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Chapter 6: Problem 12

In about \(1915,\) Henry sincosky of Philadelphia suspended himself from a rafter by gripping the rafter with the thumb of each hand on one side and the fingers on the opposite side (Fig. 6-21). sincosky's mass was \(79 \mathrm{~kg}\). If the coefficient of static friction between hand and rafter was \(0.70,\) what was the least magnitude of the normal force on the rafter from each thumb or opposite fingers? (After suspending himself, sincosky chinned himself on the rafter and then moved hand-over-hand along the rafter. If you do not think sincosky's grip was remarkable, try to repeat his stunt.)

Short Answer

Expert verified
The normal force per hand should be at least 554.14 N.

Step by step solution

01

Understand the Problem

To determine the least magnitude of the normal force on the rafter, we need to calculate the minimum force exerted by each hand required to counterbalance Henry Sincosky's weight, given the static friction coefficient. We need to consider the forces involved: gravity acting downward, his grip producing normal forces, and static friction opposing the gravitational force.
02

Analyzing Forces and Setting Up the Equation

Each hand exerts a normal force, denoted as \( F_n \), and static friction \( f_s \) arises due to this normal force. The maximum static friction force is given by \( f_s = \mu_s F_n \), where \( \mu_s = 0.70 \) is the static friction coefficient. This friction force must equal the gravitational force exerted by the weight of the body to prevent slipping.
03

Calculating the Weight Force

Calculate Henry's weight using his mass \( m = 79 \, \text{kg} \) and gravitational acceleration \( g = 9.81 \, \text{m/s}^2 \). The weight force \( W \) is: \[ W = m \cdot g = 79 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 775.79 \, \text{N}. \]
04

Equating Friction and Weight

Since each hand needs to exert an equitable portion of the total weight force with static friction, the total frictional force should equal the weight: \[ \text{Total friction force} = 2 \cdot f_s = 775.79 \, \text{N}. \]
05

Solving for the Normal Force per Hand

Since \( f_s = \mu_s F_n \), substitute for \( f_s \):\[ 2 \cdot \mu_s F_n = 775.79 \, \text{N}. \] Rearrange to solve for the normal force \( F_n \):\[ F_n = \frac{775.79 \, \text{N}}{2 \cdot 0.70} = 554.14 \, \text{N}. \]
06

Conclusion

The least magnitude of the normal force required on the rafter from each thumb or opposite fingers to keep Sincosky suspended is approximately \( 554.14 \, \text{N} \).

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Key Concepts

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

Normal Force
The normal force is an essential concept when dealing with static friction and other situations involving surfaces in contact. Think of it as the support force exerted by a surface, perpendicular to that surface, to counteract the weight of an object pressing against it.
In the exercise, each of Sincosky's hands applies a normal force against the rafter. This force ensures that he remains suspended without falling.
  • Normal force is perpendicular to the contact surface.
  • It's the reaction force to the weight or other forces pushing against a surface.
  • In this problem, the normal forces exerted by his fingers and thumbs counterbalance his weight.
The normal force here isn't determined by the material's properties like friction is; it simply responds to how much pressure or weight is applied on the surface. Be mindful that when two surfaces are in contact, and there is no movement, the normal force can be affected by angles or other external forces applied.
Gravitational Force
Gravitational force is the force of attraction between two masses. On Earth, this means it's the force pulling objects toward the center of the planet. In practical terms, we experience it as weight.
Henry Sincosky鈥檚 mass creates a gravitational force due to Earth's pull, which is why he needs to exert a specific grip force to stay suspended.
  • The gravitational force on an object is calculated by multiplying its mass by the acceleration due to gravity (\( g \), which is roughly 9.81 m/s虏 on Earth).
  • In Sincosky鈥檚 case, his weight, \( W = m imes g \).
  • This force must be counteracted by the static friction force exerted by his hands.
Understanding gravitational force is crucial in figuring out how much resistance (friction) is needed from his hands to counteract the pull from Earth. It's the balance of this force with the frictional grip that allows him to execute his remarkable feat.
Coefficient of Friction
The coefficient of friction is a number that represents how easily one object moves in contact over another. It varies depending on the nature and characteristics of surfaces in contact.
Static friction is especially relevant when there's no movement between the surfaces, as in Sincosky's gripping situation.
  • The coefficient of static friction (\(\mu_s\)) describes the ratio of the maximum static frictional force that can be exerted before motion starts to the normal force.
  • For Sincosky, \(\mu_s = 0.70\), meaning the frictional force is 70% of the normal force applied by his hands.
  • It allows us to calculate the friction needed to prevent sliding.
By understanding the coefficient of friction, we can solve why the static frictional force balances out the gravitational force, ensuring Sincosky remains suspended without slipping off. This grip is only possible if the frictional force equals the gravitational pull, which is made possible due to the specific nature of the surfaces in contact and the pressure applied.

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Most popular questions from this chapter

Calculate the ratio of the drag force on a jet flying at \(1000 \mathrm{~km} / \mathrm{h}\) at an altitude of \(10 \mathrm{~km}\) to the drag force on a propdriven transport flying at half that speed and altitude. The density of air is \(0.38 \mathrm{~kg} / \mathrm{m}^{3}\) at \(10 \mathrm{~km}\) and \(0.67 \mathrm{~kg} / \mathrm{m}^{3}\) at \(5.0 \mathrm{~km} .\) Assume that the airplanes have the same effective cross-sectional area and drag coefficient \(C\).

A house is built on the top of a hill with a nearby slope at angle \(\theta=45^{\circ}\) (Fig. 6-55). An engineering study indicates that the slope angle should be reduced because the top layers of soil along the slope might slip past the lower layers. If the coefficient of static friction between two such layers is \(0.5,\) what is the least angle \(\phi\) through which the present slope should be reduced to prevent slippage?

An amusement park ride consists of a car moving in a vertical circle on the end of a rigid boom of negligible mass. The combined weight of the car and riders is \(5.0 \mathrm{kN},\) and the circle's radius is \(10 \mathrm{~m} .\) At the top of the circle, what are the (a) magnitude \(F_{B}\) and \((\mathrm{b})\) direction (up or down) of the force on the car from the boom if the car's speed is \(v=5.0 \mathrm{~m} / \mathrm{s} ?\) What are (c) \(F_{B}\) and (d) the direction if \(v=12 \mathrm{~m} / \mathrm{s} ?\)

A banked circular highway curve is designed for traffic moving at \(60 \mathrm{~km} / \mathrm{h}\). The radius of the curve is \(200 \mathrm{~m}\). Traffic is moving along the highway at \(40 \mathrm{~km} / \mathrm{h}\) on a rainy day. What is the minimum coefficient of friction between tires and road that will allow cars to take the turn without sliding off the road? (Assume the cars do not have negative lift.)

A child weighing \(140 \mathrm{~N}\) sits at rest at the top of a playground slide that makes an angle of \(25^{\circ}\) with the horizontal. The child keeps from sliding by holding onto the sides of the slide. After letting go of the sides, the child has a constant acceleration of \(0.86 \mathrm{~m} / \mathrm{s}^{2}\) (down the slide, of course). (a) What is the coefficient of kinetic friction between the child and the slide? (b) What maximum and minimum values for the coefficient of static friction between the child and the slide are consistent with the information given here?
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