Equilibrium Analysis: Calculating the Wall Force on a Ladder

In engineering and physics, understanding the forces acting on objects is crucial, especially when dealing with scenarios involving equilibrium. This article delves into the analysis of a specific scenario involving a uniform ladder leaning against a smooth vertical wall. We will apply the principles of equilibrium to solve for the force exerted by the wall on the ladder. The ladder weighs 100 N and forms a 70° angle with the horizontal.

Given Data

The weight of the ladder is given as W 100 N. The angle with the horizontal is θ 70°. Let's analyze the forces acting on the ladder and how to calculate the force exerted by the wall.

Forces Acting on the Ladder

Weight W - Acts downward at the center of the ladder. Normal Force from the Wall F_w - Acts horizontally towards the ladder from the wall. Normal Force from the Floor N - Acts vertically upward at the base of the ladder. Frictional Force F_f - Acts horizontally at the base of the ladder opposing the motion.

Conditions for Equilibrium

The object is in equilibrium when there is no net force or net torque acting on it. This can be expressed as follows:

Sum of Vertical Forces

[N - W 0]

Sum of Horizontal Forces

[F_w - F_f 0]

Sum of Moments

For equilibrium, the sum of the moments about any point should be zero. Let's take the base of the ladder as the point of reference.

Geometry of the Ladder

The ladder length is denoted by L. The height at which the ladder touches the wall and the distance from the wall to the base of the ladder can be mathematically expressed as:

h L sin70° d L cos70°

Moment Calculation

When taking the moments about the base of the ladder, we obtain the following:

Moment due to the weight of the ladder acting at L/2: [text{Moment} W cdot left(frac{L}{2} cos70°right)] Moment due to the wall force: [text{Moment} F_w cdot L sin70°] Setting the moments equal for equilibrium: [W cdot left(frac{L}{2} cos70°right) F_w cdot L sin70°]

Simplifying the Equation

To simplify the equation, we can cancel L from both sides (assuming L is not zero).

[W cdot frac{1}{2} cos70° F_w cdot sin70°]

Substituting W 100 N:

[100 cdot frac{1}{2} cos70° F_w cdot sin70°]

[50 cos70° F_w cdot sin70°]

Solving for F_w

Now we can solve for F_w:

[F_w frac{50 cos70°}{sin70°}]

Using approximate values for cos70° ≈ 0.342 and sin70° ≈ 0.940, we can substitute these values:

[F_w frac{50 cdot 0.342}{0.940} approx frac{17.1}{0.940} approx 18.2 text{ N}]

Conclusion

The force exerted by the wall on the ladder is approximately 18.2 N. This calculation highlights the importance of understanding equilibrium principles and applying them to real-world scenarios.