Two forces F1and F2 on a particle. F1 has magnitude 5N and in direction 030°,and F2 has magnitude 8N and in direction 090°.find the magnitude and direction of their resultant.
i can't solve it so i need your help
F1 = 5 sin 30 East + 5 cos 30 North = 2.5 E + 4.33 N
F2 = 8 sin 90 East + 8 cos 90 North = 8.0 E + 0.00 N
add
F = 10.5 E + 4.33 N
|F| = sqrt (10.5^2 + 4.33^2) = sqrt (110.25 + 18.75) =sqrt (129)
= 11.4 Newtons
tan direction = E/N = 10.5 / 4.33 = 2.42
direction = 67.6 degrees compass(clockwise from North)
To find the magnitude and direction of the resultant of two forces, we can use the concept of vector addition.
Step 1: Break down each force into its horizontal and vertical components.
- Force F1 has a magnitude of 5N and is at an angle of 30° from the positive x-axis.
- The horizontal component of F1, F1x, can be found using the formula F1x = F1 * cosθ1
- F1x = 5N * cos(30°) = 4.33N (approximately)
- The vertical component of F1, F1y, can be found using the formula F1y = F1 * sinθ1
- F1y = 5N * sin(30°) = 2.5N (approximately)
- Force F2 has a magnitude of 8N and is at an angle of 90° from the positive x-axis.
- The horizontal component of F2, F2x, can be found using the formula F2x = F2 * cosθ2
- F2x = 8N * cos(90°) = 0N
- The vertical component of F2, F2y, can be found using the formula F2y = F2 * sinθ2
- F2y = 8N * sin(90°) = 8N
Step 2: Add the horizontal and vertical components of both forces together to find the resultant.
- The horizontal component of the resultant, Rx, is the sum of F1x and F2x.
- Rx = F1x + F2x = 4.33N + 0N = 4.33N (approximately)
- The vertical component of the resultant, Ry, is the sum of F1y and F2y.
- Ry = F1y + F2y = 2.5N + 8N = 10.5N (approximately)
Step 3: Use the Pythagorean theorem to find the magnitude of the resultant.
- The magnitude of the resultant, R, can be found using the formula R = √(Rx^2 + Ry^2)
- R = √(4.33N^2 + 10.5N^2) = √(18.7329N^2 + 110.25N^2) = √(129.9829N^2) = 11.40N
Step 4: Use the inverse tangent function to find the direction of the resultant.
- The direction of the resultant, θR, can be found using the formula θR = tan^(-1)(Ry/Rx)
- θR = tan^(-1)(10.5N/4.33N) = 67.3° (approximately)
Therefore, the magnitude of the resultant force is 11.40N and the direction is approximately 67.3°.