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Chapter 4: Problem 60

The drawing shows a circus clown who weighs 890 N. The coefficient of static friction between the clown鈥檚 feet and the ground is 0.53. He pulls vertically downward on a rope that passes around three pulleys and is tied around his feet. What is the minimum pulling force that the clown must exert to yank his feet out from under himself?

Short Answer

Expert verified
The clown must exert at least 471.7 N.

Step by step solution

01

Identify the forces involved

The forces acting on the clown include his weight of 890 N and the frictional force which opposes the motion. The frictional force (static friction) is what we overcome by pulling on the rope.
02

Understand static friction

The static friction force can be calculated using the formula: \( f_s = \mu_s \cdot N \), where \( f_s \) is the static friction force, \( \mu_s = 0.53 \) is the coefficient of static friction, and \( N = 890 \text{ N} \) is the normal force (equal to the clown's weight).
03

Calculate the static frictional force

Substitute the values into the formula: \( f_s = 0.53 \times 890 \text{ N} = 471.7 \text{ N} \). This means the force required to break the static friction is 471.7 N.
04

Determine the minimum pulling force

The minimum force the clown needs to apply vertically downward via the rope should equal the static friction force calculated, which is 471.7 N.

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Key Concepts

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

Force Calculation
In physics, force calculation is a fundamental concept that helps us understand the effects of forces acting upon objects. For this exercise, understanding how to calculate forces is crucial.

When the clown pulls on the rope, he exerts a force that must be strong enough to overcome static friction. To find the minimum force required, we must first determine the frictional force opposing the movement of the clown's feet.

The calculated static friction force was found using the formula:
  • Static Friction Force (\( f_s \)) = Coefficient of Static Friction (\( \mu_s \)) 脳 Normal Force (\( N \))
  • Substituting the given values, \( f_s = 0.53 \times 890 \text{ N} = 471.7 \text{ N} \)
The static friction force of 471.7 N defines the minimum pulling force the clown must exert to release his feet. This calculation is essential in understanding the setup of the problem and forms the heart of the force analysis for the system.
Pulley Systems
Pulley systems are mechanical devices comprised of wheels and ropes that change the direction and magnitude of forces used, making certain tasks easier.

Using a pulley system, the clown can efficiently exert a downward force on the rope tied around his feet by redirecting the force applied.
  • The pulley system used in this case allows the clown to pull vertically downward, applying force through a series of three pulleys.
  • The advantage of using a pulley is that it allows for a more effective force exertion while also providing mechanical advantage by reducing the effort needed.
In practical scenarios, pulley systems are used to lift heavy objects with lesser force. Here, the mechanism helps the clown apply the required 471.7 N force in an appropriate manner to overcome static friction.
Coefficient of Friction
The coefficient of friction is a crucial variable that determines the amount of frictional force between two surfaces. It is a dimensionless number that represents the ratio of the frictional force resisting the motion of two objects to the normal force pressing the objects together.

In this exercise, the coefficient of static friction (\( \mu_s \)) is given as 0.53 between the clown's feet and the ground. This means:
  • The surfaces in contact provide a moderate level of resistance against motion.
  • This value is used in calculating the total static friction force that must be overcome for movement to occur.
Understanding the coefficient of friction is critical for engineers and physicists as it allows them to design systems and structures that can predictably interact with various surfaces. It affects decisions ranging from the grip of tires on the road to the slip resistance of shoes like those worn by the clown.

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Most popular questions from this chapter

A 15-g bullet is fired from a rifle. It takes \(2.50 \times 10^{-3}\) s for the bullet to travel the length of the barrel, and it exits the barrel with a speed of \(715 \mathrm{m} / \mathrm{s} .\) Assuming that the acceleration of the bullet is constant, find the average net force exerted on the bullet.

A \(60.0-\mathrm{kg}\) crate rests on a level floor at a shipping dock. The coefficients of static and kinetic friction are 0.760 and \(0.410,\) respectively. What horizontal pushing force is required to (a) just start the crate moving and (b) slide the crate across the dock at a constant speed?

A car is traveling up a hill that is inclined at an angle \(\theta\) above the horizontal. Determine the ratio of the magnitude of the normal force to the weight of the car when (a) \(\theta=15^{\circ}\) and (b) \(\theta=35^{\circ}\).

The speed of a bobsled is increasing because it has an acceleration of \(2.4 \mathrm{m} / \mathrm{s}^{2} .\) At a given instant in time, the forces resisting the motion, including kinetic friction and air resistance, total 450N. The combined mass of the bobsled and its riders is \(270 \mathrm{kg}\) (a) What is the magnitude of the force propelling the bobsled forward? (b) What is the magnitude of the net force that acts on the bobsled?

A 6.00-kg box is sliding across the horizontal floor of an elevator. The coefficient of kinetic friction between the box and the floor is 0.360. Determine the kinetic frictional force that acts on the box when the elevator is (a) stationary, (b) accelerating upward with an acceleration whose magnitude is \(1.20 \mathrm{m} / \mathrm{s}^{2}\), and \((\mathrm{c})\) accelerating downward with an acceleration whose magnitude is \(1.20 \mathrm{m} / \mathrm{s}^{2}\).
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