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Chapter 17: Problem 1

A uniform ladder of weight \(W\) rests with one end on rough horizontal ground and with the other end against a smooth vertical wall. The ladder is at an angle \(\tan ^{-1} 2\) to the ground and is in a vertical plane perpendicular to the wall. The coefficient of friction between the ladder and the ground is \(1 .\) Find how far up the ladder a boy of weight \(2 W\) can climb without disturbing equilibrium. Find also the least horizontal force which must be applied to the foot of the ladder to enable the boy to climb to the top of the ladder without it slipping.

Short Answer

Expert verified
The boy can climb \frac{L}{6}\ up the ladder. The horizontal force needed is \frac{}{2}\ to the top of the ladder.

Step by step solution

01

Understanding Forces and Conditions

Identify all the forces acting on the ladder. Let's denote the weight of the ladder by W and the weight of the boy by 2W. The forces include the gravitational force of the ladder acting at its center, the gravitational force of the boy acting at a point we need to determine, the normal force N by the ground, the friction force f by the ground, and the normal force by the wall which will be H as the wall is smooth.
02

Expressions for Friction and Normal Force

The friction force can be expressed as: \[ f = \text{coefficient of friction} \times N = 1 \times N = N \] Using equilibrium conditions, we know: \[ f = H \]
03

Angle Calculation

Given that the angle of the ladder to the ground is \( \tan^{-1} 2 \), we can determine: \[ \theta = \tan^{-1} 2 \]
04

Derived Expressions for Normal and Vertical Forces

By resolving forces perpendicular to the ladder, we get:\[ N \times \tan( \theta) = W + 2W \] Since \( \tan( \theta) = 2 \), thus: \[ N \times 2 = 3W \] Solving for N: \[ N = \frac{3W}{2} \]
05

Torque Equations for Equilibrium

Taking torques about the point where the ladder contacts the floor: \[ H \times L \times \tan( \theta)= (2/3)L \times W + x \times 2W\] where x is the distance up the ladder the boy can climb. Using \( H = N = \frac{3W}{2} \times L \times 2 = \frac{3WL}{4} \), the torque equation simplifies to: \[ \frac{3WL}{4} = \frac{2WL}{3} + 2Wx\] Solving for x: \[ x = \frac{L}{6} \]
06

Least Horizontal Force for Stability

To find the least horizontal applied force (F) for the boy to be able to climb to the top of the ladder without slipping: \[ F = H - f \]. Since \( H = N = \frac{3W}{2} \) and \( f = N = \frac{3W}{2} \), the smallest applied F is:\[ F = 0 x 1 = \frac{3W}{2} \]
07

Final Step: Checking all Conditions

Verify equilibrium conditions for total forces and torques to ensure our solutions follow the laws of physics.

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Key Concepts

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

Torque Calculation
Torque is a measure of the rotational force applied to an object. In the case of a ladder problem, we consider the torques about the point where the ladder contacts the ground. This helps us understand how different forces impact the ladder's equilibrium. The torque \(\tau\) due to a force \(F\) applied at a distance \(r\) from the pivot point is given by: \[\tau = F * r * \sin(\theta)\] where:
  • \(F\) is the force applied
  • \(r\) is the distance from the pivot point
  • \(\theta\) is the angle between the force and the lever arm
In this ladder problem, the main forces to consider for torque calculations are the weights of the ladder and the boy, the normal force by the ground, and the horizontal force by the wall.
Friction Force
Friction is the force that opposes the sliding motion of two surfaces in contact. The friction force \(f\) between the ladder and the ground is given by: \[f = \text{coefficient of friction} \times N\] where:
  • Coefficient of friction is a measure of how 'grippy' the surfaces are (in this case, it is 1)
  • N is the normal force
In our problem, since the coefficient of friction is 1, the friction force is equal to the normal force: \[f = N\] It helps to resist the ladder's tendency to slide outwards.
Normal Force
The normal force is the perpendicular force exerted by a surface to support the weight of an object resting on it. For the ladder problem, the normal force \(N\) can be calculated using the vertical forces acting on the ladder. Given that: \[N \times \tan(\theta) = W + 2W\] and \(\tan(\theta) = 2\) the equation simplifies to: \[N \times 2 = 3W\] hence, \[N = \frac{3W}{2}\] The normal force plays a critical role in determining the friction force and maintaining equilibrium.
Ladder Problem
A classic ladder problem involves analyzing the interactions of various forces to ensure equilibrium. Understanding how far up a boy can climb without causing the ladder to slip involves:
  • Identifying all forces: gravitational forces, normal force, friction force, horizontal force by the wall
  • Calculating these forces based on given data (e.g., ladder weight, angle, coefficient of friction)
  • Using torque and force equilibrium conditions
By carefully resolving the horizontal and vertical components of forces and taking torques about the ladder foot, we determine the safe climbing distance (one-sixth of the ladder's length) and ensure no slipping occurs.

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Most popular questions from this chapter

A uniform rod \(\mathrm{AB}\) of length \(2 l\) and weight \(W\) is in limiting equilibrium at an angle of \(45^{\circ}\) to the horizontal with its end \(A\) on a rough horizontal plane and with a point \(\mathrm{C}\) in its length against a horizontal rail. This rail is at right angles to the vertical plane containing \(A B\). The coefficient of friction between the rod and the plane is \(\frac{1}{2}\) and between the rod and the rail is \(\frac{1}{3}\). Calculate: (a) the magnitude and direction of the resultant reaction at \(\mathrm{A}\), (b) the length \(\mathrm{AC}\).

A uniform cylinder of radius \(a\) and weight \(W\) rests with its curved surface in contact with two fixed planes, each of which is inclined at \(45^{\circ}\) to the horizontal, the line of intersection of the planes being horizontal and parallel to the axis of the cylinder. A couple is applied to the cylinder in a plane perpendicular to its axis. If the angle of friction between the cylinder and each plane is \(15^{\circ}\) show that the cylinder will rotate if the moment of the couple exceeds \(W a /(2 \sqrt{2})\).

A rough circular cylinder of radius \(a\) is fixed with its axis horizontal. Two uniform rods \(\mathrm{AB}\) and \(\mathrm{AC}\), each of weight \(W\) and length \(4 a\) are rigidly jointed together at \(\mathrm{A}\) to enclose an angle of \(60^{\circ}\) and rest on the cylinder in a plane perpen. dicular to the axis with A uppermost and vertically above the axis. The coefficient of friction at each contact is 3. A pull of magnitude \(P\) is applied at \(\mathrm{A}\) in the plane \(\mathrm{ABC}\) at an angle \(\theta\) with the upward drawn vertical. If the rods are about to slip, show that $$ P(3 \cos \theta+4 \sin \theta)=6 W $$ and find the magnitude and direction of the least possible pull.

A uniform rod \(A B\) of weight \(W_{1}\) is attached at \(A\) to a fixed smooth pivot and is freely hinged at \(\mathrm{B}\) to a uniform rod \(\mathrm{BC}\) of weight \(W_{2}\). The system is in equilibrium in a vertical plane with \(\mathrm{AB}\) resting on a smooth peg P below the level of \(\mathrm{A}\) and the end \(\mathrm{C}\) of the rod \(\mathrm{BC}\) on a smooth horizontal plane. The distance \(\mathrm{AP}\) is \(x\), the length \(\mathrm{AB}\) is \(2 a\) and the acute angle which \(\mathrm{AB}\) makes with the horizontal is \(\theta\). Prove that the force between the rods at B is vertical and equal to \(\frac{1}{2} W_{2}\), and find the reaction at the peg. If the reaction at \(\mathrm{A}\) is horizontal, find its magnitude in terms of \(W_{1}, W_{2}\) and \(\theta\), and prove that $$ x=\frac{2 a\left(w_{1}+W_{2}\right) \cos ^{2} \theta}{2 W_{1}+W_{2}} $$

Two uniform rods, \(\mathrm{AB}\) and \(\mathrm{BC}\) are of the same length and weigh \(3 W\) and \(W\) respectively. They are smoothly jointed at \(\mathrm{B}\) and stand in a vertical plane with \(\mathrm{A}\) and \(C\) on a rough horizontal plane. The coefficient of friction between each rod and the plane is 3. Equilibrium is about to be broken by one of the rods slipping on the plane. Find which rod will slip and calculate the angle each rod makes with the plane. Calculate also the reaction at the hinge \(\mathrm{B}\) in magnitude and direction.
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