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Chapter 11: Problem 49

Mountaineers often use a rope to lower themselves down the face of a cliff (this is called rappelling). They do this with their body nearly horizontal and their feet pushing against the cliff (Fig. \(\mathbf{P 1 1 . 4 9}\) ). Suppose that an \(82.0 \mathrm{~kg}\) climber, who is \(1.90 \mathrm{~m}\) tall and has a center of gravity \(1.1 \mathrm{~m}\) from his feet, rappels down a vertical cliff with his body raised \(35.0^{\circ}\) above the horizontal. He holds the rope \(1.40 \mathrm{~m}\) from his feet, and it makes a \(25.0^{\circ}\) angle with the cliff face. (a) What tension does his rope need to support? (b) Find the horizontal and vertical components of the force that the cliff face exerts on the climber's feet. (c) What minimum coefficient of static friction is needed to prevent the climber's feet from slipping on the cliff face if he has one foot at a time against the cliff?

Short Answer

Expert verified
a) The tension in the rope is approx. 552 N. b) The vertical and horizontal components of cliff face force are approx. 710 N and 227 N respectively. c) The minimum coefficient of static friction needed is approx. 0.32.

Step by step solution

01

Formulating the problem

First, draw a free body diagram and identify all the forces acting on the climber. Weight (\(F_g\)) acts vertically downwards from the climber's center of gravity, tension (\(T\)) in the rope acts parallel to the rope, and friction (\(f\)) and normal force (\(F_n\)) act on the feet of the climber.
02

Calculate the tension in the rope

Next, calculate the tension in the rope. This is equal to the weight component that acts parallel to the cliff face, which can be computed using the climber鈥檚 weight (\(F_g = mg\)) and the angle between the cliff face and the rope. Thus, \(T = F_g \sin(25.0^\circ)\).
03

Find the horizontal and vertical components of the force

Now, find the horizontal and vertical components of the force that the cliff face exerts on the climber's feet. This can be done by equating the sum of force components in the vertical and horizontal directions to zero (since the climber is in equilibrium). Vertical equilibrium gives the normal force on the cliff (\(F_n = F_g \cos(35.0^\circ) - T \cos(25.0^\circ)\)), and horizontal equilibrium gives the friction (\(f = T \sin(25.0^\circ)\)).
04

Calculate the coefficient of static friction

Lastly, calculate the minimum coefficient of static friction, \(\mu_s\), to prevent slipping. This is found by dividing the friction force by the normal force (\(\mu_s = f / F_n\)).

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Key Concepts

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

Rappelling Mechanics
Rappelling, also known as abseiling, involves using a rope to descend a vertical surface such as a cliff. In this technique, the climber positions their body nearly horizontal to maximize surface area contact and balance. This ensures safety during the descent. The rope provides control and support as the climber advances downward. Key elements of rappelling include:
  • Positioning: The climber must maintain a stable position with their legs extended and knees slightly bent.
  • Anchor Point: The rope is securely anchored above the climber.
  • Angles: The rope and body angles affect the forces on the climbers such as tension and friction.
Understanding the mechanics of rappelling, like proper body alignment and rope management, is essential for a safe and controlled descent.
Equilibrium of Forces
When rappelling, the climber is in a state of equilibrium. This means that all the forces acting on them are balanced, and they remain stationary or move at a constant velocity. For the mountaineer, here are the forces at play:
  • Gravitational Force (\(F_g\)): This force pulls the climber downward due to gravity.
  • Tension in the Rope (\(T\)): This force acts along the same line as the rope, counteracting the climber's weight.
  • Normal Force (\(F_n\)): The cliff face exerts an upward force on the climber鈥檚 feet.
  • Frictional Force (\(f\)): This force prevents the climber鈥檚 feet from slipping.
To maintain equilibrium, the sum of these forces should equal zero. Thus, understanding how these forces interact helps climbers make appropriate adjustments during rappelling.
Coefficient of Static Friction
The coefficient of static friction, denoted as \(\mu_s\), is key to understanding when an object will start to slide on a surface. It measures how much grip or friction there is between two surfaces before movement starts. For a climber, the coefficient of static friction must be sufficiently high to prevent the feet from slipping against the cliff. This frictional force depends on:
  • The normal force (\(F_n\)), which acts perpendicular to the surface.
  • The surface texture of the climber's footwear and the cliff.
To compute \(\mu_s\): \(\mu_s = \frac{f}{F_n}\).By understanding this coefficient, climbers can determine minimum safety requirements for different surfaces, ensuring they maintain stability while rappelling.
Free Body Diagram Analysis
A free body diagram is an essential tool in physics to analyze forces acting on an object. It visually represents all forces on a single object, showing their directions and magnitudes.For the climber scenario:
  • Gravitational force (\(F_g\)) acts downward from the climber鈥檚 center of gravity.
  • Tension forces, acting through the rope at the angle specified.
  • Normal and Frictional forces, acting at the feet where climber contacts the cliff.
Creating and analyzing a free body diagram helps to understand how various forces interact, allowing for the solving of complex problems involving multiple forces in equilibrium. With this tool, climbers can visualize all the relevant forces to ensure a safe rappel.

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Most popular questions from this chapter

A uniform steel rod has mass \(0.300 \mathrm{~kg}\) and length \(40.0 \mathrm{~cm}\) and is horizontal. A uniform sphere with radius \(8.00 \mathrm{~cm}\) and mass \(0.900 \mathrm{~kg}\) is welded to one end of the bar, and a uniform sphere with radius \(6.00 \mathrm{~cm}\) and mass \(0.380 \mathrm{~kg}\) is welded to the other end of the bar. The centers of the rod and of each sphere all lie along a horizontal line. How far is the center of grayity of the combined obiect from the center of the rod?

The left-hand end of a light rod of length \(L\) is attached to a vertical wall by a frictionless hinge. An object of mass \(m\) is suspended from the rod at a point a distance \(\alpha L\) from the hinge, where \(0<\alpha \leq 1.00 .\) The rod is held in a horizontal position by a light wire that runs from the right- hand end of the rod to the wall. The wire makes an angle \(\theta\) with the rod. (a) What is the angle \(\beta\) that the net force exerted by the hinge on the rod makes with the horizontal? (b) What is the value of \(\alpha\) for which \(\beta=\theta ?\) (c) What is \(\beta\) when \(\alpha=1.00 ?\)

In lab tests on a \(9.25 \mathrm{~cm}\) cube of a certain material, a force of \(1375 \mathrm{~N}\) directed at \(8.50^{\circ}\) to the cube (Fig. E11.38) causes the cube to deform through an angle of \(1.24^{\circ} .\) What is the shear modulus of the material?

You need to measure the mass \(M\) of a \(4.00-\mathrm{m}\) -long bar. The bar has a square cross section but has some holes drilled along its length, so you suspect that its center of gravity isn't in the middle of the bar. The bar is too long for you to weigh on your scale. So, first you balance the bar on a knife-edge pivot and determine that the bar's center of gravity is \(1.88 \mathrm{~m}\) from its left-hand end. You then place the bar on the pivot so that the point of support is \(1.50 \mathrm{~m}\) from the left-hand end of the bar. Next you suspend a \(2.00 \mathrm{~kg}\) mass \(\left(m_{1}\right)\) from the bar at a point \(0.200 \mathrm{~m}\) from the left-hand end. Finally, you suspend a mass \(m_{2}=1.00 \mathrm{~kg}\) from the bar at a distance \(x\) from the left-hand end and adjust \(x\) so that the bar is balanced. You repeat this step for other values of \(m_{2}\) and record each corresponding value of \(x\). The table gives your results. $$ \begin{array}{l|llllll} m_{2}(\mathrm{~kg}) & 1.00 & 1.50 & 2.00 & 2.50 & 3.00 & 4.00 \\ \hline x(\mathrm{~m}) & 3.50 & 2.83 & 2.50 & 2.32 & 2.16 & 2.00 \end{array} $$ (a) Draw a free-body diagram for the bar when \(m_{1}\) and \(m_{2}\) are suspended from it. (b) Apply the static equilibrium equation \(\Sigma \tau_{z}=0\) with the axis at the location of the knife-edge pivot. Solve the equation for \(x\) as a function of \(m_{2}\). (c) Plot \(x\) versus \(1 / m_{2}\). Use the slope of the best-fit straight line and the equation you derived in part (b) to calculate that bar's mass \(M .\) Use \(g=9.80 \mathrm{~m} / \mathrm{s}^{2} .\) (d) What is the \(y\) -intercept of the straight line that fits the data? Explain why it has this value.

A claw hammer is used to pull a nail out of a board (see Fig. \(\mathrm{P} 11.45\) ). The nail is at an angle of \(60^{\circ}\) to the board, and a force \(\overrightarrow{\boldsymbol{F}}_{1}\) of magnitude \(400 \mathrm{~N}\) applied to the nail is required to pull it from the board. The hammer head contacts the board at point \(A,\) which is \(0.080 \mathrm{~m}\) from where the nail enters the board. A horizontal force \(\vec{F}_{2}\) is applied to the hammer handle at a distance of \(0.300 \mathrm{~m}\) above the board. What magnitude of force \(\overrightarrow{\boldsymbol{F}}_{2}\) is required to apply the required \(400 \mathrm{~N}\) force \(\left(F_{1}\right)\) to the nail? (Ignore the weight of the hammer.)
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