The resultant displacement is approximately 77.9 mm at an angle of approximately -101.7° from the positive x-axis.
To find the resultant displacement and its direction, we can add the individual displacements vectorially.
Given the following displacements in the x, y plane:
1. 60 mm in the +y direction
2. 30 mm in the -x direction
3. 40 mm at 150° (counterclockwise from the positive x-axis)
4. 50 mm at 240° (counterclockwise from the positive x-axis)
Let's represent each displacement vector as follows:
1. Vector A: 60 mm in the +y direction, denoted as A = 0î + 60ĵ mm
2. Vector B: 30 mm in the -x direction, denoted as B = -30î + 0ĵ mm
3. Vector C: 40 mm at 150°, denoted as C = 40cos(150°)î + 40sin(150°)ĵ mm
4. Vector D: 50 mm at 240°, denoted as D = 50cos(240°)î + 50sin(240°)ĵ mm
Now, we can add these vectors to find the resultant displacement:
Resultant displacement = A + B + C + D
Calculating the individual components:
Resultant displacement in the x-direction: x-component = 0 + (-30) + (40cos(150°)) + (50cos(240°))
Resultant displacement in the y-direction: y-component = 60 + 0 + (40sin(150°)) + (50sin(240°))
Using trigonometric identities and evaluating the expressions:
x-component ≈ -10.3 mm
y-component ≈ -76.8 mm
To find the magnitude of the resultant displacement, we use the Pythagorean theorem:
Magnitude = √((-10.3)^2 + (-76.8)^2) ≈ 77.9 mm
To find the direction of the resultant displacement, we use the inverse tangent function:
Direction = tan^(-1)(y-component/x-component) ≈ -101.7°
Therefore, the resultant displacement is approximately 77.9 mm at an angle of approximately -101.7° from the positive x-axis.