Aladder is leaning against a vertical wall, and both ends of the ladder are at the point of slipping. the coefficient of friction between the ladder and the horizontal surface is =0.115μ1=0.115 and the coefficient of friction between the ladder and the wall is =0.253μ1=0.253 .

13.1 degrees

Explanation:

Let N be the normal reaction of the floor on the ladder

Let R be the normal reaction of the wall on the ladder.

Let W be the weight of the (assumed uniform) ladder.

Let L be the length of the ladder

Summing horizontal forces to zero

0.115N = R

N = R / 0.115

Summing vertical forces to zero

N + 0.253 R = W

R / 0.115 + 0.253R = W

Sum moments about the floor contact to zero

W[(L/2)sinθ] - R[Lcosθ] - 0.253R[Lsinθ] = 0

Divide all terms by L and sub for W

(R / 0.115 + 0.253R)[(1/2)sinθ] - R[cosθ] - 0.253R[sinθ] = 0

expand

(Rsinθ / (2)0.115 + (0.253Rsinθ)/2) - R[cosθ] - 0.253R[sinθ] = 0

Divide all terms by R and combine

sinθ / 0.23 + 0.1265sinθ - cosθ - 0.253sinθ = 0

sinθ / 0.23 + 0.1265sinθ - 0.253sinθ = cosθ

4.2393sinθ = cosθ

tanθ = 1 / 4.2393

θ = 13.3 degrees

answer: 0.20

explanation: this can only be the answew because all the other measuremnets are not close

measure their densities same density, same sub