, we need to calculate the resultant of the three forces.
To do this, we first resolve each force into its horizontal and vertical components:
1) The first force, 30N at 0°, has a horizontal component of 30N * cos(0°) = 30N * 1 = 30N and a vertical component of 30N * sin(0°) = 30N * 0 = 0N.
2) The second force, 40N at 90°, has a horizontal component of 40N * cos(90°) = 40N * 0 = 0N and a vertical component of 40N * sin(90°) = 40N * 1 = 40N.
3) The third force, 50N at 217°, has a horizontal component of 50N * cos(217°) = -50N * 0.697 = -34.85N (approximately) and a vertical component of 50N * sin(217°) = 50N * 0.717 = 35.85N (approximately).
Now, sum up the horizontal and vertical components of the forces:
Horizontal component: 30N + 0N - 34.85N = -4.85N (approximately).
Vertical component: 0N + 40N + 35.85N = 75.85N (approximately).
Using the Pythagorean theorem, we can find the magnitude of the net force:
Net force = (horizontal component)^2 + (vertical component)^2
= (-4.85N)^2 + (75.85N)^2
= 23.52N^2 + 5742.02N^2
= 5765.54N^2
Taking the square root of 5765.54N^2, we get approximately 75.94N.
Therefore, the net force acting on the point is approximately 75.94N at an angle of atan(75.85N / -4.85N) = -86.23° (approximately) from the positive x-axis.