Answer:
17.64 N
Explanation:
Draw a free body diagram (see attached). At the center of the ladder is weight pulling down. At point C, we have a normal force pushing perpendicular to the ladder. At point A, we have reaction forces in the x and y directions.
For simplicity, we can divide the normal force into x and y components. The slope of the ladder is -80/60, so the slope of the normal force is 60/80. Therefore:
Ny / Nx = 60 / 80
Ny / Nx = 3/4
Next, we'll take the sum of moments at point A:
∑τ = Iα
-Wd + Ny (60) + Nx (80) = 0
d is the horizontal distance between A and the center of the ladder. We can find it using similar triangles. From Pythagorean theorem, we know the distance between A and C is 100 cm. So:
60 / 100 = d / 60
d = 36
-(5)(9.8)(36) + Ny (60) + Nx (80) = 0
60 Ny + 80 Nx = 1764
We now have two equations and two variables. Solving:
60 (3/4 Nx) + 80 Nx = 1764
45 Nx + 80 Nx = 1764
125 Nx = 1764
Nx = 14.112
Ny = 3/4 Nx
Ny = 10.584
Using Pythagorean theorem to find N:
N = √(Nx² + Ny²)
N = 17.64
The magnitude of the normal force is 17.64 N.
