Answer to Question #101403 in Physics for Nyx

Question

A ladder rests against a vertical wall. There
is no friction between the wall and the ladder.
The coefficient of static friction between the
ladder and the ground is µ = 0.607 . Consider the following expressions:
A1: f = Fw
A2: f = Fw sin θ
B1: N =W/2
B2: N = W
C1: ℓ Fw sin θ = 2 Fw cos θ
C2: ℓ Fw sin θ = ℓ W cos θ
C3: ℓ Fw sin θ =1/2 ℓ W cos θ ,
where
f: force of friction between the ladder and
the ground,
Fw: normal force on the ladder due to the
wall,
θ: angle between the ladder and the ground,
N: normal force on the ladder due to the
ground,
W: weight of the ladder, and
ℓ: length of the ladder.
Identify the set of equations which is correct.
1. A2, B1, C3
2. A1, B2, C1
3. A2, B1, C1
4. A1, B2, C3
5. A1, B1, C2
6. A1, B2, C2
7. A1, B1, C1
8. A2, B1, C2
9. A2, B2, C1
10. A1, B1, C3

Determine the smallest angle θ for which the
ladder remains stationary.
Answer in units of ◦. Don't round answer.

Answer 1

The force equilibrium along horizontal requires

$$f=F_W$$

The force equilibrium along vertical requires

$$N=W$$

The torque equilibrium requires

$$F_W l \sin\theta=W\frac{l}{2}\cos\theta$$

So, the right answer is (4) A1, B2, C3.

The equations above give

$$f\sin\theta=W\cos\theta/2$$

Since

$$f=\mu N=\mu W$$

we get

$$\tan \theta=\frac{1}{2\mu}=\frac{1}{2\times 0.607}=0.824$$

$$\theta=39.5^{\circ}$$