Find the resultant, in magnitude and direction of a force of 40 N which makes an angle with the + X axis of 38o and a force of 20 N acting along the + X axis. Find the x and y components of the resultant and express them as i and j vectors.
40N at 38° = 31.52i + 25.86j
20N at 0° = 20.00i + 0j
sum = 51.52i + 25.86j
To find the resultant force in magnitude and direction, we can use vector addition. First, we need to find the x and y components of each force.
Given:
Force F1 = 40 N at an angle of 38° with the +X axis
Force F2 = 20 N along the +X axis
Step 1: Find the x and y components of F1.
The x component of F1 (F1x) can be found using the formula:
F1x = F1 * cos(theta)
where theta is the angle with the +X axis.
F1x = 40 N * cos(38°)
F1x ≈ 30.53 N
The y component of F1 (F1y) can be found using the formula:
F1y = F1 * sin(theta)
where theta is the angle with the +X axis.
F1y = 40 N * sin(38°)
F1y ≈ 24.31 N
Step 2: Find the x and y components of F2.
Since F2 is along the +X axis, its y component (F2y) would be zero.
F2y = 0 N
The x component of F2 (F2x) is equal to the magnitude of F2.
F2x = 20 N
Step 3: Add the x and y components of both forces to find the resultant force components.
Resultant x component (Rx) = F1x + F2x
Rx = 30.53 N + 20 N
Rx = 50.53 N
Resultant y component (Ry) = F1y + F2y
Ry = 24.31 N + 0 N
Ry = 24.31 N
Step 4: Express the resultant force in magnitude and direction.
To find the magnitude of the resultant force (R), we can use the Pythagorean theorem:
R = sqrt(Rx^2 + Ry^2)
R = sqrt((50.53 N)^2 + (24.31 N)^2)
R ≈ 56.38 N
To find the direction of the resultant force (theta_r), we can use the inverse tangent function:
theta_r = arctan(Ry / Rx)
theta_r = arctan(24.31 N / 50.53 N)
theta_r ≈ 26.31°
Therefore, the resultant force has a magnitude of approximately 56.38 N and is at an angle of approximately 26.31° with the +X axis.
Finally, expressing the x and y components of the resultant force as i and j vectors, we have:
Resultant force (R) = Rx i + Ry j
R ≈ (50.53 N)i + (24.31 N)j