Question

A ladder 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 μ₁ = 0.115 and the coefficient of friction between the ladder and the wall is μ₂ = 0.103. Determine the maximum angle with the vertical the ladder can make without falling on the ground.

Answer 1

Answer:

$\theta = 1.73 degree$

Explanation:

Let the ladder makes some angle with the horizontal

now by force balance in x direction we have

$F_{f1} = N_2$

in Y direction force balance is given as

$F_{f2} + N_1 = mg$

now we know that

$F_{f1} = 0.115 N_1$

$F_{f2} = 0.103 N_2$

also by torque balance we have

$N_2(Lcos\theta) + 0.103N_2(Lsin\theta) = mg(\frac{L}{2} cos\theta)$

$0.115 N_1 = N_2$

$0.103 N_2 + N_1 = mg$

$(0.103)(0.115 N_1) + N_1 = mg$

$1.012 N_1 = mg$

$N_1 = 0.988 mg$

$N_2 = 0.113 mg$

now from above equation

$(0.113 mg)cos\theta + (0.103)(0.113mg) sin\theta = \frac{mg}{2} cos\theta$

$0.113 + 0.0117 tan\theta = 0.5$

$\theta = 88.3 degree$

so angle with the vertical is given as

$\theta = 1.73 degree$