Question

A uniform ladder \(10.0 \mathrm{~m}\) long is leaning against a frictionless wall at an angle of \(60.0^{\circ}\) above the horizontal. The weight of the ladder is \(20.0 \mathrm{lb} . \mathrm{A} 61.0-\mathrm{lb}\) boy climbs \(4.00 \mathrm{~m}\) up the ladder. What is the magnitude of the frictional force exerted on the ladder by the floor?

Short answer

Answer: The magnitude of the frictional force exerted on the ladder by the floor is 34.4 lb.

Step-by-step solution

Identify the forces acting on the ladder

There are four forces acting on the ladder: 1. The weight of the ladder (\(w_l\)) acting downward from the center of the ladder. 2. The weight of the boy (\(w_b\)) acting downward at a point 4m up the ladder. 3. The normal force from the floor (\(N_f\)) acting perpendicular to the floor at the bottom of the ladder. 4. The frictional force from the floor (\(f_f\)) acting horizontally at the bottom of the ladder in the direction opposite the wall.

Setup the equations for the torques

Let's choose the bottom of the ladder as the pivot point. From the problem, the angle between ladder and the horizontal is 60 degrees. Then, we find the magnitudes of the perpendicular components of the forces due to the weight of the ladder, weight of the boy, and the normal force from the wall that cause a torque about the bottom of the ladder. \({w_l}_\perp = w_l\sin{(60)} = 20\sin{(60)}\) \({w_b}_\perp = w_b\sin{(60)} = 61\sin{(60)}\) Since the ladder is not rotating, the sum of the torques about the pivot point must be zero. \(\sum \tau = {w_l}_\perp \cdot (\frac{1}{2}L) + {w_b}_\perp\cdot(4m) - f_f\cdot(L\sin{(60^{\circ})}) = 0\) Substitute the expressions we found for \({w_l}_\perp\) and \({w_b}_\perp\), and we get, \(20\sin{(60)} \cdot (\frac{1}{2}\cdot10) + 61\sin{(60)}\cdot(4) - f_f\cdot(10\sin{(60^{\circ})}) = 0\)

Solve for the frictional force\( f_f\)

Now, we just need to solve the equation for the frictional force \(f_f\). \(100\sin{(60)}+244\sin{(60)}- f_f\cdot(10\sin{(60^{\circ})}) = 0\) \( f_f = \frac{100\sin(60) + 244\sin(60)}{10\sin(60)}\) \(f_f = \frac{344\sin(60)}{10\sin(60)}\) \(f_f = 34.4 \thinspace lb\) Therefore, the magnitude of the frictional force exerted on the ladder by the floor is 34.4 lb.

Key concepts

Torques and Equilibrium

In physics, a torque is a measure of the force that can cause an object to rotate about an axis. Just like forces, torques can be summed to determine the rotational motion of an object. When the sum of all torques acting on a system is zero, the system is said to be in equilibrium, and there is no change in the rotational motion of the object. For the ladder leaning against a wall, considering torques helps us understand how different forces are balanced so that the ladder does not rotate or fall over.

Equilibrium is a key concept not only in static situations, as with the ladder, but also in a variety of mechanical systems. It ensures that if an object is at rest, it remains at rest unless acted upon by an external force or torque. Analyzing a physics problem involving equilibrium typically requires identifying all the forces, calculating their torques, and then solving equations to find unknowns, just like in the provided ladder example.

Static Friction

Static friction is the force that prevents two surfaces from sliding past each other when at rest. It is the force that must be overcome to start moving an object that is initially stationary. The magnitude of static friction depends on the nature of the surfaces involved and the normal force pressing them together. It is usually represented by the equation
\( f_{s} = \thinspace\thinspace \mu_s N \)
where \(f_{s}\) is the static frictional force, \(\thinspace\mu_s\) is the coefficient of static friction, and \(N\) is the normal force.

In many physics problems, including ladder scenarios, static friction plays a crucial role in preventing slipping. It acts in the direction opposite to what would be the motion of the ladder if it were to slide. If the frictional force were not enough, the ladder would slip, potentially leading to dangerous consequences.

Forces in Physics

In the realm of physics, forces are pushes or pulls that can cause an object to accelerate. They are vectors, which means they have both magnitude and direction. Understanding forces is crucial for solving a wide range of physics problems, from simple ones involving gravity to more complex scenarios with multiple forces acting simultaneously.

For any given situation, drawing a free-body diagram helps visualize all the forces at play. This includes gravitational forces, normal forces, frictional forces, tension, and more. In the case of the ladder problem, we consider the gravitational forces due to the weights of the ladder and the boy, and the reaction forces from the wall and the floor, which include the normal and frictional forces, respectively.

Ladder Physics Problem

Ladder problems are a common example of a static equilibrium scenario often explored in physics. They involve a ladder leaning against a wall, where the forces of gravity, friction, and the normal forces by the wall and floor are acting. The aim is to calculate one of the forces given certain dimensions, weight, and angles. It's essential to understand that the ladder, wall, and floor create a right-triangle system.

The ladder problem is a perfect candidate to apply your knowledge of forces, torques, and equilibrium to find an unknown quantity. Most importantly, in ladder problems, it is always assumed that no slipping occurs—this requires that the frictional force is large enough to counteract any potential sliding force caused by the weight of the ladder and any additional loads, such as a person climbing the ladder.