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Chapter 12: Problem 7

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) 632 N, (b) 833 N, (c) 37.1掳

Step by step solution

01

Analyze forces acting on the ladder

The ladder is in equilibrium, meaning all forces and moments must balance. Forces include the gravitational forces on the cleaner and ladder, normal forces at both ends of the ladder, and the force exerted by the window on the ladder (force on window).
02

Identify ladder and cleaner forces

The weight of the cleaner \[F_{c} = m_cg = 75 ext{ kg} \times 9.8 ext{ m/s}^2 = 735 ext{ N}\]and the weight of the ladder\[F_{l} = m_lg = 10 ext{ kg} \times 9.8 ext{ m/s}^2 = 98 ext{ N}.\]
03

Determine forces exerted at each end of the ladder

Let the force of the window on the ladder be \(F_w\) (horizontal), and of the ground on the ladder be \(N\) (normal upwards) and \(F_h\) (horizontal). The reaction components at each end must balance these forces.
04

Calculate moments about the base

Taking moments about the base to eliminate normal force at the ground, we can use the equation:\[nF_h \times 5.0 - 735 \times 2 = F_w \times 5.0 \]Solving for \(F_w\), considering cleaner at 3m:\[735 \times 3.0 + 98 \times 2.5 = F_w 5.0\] \[F_w = \frac{735 \times 3.0 + 98 \times 2.5}{5.0} = 632 ext{ N}\]
05

Calculate ground reaction forces

Using vertical and horizontal force balance:\[N - 735 - 98 = 0\]\[N = 833 ext{ N}\]\[F_h = F_w = 632 ext{ N}\]
06

Determine the angle of force on the ladder

The angle \(\theta\) is given by:\[\tan \theta = \frac{F_h}{N}\]\[\theta = \arctan \left(\frac{632}{833}\right) \approx 37.1^\circ\]

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Key Concepts

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

Force Analysis
In mechanics, force analysis is crucial for understanding how objects interact with each other. When analyzing forces, the first step is to identify all the forces acting on each object in the system. For the ladder problem, various forces are acting on the ladder, which is a rigid body held between a cracked window and the ground.
  • Firstly, the weight of the window cleaner applies a downward force on the ladder. This force, called gravitational force, acts vertically downward through the cleaner's center of mass.
  • Next, there is the gravitational force of the ladder itself, which also acts vertically downward through its center of mass.
  • Additionally, the window exerts a horizontal force on the ladder, which we call the normal force. This is because the window provides a reactive force perpendicular to the ladder.
  • Finally, at the base, the ground provides support through an upward normal force and a horizontal frictional force, preventing the ladder from slipping backward.
Understanding these forces helps in calculating the ladder's equilibrium conditions, ensuring that they balance out and the ladder remains stationary until the window breaks.
Gravitational Forces
Gravitational forces are central to this exercise because they dictate how the window cleaner and ladder interact with the Earth. Every object with mass experiences a gravitational pull towards the center of the Earth. The strength of this force can be calculated using the formula: \( F = mg \), where \( m \) is the mass and \( g \) is the acceleration due to gravity, approximately \( 9.8 \, \text{m/s}^2 \) on Earth.For this scenario:
  • The gravitational force acting on the window cleaner is calculated as \( 735 \, \text{N} \), resulting from his mass \( 75 \, \text{kg} \).
  • The ladder, with a mass of \( 10 \, \text{kg} \), experiences a gravitational force of \( 98 \, \text{N} \).
These forces are particularly important because they contribute to the total load the window needs to support. Consequently, understanding gravitational forces is key to resolving the balance of forces and moments in this mechanical problem.
Moment Calculation
Moment calculation is crucial in determining whether the ladder remains stable or tips over. A moment is essentially a measure of the turning effect a force has on an object. It is calculated as the product of the force and the distance from the point of rotation, expressed as \( M = F \times d \).For the ladder problem:
  • Moments about the base of the ladder are considered for ease of calculation, as it lets us focus on horizontal and vertical forces without initially considering the base's normal force.
  • The force the cleaner applies at the third meter of the 5-meter ladder gives a moment of \( 735 \, \text{N} \times 3.0 \, \text{m} \).
  • The force exerted by the ladder's weight produces another moment of \( 98 \, \text{N} \times 2.5 \, \text{m} \).
  • Solving the balance of moments lets us find the horizontal force the wall (window) exerts back, noted as \( F_w \).
By resolving these moments, we ensure equilibrium so every force and moment works in harmony to keep the ladder in balance without undue stress on the window.

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Most popular questions from this chapter

A uniform cube of side length \(8.0 \mathrm{~cm}\) rests on a horizontal floor. The coefficient of static friction between cube and floor is \(\mu .\) A horizontal pull \(\vec{P}\) is applied perpendicular to one of the vertical faces of the cube, at a distance \(7.0 \mathrm{~cm}\) above the floor on the vertical midline of the cube face. The magnitude of \(\vec{P}\) is gradually increased. During that increase, for what values of \(\mu\) will the cube eventually (a) begin to slide and (b) begin to tip? (Hint: At the onset of tipping, where is the normal force located?)

A makeshift swing is constructed by making a loop in one end of a rope and tying the other end to a tree limb. A child is sitting in the loop with the rope hanging vertically when the child's father pulls on the child with a horizontal force and displaces the child to one side. Just before the child is released from rest, the rope makes an angle of \(15^{\circ}\) with the vertical and the tension in the rope is \(280 \mathrm{~N}\). (a) How much does the child weigh? (b) What is the magnitude of the (horizontal) force of the father on the child just before the child is released? (c) If the maximum horizontal force the father can exert on the child is \(93 \mathrm{~N}\), what is the maximum angle with the vertical the rope can make while the father is pulling horizontally?

A particle is acted on by forces given, in newtons, by \(\vec{F}_{1}=\) \(8.40 \hat{\mathrm{i}}-5.70 \hat{\mathrm{j}}\) and \(\vec{F}_{2}=16.0 \hat{\mathrm{i}}+4.10 \hat{\mathrm{j}} .\) (a) What are the \(x\) component and (b) \(y\) component of the force \(\vec{F}_{3}\) that balances the sum of these forces? (c) What angle does \(\vec{F}_{3}\) have relative to the \(+x\) axis?

A rope of negligible mass is stretched horizontally between two supports that are 3.44 m apart. When an object of weight 3160 N is hung at the center of the rope, the rope is observed to sag by 35.0 cm. What is the tension in the rope?

A crate, in the form of a cube with edge lengths of \(1.2 \mathrm{~m}\), contains a piece of machinery; the center of mass of the crate and its contents is located \(0.30 \mathrm{~m}\) above the crate's geometrical center. The crate rests on a ramp that makes an angle \(\theta\) with the horizontal. As \(\theta\) is increased from zero, an angle will be reached at which the crate will either tip over or start to slide down the ramp. If the coefficient of static friction \(\mu_{s}\) between ramp and crate is \(0.60,(\) a) does the crate tip or slide and (b) at what angle \(\theta\) does this occur? If \(\mu_{s}=0.70\), (c) does the crate tip or slide and (d) at what angle \(\theta\) does this occur? (Hint: At the onset of tipping, where is the normal force located?)
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