Answer:
The normal force N1 exerted by the floor is [tex]N_1 = 951 \ N[/tex]
The normal force N2 exerted by the wall is [tex]N_2= 616.43 \ N[/tex]
The frictional force exerted by the wall is [tex]f = N_2 = 616.43 \ N[/tex]
Explanation:
From the question we are told that
The length of the ladder is [tex]L = 4.6 \ m[/tex]
The weight of the ladder is
The distance of the ladder position on the wall from the floor is [tex]D = 3.75 \ m[/tex]
The mass of the person is [tex]m = 90 kg[/tex]
Applying Pythagoras theorem
The length of the position the ladder on the ground from the base of the wall is
[tex]A = \sqrt{L^ 2 - D^2}[/tex]
substituting values
[tex]A = \sqrt{(4.6^2)-(3.75^2)}[/tex]
[tex]A = 2.66 \ m[/tex]
In order the for the ladder not to shift from the ground the sum of the moment about the position of the ladder on the ground must be equal to zero this is mathematically represented as
[tex]\sum M = 0 = N_2 * D - [\frac{1}{2} * W_L ] * [(mg) *A ][/tex]
[tex]\sum M = 0 = N_2 * 3.75 - [\frac{1}{2} * 69.0 ] * [(90*9.8) * \frac{4.6}{2.66} ][/tex]
[tex]N_2 * 3.75 =2311.62[/tex]
[tex]N_2 * 3.75 =2311.62[/tex]
[tex]N_2= 616.43 \ N[/tex]
Now the force exerted by the floor on the ladder is mathematically represented as
[tex]N_1 = W_L + (m * g )[/tex]
substituting values
[tex]N_1 = 951 \ N[/tex]
Now the horizontal forces acting on the ladder are [tex]N_2 \ and \ f[/tex] and they are in opposite direction so
[tex]f = N_2 = 616.43 \ N[/tex]
