Question

a ladder is resting as shown in the picture. if the mass of ladder is 5 kg and the length is 120 cm. Determine the Normal Force act on C​

Answer 1

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.