**Educational Exercise: Calculating Mass with Friction****Problem Statement:**A man is supporting his weight through friction while pressing into rock walls with his feet and back. Assume friction is completely supporting his weight – he might use the rope for balance without pulling. The normal force on his feet (combined from both feet and back) is 743.2 N. Given a coefficient of static friction of 0.54, calculate his mass if he is just barely not slipping.**Question:**What is the man's mass?- **Normal Force:** 743.2 N- **Coefficient of Static Friction:** 0.54**Diagram:**The image shows a man in a climbing position, using friction against rock walls to maintain his stance.**Calculation:**We use the equation for friction: \[ f = \mu \cdot N \] where \( f \) is the frictional force, \( \mu \) is the coefficient of static friction, and \( N \) is the normal force.To find the man's mass (\( m \)), use the balance between gravitational force and frictional force:\[ m \cdot g = \mu \cdot N \]Solving for \( m \):\[ m = \frac{\mu \cdot N}{g} \]where \( g \approx 9.81 \, \text{m/s}^2 \) (acceleration due to gravity).**Calculation Attempt:**An initial calculation of 40.95 kg was attempted but deemed incorrect. Ensure correct substitution and calculation for accuracy.**Previous Submissions:**- 40.95 kg (incorrect)**Answer Input Box:**- [Mass: _____ kg] [Incorrect]### Conclusion:Perform the calculation carefully, checking each step to find the correct mass that satisfies the given conditions.

The upward friction force is μN on both feet and back. This force must equal the weight of person.…

The man is supporting his weight with friction. He's pressing into the rock walls with his feet and back. Suppose friction is supporting all his weight - he's perhaps steadying himself with the rope but not pulling up. The normal force on his feet (both feet together) and back are the same, 743.2 N. The coefficient of static friction is 0.54. If he's just barely not slipping down, what must be his mass?

The man is supporting his weight with friction. He's pressing into the rock walls with his feet and back. Suppose friction is supporting all his weight - he's perhaps steadying himself with the rope but not pulling up. The normal force on his feet (both feet together) and back are the same, 743.2 N. The coefficient of static friction is 0.54. If he's just barely not slipping down, what must be his mass?

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

See similar textbooks

Similar questions

A 4.15 kg4.15 kg box is suspended from a system of massless wires attached to a ceiling. The first wire makes an angle of ?1=63.0∘θ1=63.0∘ with the ceiling, whereas the second wire makes an angle of ?2=41.5∘θ2=41.5∘ with the ceiling. Determine the magnitudes of the tensions in the first and second wires, ?1T1 and ?2,T2, respectively.
The crate A weighs 780 lb. Between all contacting surfaces, μs = 0.34 and μk = 0.29. Neglect the weights of the wedges. What force F is required to move A to the right at a constant rate? Wedge angle is 6 degrees.
Suppose you have a stack of 17 identical steel bars (steel bars have a mass of 1.00kg each). You are required to pull out bar number 16 from the stack (counting from the top of the stack down). Assuming that your assistant holds all other bars in place. If the coefficient of friction between steel and steel is 0.750, how hard do you have to pull (in N)?
A 1am playing tug-o-war with my three nephews. The youngest in pulling with a force of 189 N. The middle child is pulling with a force of 334 N. The oldest is pulling with a force of 657 N. I am on a team with the youngest child. If I want to keep the four of us in equilibrium, how hard (in newtons) must I pull?
Two masses m1=5 Kg and m2=12 Kg are connected with string and pulled with a force F=25 N that makes an angle of 53° with a smooth horizontal surface as shown in the figure. Find the tension in the string in a unit of Newton. F A. 6.258 B. 8.85 m2 m1 0 C. 2213 D. None E. 4.425
The steel I-beam in the drawing has a weight of 1.51 × 103 N and is being lifted at a constant velocity. What is the tension in each cable attached to its ends?
A 30.6 kg mass (m) is suspended by the cable assembly as shown in the figure. The cables have no mass of their own. The cable to the left (T|) of the mass makes an angle of 0.00° with the horizontal, and the cable to the right (T2) makes an angle (02) of 36.4°. If the mass is at rest, what is the tension in T each of the cables, T¡ and T,? The acceleration due to gravity is g = 9.81 m/s². T = N m T2 = N
A mountain climber, in the process of crossing between two cliffs by the rope, pauses to rest. She weight 535 N. She is closer to the left cliff than to the right cliff. The left rope makes 600 and the right rope makes 800 with the vertical. Find the tensions in the rope to the left and to the right of the mountain climber.