/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 7 A \(75 \mathrm{~kg}\) window cle... [FREE SOLUTION] | 91Ó°ÊÓ

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A \(75 \mathrm{~kg}\) window cleaner uses a \(10 \mathrm{~kg}\) ladder that is \(5.0 \mathrm{~m}\) long. He places one end on the ground \(2.5 \mathrm{~m}\) from a wall, rests the upper end against a cracked window, and climbs the ladder. He is \(3.0 \mathrm{~m}\) up along the ladder when the window breaks. Neglect friction between the ladder and window and assume that the base of the ladder does not slip. When the window is on the verge of breaking, what are (a) the magnitude of the force on the window from the ladder, (b) the magnitude of the force on the ladder from the ground, and (c) the angle (relative to the horizontal) of that force on the ladder?

Short Answer

Expert verified
(a) 981 N, (b) 1288.5 N, (c) 40.1°

Step by step solution

01

Determine the Forces Acting on the Ladder

The ladder has four forces acting on it: (1) the gravitational force of the ladder (\( F_{ladder} = mg = 10 \times 9.81 \, \text{N} \)), (2) the gravitational force of the window cleaner (\( F_{cleaner} = mg = 75 \times 9.81 \, \text{N} \)), (3) the normal force from the ground (\( N \)), and (4) the normal force from the window (\( F \)).
02

Set the Conditions for Static Equilibrium

For static equilibrium, the sum of horizontal and vertical forces must be zero, and the sum of the moments about any point must also be zero.
03

Apply the Equilibrium Condition for Moments

Choose the bottom of the ladder as the point of rotation. Calculate the moments from the forces:- \( F_{ladder} \) acts at the midpoint of the ladder (2.5 m from the bottom), making a moment \( F_{ladder} \cdot 2.5 \). - \( F_{cleaner} \) acts 3.0 m from the bottom, adding a moment \( F_{cleaner} \cdot 3.0 \).- \( F \), the force of the window, creates a counter-moment \( F \cdot 5.0 \cdot \sin \theta \).Set the sum of these moments to zero: \[F \cdot 5.0 \cos \theta = F_{ladder} \cdot 2.5 + F_{cleaner} \cdot 3.0.\]
04

Solve for the Force on the Window

The angle \( \theta \) can be found using the triangle formed by the ladder. We know:\( \cos \theta = \frac{2.5}{5.0} = 0.5 \). Thus,\[F \cdot 5.0 \cdot 0.5 = 10 \times 9.81 \cdot 2.5 + 75 \times 9.81 \cdot 3.0.\]Solving this gives us:\[F = \frac{245.25 + 2207.25}{2.5} \approx 981 \, \text{N}.\]
05

Determine the Forces from the Ground

The vertical component is the sum of the weights: \( N_y = 10 \times 9.81 + 75 \times 9.81 = 833.85 \, \text{N} \).For horizontal equilibrium:\( N_x = F = 981 \, \text{N} \).The total force from the ground:\[N = \sqrt{N_x^2 + N_y^2} = \sqrt{981^2 + 833.85^2} \approx 1288.5 \, \text{N}.\]
06

Calculate the Angle of the Force from the Ground

The angle \( \phi \) that the force makes with the horizontal can be found with:\[\tan \phi = \frac{N_y}{N_x} = \frac{833.85}{981}.\]Thus,\[\phi = \arctan\left(\frac{833.85}{981}\right) \approx 40.1^\circ.\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Static Equilibrium
Static equilibrium is a state where an object is at rest and all the forces and moments acting on it are balanced. In physics problems involving static equilibrium, like our window cleaner scenario, the key is that both the sum of all horizontal and vertical forces must equal zero. This ensures that there is no net force causing the object to move.

In our exercise, the ladder experiences several forces: the gravitational force from both the ladder and the window cleaner, a normal force from the ground, and a force from the window. To achieve static equilibrium:
  • All vertical and horizontal forces must cancel one another out.
  • The rotational forces (moments) need to be in balance, meaning the ladder doesn't rotate about any point.
By considering these factors and solving for the unknowns, we determine the forces that prevent the system from moving or rotating.
Moment Calculation
Moment calculation is crucial when analyzing scenarios involving rotational forces, especially in problems like determining when the window cracks under the ladder's pressure. A moment, or torque, is the rotational equivalent of a linear force and is calculated by multiplying the force by the perpendicular distance from the point of rotation.

In this problem, to maintain equilibrium and prevent rotation:
  • We choose the base of the ladder as the pivot point.
  • The moment from the gravitational forces of both the ladder and window cleaner are calculated based on their distances from this pivot.
  • The force on the window creates a counteracting moment, as it is the reaction force attempting to balance out the ladder and cleaner's weight.
  • The requirement is \[ F_{ladder} \times 2.5 + F_{cleaner} \times 3.0 = F \times 5.0 \times \cos \theta \]
Using these moment calculations, we confirm the ladder doesn't rotate, keeping the scenario in equilibrium.
Forces on an Inclined Plane
When dealing with forces on an inclined plane, like the ladder in this exercise, it's important to analyze how gravity and other forces interact with the plane, considering both parallel and perpendicular components. The ladder forms an inclined plane with the wall and ground, making understanding these force components critical.

In our scenario, the gravitational forces from the window cleaner and ladder can be split into components. The normal force from the ground and any force from the window can also be expressed in terms of perpendicular components to the inclined plane.
  • The angle \( \theta \) of the incline, determined by the triangle of the ladder, defines these components.
  • We use trigonometric relationships, such as \( \cos \theta \) and \( \sin \theta \), to find the size of the force components.
  • By applying these components, we ensure forces are balanced without causing movement down the slope or perpendicular collapse.
This understanding of forces on an inclined plane helps ensure the scenario is accurately analyzed, resulting in precise determinations of necessary forces and their exact angles.

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