Answer :
Since X is negative, it means the point of application of the resultant force is to the left of the frame's centerline. The magnitude of the resultant force is approximately 874 lb, and its point of application is approximately 0.0612 ft to the left of the frame's centerline.
To find the magnitude and point of application of the resultant of the four wind forces, we can use the concept of vector addition. Let's represent the forces with their respective magnitudes and directions.
Given wind forces:
A = 175 lb (horizontal force)
B = 195 lb (horizontal force)
C = 244 lb (horizontal force)
D = 260 lb (horizontal force)
Distances from the frame to the point of application of each force:
a = 1 ft
b = 12 ft
To find the resultant force, we need to add all the forces vectorially:
Resultant force (R) = A + B + C + D
Magnitude of the resultant force (R_mag) = |R| = √(R_x^2 + R_y^2)
To find the point of application (X) of the resultant force, we can use the concept of moments. The total moment of the forces about a point X must be zero for the frame to be in equilibrium.
The total moment about point X is given by:
Total Moment = (Magnitude of force A) * (Distance of force A from point X)
+ (Magnitude of force B) * (Distance of force B from point X)
+ (Magnitude of force C) * (Distance of force C from point X)
+ (Magnitude of force D) * (Distance of force D from point X)
Setting the total moment to zero:
Total Moment = (A * a) + (B * a) + (C * b) + (D * b) = 0
Now, let's calculate the resultant magnitude and point of application:
R_mag = √((175 + 195 + 244 + 260)^2)
R_mag = √(874^2)
R_mag ≈ 874 lb
To find the point of application (X), we'll use the moment equation:
(175 * 1) + (195 * 1) + (244 * 12) + (260 * 12) = 0
175 + 195 + 2928 + 3120 = 0
We need to solve for X:
(2928 + 3120)X = -(175 + 195)
6048X = -370
X = -370 / 6048
X ≈ -0.0612 ft
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