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Chapter 8: Problem 34

A mountain climber is rappelling down a vertical wall. The rope attaches to a buckle strapped to the climber's waist \(15 \mathrm{cm}\) to the right of his center of gravity. If the climber weighs \(770 \mathrm{N},\) find (a) the tension in the rope and (b) the magnitude and direction of the contact force exerted by the wall on the climber's feet.

Short Answer

Expert verified
Answer: (a) The tension in the rope is \(770 N\). (b) The magnitude of the contact force exerted by the wall on the climber's feet is \(1089.6 N\) and its direction is \(45^\circ\) above the horizontal.

Step by step solution

01

Draw a free body diagram representing all forces acting on the climber

Draw a free body diagram that represents all the forces acting on the mountain climber, including the climber's weight (W) acting downward at his center of gravity, the tension (T) in the rope acting \(15 cm\) to the right of his center of gravity, and contact force (F) exerted by the wall on the climber's feet.
02

Apply the equilibrium conditions for forces

To satisfy the condition of equilibrium, the sum of the forces in the horizontal and vertical directions should be equal to zero. Therefore, we can write: Horizontal forces: \(F_x - T = 0\) Vertical forces: \(F_y - W = 0\)
03

Apply the equilibrium condition for torques

Next, we will apply the equilibrium conditions for the torques. Since the climber is in equilibrium, the net torque about his center of gravity equals zero. We will take counterclockwise torques as positive. Let's consider the torque created by tension force T and contact force F: Torque by T: \(\tau_T = T \times d\), where \(d=15 cm = 0.15 m\) Torque by F: \(\tau_F = F_x \times L\), where L is the distance between the climber's center of gravity and his feet. Therefore, we can write the equilibrium condition for torques as: \(\tau_T - \tau_F = 0 \Rightarrow T\times0.15 - F_x\times L = 0\)
04

Solve for the tension T in the rope

From the previous step, we have the torque equilibrium condition: \(T\times0.15 - F_x\times L = 0 \Rightarrow T\times0.15 = F_x\times L\) Recalling the horizontal force equation, \(F_x = T\). Therefore, \(T\times0.15 = T\times L \rightarrow L = 0.15\) Now we know the distance L from the climber's center of gravity to his feet. Since T is equal to F_x and is the only horizontal force, we get: Tension in the rope, T = \(F_x = 770 N\)
05

Finding the contact force F and its direction

Now that we have T, we can find F_y using the vertical force equation: \(F_y - W = 0 \Rightarrow F_y = W = 770 N\) Finally, to find the magnitude of the contact force F and its angle (θ) with the horizontal axis, we can use the Pythagorean theorem and tangent function: Magnitude of the contact force, \(F = \sqrt{F_x^2 + F_y^2} = \sqrt{(770N)^2 + (770N)^2} = 1089.6 N\) Angle of contact force with the horizontal, \(\theta = \tan^{-1}\frac{F_y}{F_x} = \tan^{-1}\frac{770N}{770N} = 45^\circ\)
06

Final Answer

(a) The tension in the rope is \(770 N\). (b) The magnitude of the contact force exerted by the wall on the climber's feet is \(1089.6 N\) and its direction is \(45^\circ\) above the horizontal.

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