Two forces of magnitude 4.ON ans 7.0N act on a small object b. The angle btw the direction along which the force is 50 degree.determine the magnitude *nd direction of d resultant force on d body b.
Two forcws of magnitud4.0N and 7.0N act on a small object B. The angle between the directions along which forces act is 50°
Determine the magnitude and direction of the resultant force on tha body B by means of a scale diagram
To find the magnitude and direction of the resultant force on object B, you can use vector addition.
1. Begin by breaking down each force into its x and y components.
Force 1 (4.0N):
- Horizontal Component: 4.0N * cos(50°)
- Vertical Component: 4.0N * sin(50°)
Force 2 (7.0N):
- Horizontal Component: 7.0N * cos(0°)
- Vertical Component: 7.0N * sin(0°)
Note: The angle for Force 1 is given as 50°, and for Force 2 as 0°, representing the direction of the forces.
2. Sum up the horizontal and vertical components of the forces separately.
Horizontal Component total: Horizontal Component of Force 1 + Horizontal Component of Force 2
Vertical Component total: Vertical Component of Force 1 + Vertical Component of Force 2
3. Find the magnitude of the resultant force using the Pythagorean theorem.
Magnitude of Resultant Force = √(Horizontal Component total^2 + Vertical Component total^2)
4. Determine the direction of the resultant force using inverse trigonometric functions.
Direction θ = tan^(-1)(Vertical Component total / Horizontal Component total)
This will give you the magnitude and direction of the resultant force.
You have to add the forces. You did not say how the forces are acting, are both pushing, or not?
Assume pushing on the object..
let the 4N be on the x axis, toward the center.
f1=-4N x
F2=-(7N*cos50 x )-(7N*sin50)y
add them.