three forces of 10N,30N and 50N act at a point at 120° to each other.Find their resultant.
Assuming the 10N force is along the positive x-axis.
Then the three forces in (magnitude, angle °) are
P(10,0), Q(30,120), R(50,240)
or in component form:
P<10,0>
Q<30cos(120),30sin(120)>=<-15, 25.9808>
R<50cos(240),50cos(240)>=<-25, -43.301>
The resultant is therefore the sum of the three vectors in component form:
Resultant=<Fx,Fy>=<-30,-17.32>
The magnitude can be found using
|F|=√(Fx²+Fy²)
and direction using
atan2(Fy/Fx), or
atan(Fy/Fx) [+180 if Fx<0]
Fr = 10N.[0o] + 30N.[120o] + 50N.[240o]
X = 10 + 30*Cos120 + 50*Cos240 = -30 N.
Y = 30*sin120 + 50*sin240 = -17.3 N.
Tan A = Y/X = -17.3/-30 = 0.57735.
A=30o S. of W. = 210o CCW from +x-axis.
Fr = -30/Cos210 = 34.64 N. @ 210o CCW.
To find the resultant of the three forces, we can use the triangle law of vector addition.
Step 1: Draw a sketch of the forces to scale.
- Draw a reference point, labeled as "O."
- Draw three arrows representing the forces.
- Label the first force as "F1" with a magnitude of 10N.
- Label the second force as "F2" with a magnitude of 30N.
- Label the third force as "F3" with a magnitude of 50N.
- The angle between the first and second forces is 120°.
Step 2: Resolve the forces into their components.
- Since the angles between the forces are given, we can resolve each force into its horizontal and vertical components.
- For F1, the horizontal component F1x is given by F1x = F1 * cos(0°), and the vertical component F1y is given by F1y = F1 * sin(0°).
- For F2, the horizontal component F2x is given by F2x = F2 * cos(120°), and the vertical component F2y is given by F2y = F2 * sin(120°).
- For F3, the horizontal component F3x is given by F3x = F3 * cos(240°), and the vertical component F3y is given by F3y = F3 * sin(240°).
Step 3: Calculate the horizontal and vertical components of the resultant.
- Add up all the horizontal components and vertical components separately.
- Let Rx be the resultant's horizontal component, given by Rx = F1x + F2x + F3x.
- Let Ry be the resultant's vertical component, given by Ry = F1y + F2y + F3y.
Step 4: Calculate the magnitude and direction of the resultant.
- The magnitude of the resultant R is given by R = sqrt(Rx^2 + Ry^2).
- The direction of the resultant θ is given by θ = tan^(-1)(Ry / Rx).
Step 5: Substitute the given values into the above equations and solve.
- F1 = 10N
- F2 = 30N
- F3 = 50N
After substituting the values and performing the calculations, the resultant comes out to be:
- Rx = -47.5N (approximately)
- Ry = 32.5N (approximately)
The magnitude of the resultant R is:
- R = sqrt((-47.5)^2 + (32.5)^2) ≈ 57.38N
The direction of the resultant θ is:
- θ = tan^(-1)(32.5 / -47.5) ≈ -34.47°
Therefore, the resultant of the three forces is approximately 57.38N in magnitude, and it acts at an angle of approximately -34.47° with respect to the positive x-axis.
To find the resultant of the three forces, we'll need to use the concept of vector addition. We can break down each force into its horizontal and vertical components and then add them together.
Step 1: Find the horizontal and vertical components of each force.
For the 10N force:
Horizontal component = 10N * cos(120°) = -5N (Since the angle is 120°, and the x-component has a negative value)
Vertical component = 10N * sin(120°) = 8.66N
For the 30N force:
Horizontal component = 30N * cos(120°) = -15N
Vertical component = 30N * sin(120°) = -25.98N
For the 50N force:
Horizontal component = 50N * cos(120°) = 25N
Vertical component = 50N * sin(120°) = 43.30N
Step 2: Add up the horizontal and vertical components separately.
Horizontal component of the resultant = -5N + (-15N) + 25N = 5N
Vertical component of the resultant = 8.66N + (-25.98N) + 43.30N = 25.98N
Step 3: Use the Pythagorean theorem to find the magnitude of the resultant.
Magnitude of the resultant = √[(horizontal component)^2 + (vertical component)^2]
Magnitude of the resultant = √[(5N)^2 + (25.98N)^2]
Magnitude of the resultant ≈ 26.99N
Step 4: Find the angle of the resultant vector with respect to the horizontal axis.
Angle = arctan(vertical component / horizontal component)
Angle = arctan(25.98N / 5N)
Angle ≈ 79.4°
Therefore, the resultant of the three forces is approximately 26.99N at an angle of 79.4° with the horizontal axis.