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

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

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