Two force 10N and 20N are inclined at angle 60degree to each other. Find the magnitude and direction of the resultant
Force
We can use vector addition to find the magnitude and direction of the resultant force:
- Draw a diagram representing the two forces as vectors. Label one force as F1 = 10N and the other as F2 = 20N. The angle between them is 60 degrees.
- Use trigonometry to find the components of each force in the x and y directions. For F1, the x-component is 10cos(60) = 5N and the y-component is 10sin(60) = 8.66N. For F2, the x-component is 20cos(120) = -10N and the y-component is 20sin(120) = 17.32N (note the negative sign for the x-component due to the angle being greater than 90 degrees).
- Add the x-components and y-components separately to find the components of the resultant force. Rx = 5N - 10N = -5N and Ry = 8.66N + 17.32N = 25.98N.
- Use the Pythagorean theorem to find the magnitude of the resultant force: R = sqrt((-5)^2 + (25.98)^2) = 26.87N.
- Use trigonometry again to find the angle that the resultant force makes with the x-axis. tan(theta) = Ry/Rx = -25.98/5, so theta = -80.54 degrees (note the negative sign due to the resultant vector being in the third quadrant). Therefore, the direction of the resultant force is 80.54 degrees below the negative x-axis.
To find the magnitude and direction of the resultant force, we can use the concept of vector addition.
Step 1: Resolve the forces into their x and y components.
The force of 10N can be split into two components:
- The x component is given by F1x = 10N * cos(60°)
- The y component is given by F1y = 10N * sin(60°)
The force of 20N can also be split into two components:
- The x component is given by F2x = 20N * cos(60°)
- The y component is given by F2y = 20N * sin(60°)
Step 2: Sum the x and y components separately.
Fx = F1x + F2x
Fy = F1y + F2y
Step 3: Calculate the magnitude of the resultant force.
The magnitude of the resultant force, Fr, can be calculated using the Pythagorean theorem:
Fr = √(Fx^2 + Fy^2)
Step 4: Calculate the direction of the resultant force.
The direction of the resultant force, θ, can be found using the inverse tangent function (tan^-1):
θ = tan^-1 (Fy/Fx)
Now let's substitute the values and calculate the magnitude and direction of the resultant force.
F1x = 10N * cos(60°) = 5N
F1y = 10N * sin(60°) = 8.66N
F2x = 20N * cos(60°) = 10N
F2y = 20N * sin(60°) = 17.32N
Fx = F1x + F2x = 5N + 10N = 15N
Fy = F1y + F2y = 8.66N + 17.32N = 25.98N
Fr = √(Fx^2 + Fy^2) = √(15N^2 + 25.98N^2) = √(225N^2 + 675N^2) = √900N^2 = 30N
θ = tan^-1 (Fy/Fx) = tan^-1 (25.98N/15N) = tan^-1 (1.73) = 60°
Therefore, the magnitude of the resultant force is 30N, and the direction is 60°.