Two forces F1 and F2 act on a particle. F1 has a magnitude of 6N and in direction 045°, and F2 has a magnitude of 7N and in direction 154°. Find the magnitude and direction of their resultant
To find the magnitude and direction of the resultant force, we can use vector addition. Here are the steps:
Step 1: Break down the forces F1 and F2 into their respective components along the x-axis and y-axis.
F1x = F1 * cos(theta1)
F1y = F1 * sin(theta1)
F2x = F2 * cos(theta2)
F2y = F2 * sin(theta2)
where F1 is the magnitude of force F1 (6N), F2 is the magnitude of force F2 (7N), theta1 is the direction of force F1 (45°), and theta2 is the direction of force F2 (154°).
Step 2: Add the x-components and the y-components separately.
Rx = F1x + F2x
Ry = F1y + F2y
Step 3: Calculate the magnitude of the resultant force using the Pythagorean theorem.
R = sqrt(Rx^2 + Ry^2)
Step 4: Calculate the direction of the resultant force using the inverse tangent function.
theta = atan(Ry / Rx)
Step 5: Convert the direction from radians to degrees.
theta_degrees = theta * (180 / pi)
Let's substitute the given values and calculate the magnitude and direction of the resultant force:
F1x = 6N * cos(45°) = 6N * 0.7071 = 4.2426N
F1y = 6N * sin(45°) = 6N * 0.7071 = 4.2426N
F2x = 7N * cos(154°) = 7N * (-0.9848) = -6.8936N
F2y = 7N * sin(154°) = 7N * 0.1736 = 1.2152N
Rx = 4.2426N - 6.8936N = -2.651N
Ry = 4.2426N + 1.2152N = 5.4578N
R = sqrt((-2.651N)^2 + (5.4578N)^2)
R = sqrt(7.0408N^2 + 29.8402N^2)
R = sqrt(36.881N^2)
R = 6.0764N
theta = atan((5.4578N) / (-2.651N))
theta = -63.13°
Note: The negative sign indicates that the direction of the resultant force is opposite to the x-axis.
Therefore, the magnitude of the resultant force is approximately 6.0764N, and the direction is approximately -63.13°.
To find the magnitude and direction of the resultant force, we can use the concept of vector addition.
First, let's find the horizontal and vertical components of each force.
F1:
Magnitude: 6N
Direction: 045°
Horizontal component F1x:
F1x = 6N * cos(45°) ≈ 4.24N
Vertical component F1y:
F1y = 6N * sin(45°) ≈ 4.24N
F2:
Magnitude: 7N
Direction: 154°
Horizontal component F2x:
F2x = 7N * cos(154°) ≈ -5.64N
Vertical component F2y:
F2y = 7N * sin(154°) ≈ 4.89N
Now, let's find the sum of the horizontal and vertical components for the resultant force.
Sum of horizontal components:
Fresx = F1x + F2x ≈ 4.24N + (-5.64N) ≈ -1.4N
Sum of vertical components:
Fresy = F1y + F2y ≈ 4.24N + 4.89N ≈ 9.13N
Next, let's find the magnitude and direction of the resultant force.
Magnitude of the resultant force:
Fres = sqrt(Fresx^2 + Fresy^2) ≈ sqrt((-1.4N)^2 + (9.13N)^2) ≈ 9.32N
Direction of the resultant force:
tanθ = Fresy / Fresx
θ = arctan(Fresy / Fresx) ≈ arctan(9.13N / -1.4N) ≈ -80.6°
The magnitude of the resultant force is approximately 9.32N, and its direction is approximately -80.6°.