A weightless ladder, 20 ft long rests against a frictionless wall at an angle of 60° with the horizontal. A 150 pound man is 4 ft from the top of the ladder. A horizontal force is needed to prevent it from slipping. Choose the correct magnitude from the following

The correct option is 70 lb.

Let's analyze the problem step-by-step using the principles of static equilibrium.

1. Problem Setup and Coordinates:
Let the ground be the horizontal plane (along the x-axis) and the wall be the vertical plane (along the y-axis).
The ladder has a length $L=20\text{ft}$ and makes an angle $\theta ={60}^{\circ }$ with the horizontal ground.
Let the point where the ladder touches the ground be $A$ (origin, $(0,0)$) and the point where it rests against the wall be $B$.
Since the ladder is weightless, we only consider the forces acting on the ladder:
- The normal force from the ground acting vertically upwards at the bottom $A$: $R_y$.
- The horizontal force $F$ applied at the bottom $A$ to prevent slipping: $F$ (acting horizontally towards the wall).
- The normal reaction from the frictionless vertical wall acting horizontally at the top $B$: $N_B$.
- The downward weight of the man $W=150\text{lb}$ acting at a distance of 4 ft from the top of the ladder (which means $20-4=16\text{ft}$ from the bottom of the ladder $A$).

2. Finding the Coordinates and Distances:
Let's find the horizontal distances (lever arms) from the bottom of the ladder at $A$:
- The top of the ladder $B$ is at a height of:
$h=L\sin \left({60}^{\circ }\right)=20\times \frac{\sqrt{3}}{2}=10\sqrt{3}\text{ft}$
- The horizontal distance from $A$ to the wall is:
$d=L\cos \left({60}^{\circ }\right)=20\times 0.5=10\text{ft}$
- The man is $16\text{ft}$ along the ladder from the bottom. The horizontal distance from $A$ to the line of action of the man's weight is:
$x_{\text{man}}=16\cos \left({60}^{\circ }\right)=16\times 0.5=8\text{ft}$

3. Applying the Equilibrium Equations:
First, the sum of horizontal forces must be zero:
$\sum F_x=0\Rightarrow F-N_B=0\Rightarrow F=N_B$
Second, taking the sum of moments about the base of the ladder $A$ to eliminate the forces acting at $A$ ($R_y$ and $F$):
$\sum M_A=0$
The moment due to the man's weight creates a clockwise rotation about $A$:
$M_{\text{man}}=-W\times x_{\text{man}}=-150\times 8=-1200\text{lb}·\text{ft}$
The moment due to the wall's normal force $N_B$ acts horizontally at height $h$ and creates a counterclockwise rotation about $A$:
$M_{N_B}=N_B\times h=N_B\times 10\sqrt{3}$
Setting the sum of moments to zero:
$N_B\times 10\sqrt{3}-1200=0$
$N_B=\frac{1200}{10\sqrt{3}}=\frac{120}{\sqrt{3}}$
Using $\sqrt{3}\approx 1.732$:
$N_B=40\sqrt{3}\approx 40\times 1.732=69.28\text{lb}$

Rounding to the nearest option, we find that the magnitude of the horizontal force $F$ required to prevent slipping is approximately 70 lb.