two forces of F1=150N,bearing of 65°and F2=200N,at bearing of 310° acts on an abject.use scale (10mm=10N) to determine the resultant force
Using vectors:
R = (150cos65,150sin65) + (200cos310,200sin310)
= (63.39, 135.95) + (128.56, -153.21)
= (191.95, -17.26)
|R| = √(191.95^2 + (-17.26)^2) = appr 192.72 N
From my sketch:
angle = tan^-1 (-17.26/191.95) = appr 354.86°
To determine the resultant force of the two forces acting on an object, we need to resolve each force into its horizontal and vertical components. Then, we can add the horizontal components and vertical components separately to find the resultant horizontal and vertical forces. Finally, we can combine the horizontal and vertical forces to find the magnitude and direction of the resultant force.
Here's how to do it step by step:
Step 1: Resolve the forces into horizontal and vertical components.
- For F1, since the bearing is 65°, we need to find the horizontal and vertical components. The horizontal component (F1x) can be calculated as F1 * cos(65°), and the vertical component (F1y) can be calculated as F1 * sin(65°).
- For F2, since the bearing is 310°, we need to find the horizontal and vertical components. The horizontal component (F2x) can be calculated as F2 * cos(310°), and the vertical component (F2y) can be calculated as F2 * sin(310°).
Step 2: Add the horizontal components (F1x and F2x) to find the resultant horizontal force (Rx).
- Rx = F1x + F2x
Step 3: Add the vertical components (F1y and F2y) to find the resultant vertical force (Ry).
- Ry = F1y + F2y
Step 4: Calculate the magnitude of the resultant force (R) using Pythagoras' theorem.
- R = √(Rx^2 + Ry^2)
Step 5: Calculate the direction of the resultant force (θ) using the inverse tangent.
- θ = tan^(-1)(Ry/Rx)
Step 6: Convert the resultant force magnitude from Newtons (N) to the scale provided (10mm = 10N) by multiplying by the scaling factor.
- R_scale = R * scaling factor
Now, let's plug in the values and calculate:
F1 = 150N
F2 = 200N
Scaling factor = 10mm/10N = 1mm/N
Step 1:
F1x = F1 * cos(65°)
= 150N * cos(65°)
≈ 64.34N
F1y = F1 * sin(65°)
= 150N * sin(65°)
≈ 135.14N
F2x = F2 * cos(310°)
= 200N * cos(310°)
≈ 172.02N
F2y = F2 * sin(310°)
= 200N * sin(310°)
≈ -107.87N (since the vertical component is negative)
Step 2:
Rx = F1x + F2x
= 64.34N + 172.02N
≈ 236.36N
Step 3:
Ry = F1y + F2y
= 135.14N + (-107.87N)
≈ 27.27N
Step 4:
R = √(Rx^2 + Ry^2)
= √((236.36N)^2 + (27.27N)^2)
≈ 239.66N
Step 5:
θ = tan^(-1)(Ry/Rx)
= tan^(-1)(27.27N/236.36N)
≈ 6.585°
Step 6:
R_scale = R * scaling factor
= 239.66N * 1mm/N
≈ 239.66mm
Therefore, the resultant force is approximately 239.66N at a direction of approximately 6.585°.