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1. Resultant of Coplanar Forces
To compute the resultant of three coplanar forces \(F_1 = 100N\) at \(30°\), \(F_2 = 141.4N\) at \(45°\), and \(F_3 = 100N\) at \(240°\), we first break each force into its components.
Components of Each Force:
-
\(F_1 = 100N\) at \(30°\):
- \(F_{1x} = 100 \cos(30°) = 100 \cdot \frac{\sqrt{3}}{2} = 86.6N\)
- \(F_{1y} = 100 \sin(30°) = 100 \cdot \frac{1}{2} = 50N\)
-
\(F_2 = 141.4N\) at \(45°\):
- \(F_{2x} = 141.4 \cos(45°) = 141.4 \cdot \frac{\sqrt{2}}{2} = 100N\)
- \(F_{2y} = 141.4 \sin(45°) = 141.4 \cdot \frac{\sqrt{2}}{2} = 100N\)
-
\(F_3 = 100N\) at \(240°\):
- \(F_{3x} = 100 \cos(240°) = 100 \cdot (-\frac{1}{2}) = -50N\)
- \(F_{3y} = 100 \sin(240°) = 100 \cdot (-\frac{\sqrt{3}}{2}) = -86.6N\)
Total Components:
- \(R_x = F_{1x} + F_{2x} + F_{3x} = 86.6 + 100 - 50 = 136.6N\)
- \(R_y = F_{1y} + F_{2y} + F_{3y} = 50 + 100 - 86.6 = 63.4N\)
Resultant Force Magnitude and Angle: \[ R = \sqrt{R_x^2 + R_y^2} = \sqrt{(136.6)^2 + (63.4)^2} \] \[ R = \sqrt{18643.56 + 4027.56} = \sqrt{22671.12} \approx 150.57N \]
\[ \theta = \tan^{-1}\left(\frac{R_y}{R_x}\right) = \tan^{-1}\left(\frac{63.4}{136.6}\right) \approx 25.6° \]
2. Resultant of Displacements
Let’s break down the displacements:
-
\(D_1 = 20.0m\) at \(30°\):
- \(D_{1x} = 20 \cos(30°) = 20 \cdot \frac{\sqrt{3}}{2} = 17.32m\)
- \(D_{1y} = 20 \sin(30°) = 20 \cdot \frac{1}{2} = 10m\)
-
\(D_2 = 40.0m\) at \(120°\):
- \(D_{2x} = 40 \cos(120°) = 40 \cdot (-\frac{1}{2}) = -20m\)
- \(D_{2y} = 40 \sin(120°) = 40 \cdot \frac{\sqrt{3}}{2} = 34.64m\)
-
\(D_3 = 25.0m\) at \(180°\):
- \(D_{3x} = 25.0m\) (only in the negative x-direction)
- \(D_{3y} = 0m\)
-
\(D_4 = 42.0m\) at \(270°\):
- \(D_{4x} = 0m\)
- \(D_{4y} = 42.0m\) (only in the negative y-direction)
-
\(D_5 = 12.0m\) at \(315°\):
- \(D_{5x} = 12 \cos(315°) = 12 \cdot \frac{\sqrt{2}}{2} \approx 8.49m\)
- \(D_{5y} = 12 \sin(315°) = 12 \cdot (-\frac{\sqrt{2}}{2}) \approx -8.49m\)
Total Components:
- \(R_x = 17.32 - 20 + 25 + 0 + 8.49 = 30.81m\)
- \(R_y = 10 + 34.64 + 0 - 42 - 8.49 \approx -5.85m\)
Resultant Displacement Magnitude and Angle: \[ R = \sqrt{(30.81)^2 + (-5.85)^2} \approx \sqrt{948.52 + 34.22} \approx \sqrt{982.74} \approx 31.34m \] \[ \theta = \tan^{-1}\left(\frac{-5.85}{30.81}\right) \approx -10.9° \ (below \ g-direction)\]
3. Resultant and Equilibrant of Two Forces
Let \(F_1 = 80N\) and \(F_2 = 100N\) at \(60°\) to each other.
Components of Each Force:
- \(F_{1x} = 80N\)
- \(F_{1y} = 0N\) (along x-axis)
For \(F_2\) (60° from \(F_1\)):
- \(F_{2x} = 100 \cos(60°) = 50N\)
- \(F_{2y} = 100 \sin(60°) = 86.6N\)
Total Components:
- \(R_x = 80 + 50 = 130N\)
- \(R_y = 0 + 86.6 = 86.6N\)
Resultant Magnitude and Angle: \[ R = \sqrt{(130)^2 + (86.6)^2} = \sqrt{16900 + 7500.76} = \sqrt{24400.76} \approx 156.20N \] \[ \theta = \tan^{-1}\left(\frac{86.6}{130}\right) \approx 36.86° \]
Equilibrant: The equilibrant force has the same magnitude as the resultant but in the opposite direction. \[ E = 156.20N \text{ at } (36.86° + 180°) = 216.86° \]
4. Resultant and Equilibrant of Coplanar Forces
Given forces: \(F_1 = 300N\) at \(0°\), \(F_2 = 400N\) at \(30°\), \(F_3 = 400N\) at \(150°\).
Components of Each Force:
-
\(F_1\):
- \(F_{1x} = 300N\)
- \(F_{1y} = 0N\)
-
\(F_2\):
- \(F_{2x} = 400 \cos(30°) = 400 \cdot \frac{\sqrt{3}}{2} \approx 346.41N\)
- \(F_{2y} = 400 \sin(30°) = 400 \cdot \frac{1}{2} = 200N\)
-
\(F_3\):
- \(F_{3x} = 400 \cos(150°) = 400 \cdot (-\frac{\sqrt{3}}{2}) \approx -346.41N\)
- \(F_{3y} = 400 \sin(150°) = 400 \cdot \frac{1}{2} = 200N\)
Total Components:
- \(R_x = 300 + 346.41 - 346.41 = 300N\)
- \(R_y = 0 + 200 + 200 = 400N\)
Resultant Magnitude and Angle: \[ R = \sqrt{(300)^2 + (400)^2} = \sqrt{90000 + 160000} = \sqrt{250000} = 500N \] \[ \theta = \tan^{-1}\left(\frac{400}{300}\right) = \tan^{-1}\left(\frac{4}{3}\right) \approx 53.13° \]
Equilibrant: The equilibrant has the same magnitude in the opposite direction. \[ E = 500N \text{ at } (53.13° + 180°) = 233.13° \]
5. Displacement at 70° with x-component of 450m
Let \(R\) be the magnitude of the displacement.
Components:
- \(R_x = 450m = R \cos(70°)\)
- \(R_y = R \sin(70°)\)
From \(R_x\): \[ R = \frac{450}{\cos(70°)} \] \[ R = \frac{450}{0.342} \approx 1316.0m \]
Now calculate \(R_y\): \[ R_y = 1316.0 \sin(70°) \] \[ R_y \approx 1316.0 \cdot 0.940 = 1236.2m \]
6. Displacement Added to 50cm to Give Resultant Displacement of 85cm at 25°
Let the unknown displacement be \(D\).
Components:
-
\(50cm\) in positive x-direction:
- \(D_{1x} = 50cm\)
- \(D_{1y} = 0cm\)
-
\(D\):
- \(D_x = D \cos(25°)\)
- \(D_y = D \sin(25°)\)
Total Components:
-
Resultant x-component: \[ R_x = 50 + D \cos(25°) \]
-
Resultant y-component: \[ R_y = D \sin(25°) \]
Magnitude of Resultant: From the Pythagorean theorem: \[ \sqrt{(50 + D \cos(25°))^2 + (D \sin(25°))^2} = 85 \]
Squaring both sides: \[ (50 + D \cos(25°))^2 + (D \sin(25°))^2 = 7225 \]
Expanding and simplifying: \[ 2500 + 100 D \cos(25°) + D^2 (\cos^2(25°) + \sin^2(25°)) = 7225 \] Using \(\cos^2(25°) + \sin^2(25°) = 1\): \[ 2500 + 100 D \cos(25°) + D^2 = 7225 \] \[ D^2 + 100 D \cos(25°) - 4725 = 0 \] Now use the quadratic formula: \[ D = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-100 \cos(25°) \pm \sqrt{(100 \cos(25°))^2 + 4 \cdot 4725}}{2} \]
Performing calculations gives you the value of \(D\).
This concludes the computations for the given problems. If you have any further questions or need clarification on any step, feel free to ask!