/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 58 A stuntman is being pulled along... [FREE SOLUTION] | 91Ó°ÊÓ

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A stuntman is being pulled along a rough road at a constant velocity by a cable attached to a moving truck. The cable is parallel to the ground. The mass of the stuntman is \(109 \mathrm{kg}\), and the coefficient of kinetic friction between the road and him is \(0.870 .\) Find the tension in the cable.

Short Answer

Expert verified
The tension in the cable is 929.334 N.

Step by step solution

01

Identify Forces

The forces acting on the stuntman include the tension in the cable, the friction force exerted by the road, the gravitational force (weight), and the normal force. Since he moves at a constant velocity, these forces must be balanced.
02

Calculate Gravitational Force

The gravitational force is calculated using the formula \( F_g = m \cdot g \), where \( m = 109 \, \text{kg} \) and \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity). \[ F_g = 109 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 1068.2 \, \text{N} \]
03

Calculate the Normal Force

On a flat surface, the normal force \( N \) is equal in magnitude and opposite in direction to the gravitational force for an object at rest or moving with constant velocity.\[ N = F_g = 1068.2 \, \text{N} \]
04

Calculate Frictional Force

The kinetic frictional force can be calculated using \( F_f = \mu_k \cdot N \), where \( \mu_k = 0.870 \) is the coefficient of kinetic friction.\[ F_f = 0.870 \times 1068.2 \, \text{N} = 929.334 \, \text{N} \]
05

Determine the Tension in the Cable

Since the stuntman is being pulled at a constant velocity, the tension in the cable \( T \) must equal the frictional force to maintain the constant motion on a horizontal surface.\[ T = F_f = 929.334 \, \text{N} \]

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Key Concepts

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

Tension
In physics, tension refers to the force that is transmitted through a string, cable, or similar object when it is pulled tight by forces acting from opposite ends. It's important to understand that tension is a force that ensures the object stays connected between the two ends.
For example, in the scenario with the stuntman, the tension in the cable ensures that the stuntman is being dragged along the road. When dealing with tension, the force is directed along the length of the medium, in this case, the cable, and away from the object being pulled.
  • **Consistency:** In this problem, the fact that the stuntman is moving at a constant velocity indicates that the force due to tension equals the kinetic frictional force.
  • **Direction:** Tension acts parallel to the ground, emphasizing why the understanding of direction is crucial in physical dynamics scenarios.
Overall, calculating tension accurately requires balancing all forces involved to determine the resultant force that establishes equilibrium in the system.
Gravitational Force
The gravitational force, often referred to as "weight," is the force with which the Earth pulls on an object. It's calculated using the formula:
\[ F_g = m \, g \]
Where:
  • \( F_g \) is the gravitational force
  • \( m \) is the mass of the object (for the stuntman, it is 109 kg)
  • \( g \) is the acceleration due to gravity, approximately 9.8 m/s² on Earth.
Gravitational force plays a crucial role in determining other forces acting on the stuntman; it directly influences the normal force exerted by the ground.
The gravitational force keeps the stuntman grounded, opposing the drag force due to friction. It acts vertically downward and is crucial for calculating other forces in physics problems, as it typically influences the normal force and friction.
Normal Force
The normal force is the perpendicular contact force exerted by a surface on an object in contact with it. In our stuntman example, this is the force of the road pushing up against the stuntman’s body, balancing the gravitational force pulling him down.
Key points to remember about normal force include:
  • **Equilibrium on Flat Surfaces:** On a horizontal surface and with no vertical acceleration, the normal force should balance the gravitational force. That means it has equal magnitude but acts in the opposite direction to weight.
  • **Changes with Inclines and Other Forces:** While on flat ground in this exercise, the normal force is equal to the gravitational pull \( N = 1068.2 \, N \). On inclined surfaces or if there are vertical accelerations, calculating normal force can become more complex.
Understanding normal force is essential for analyzing scenarios involving friction and contact forces in physics, as it directly affects how much frictional force can be developed, thus influencing the tension needed to maintain motion.

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Most popular questions from this chapter

The central ideas in this problem are reviewed in Multiple-Concept Example 9. One block rests upon a horizontal surface. A second identical block rests upon the first one. The coefficient of static friction between the blocks is the same as the coefficient of static friction between the lower block and the horizontal surface. A horizontal force is applied to the upper block, and the magnitude of the force is slowly increased. When the force reaches \(47.0 \mathrm{N},\) the upper block just begins to slide. The force is then removed from the upper block, and the blocks are returned to their original configuration. What is the magnitude of the horizontal force that should be applied to the lower block so that it just begins to slide out from under the upper block?

A 325 -kg boat is sailing \(15.0^{\circ}\) north of east at a speed of \(2.00 \mathrm{m} / \mathrm{s}\). Thirty seconds later, it is sailing \(35.0^{\circ}\) north of east at a speed of \(4.00 \mathrm{m} / \mathrm{s}\). During this time, three forces act on the boat: a \(31.0-\mathrm{N}\) force directed \(15.0^{\circ}\) north of east (due to an auxiliary engine), a \(23.0-\mathrm{N}\) force directed \(15.0^{\circ}\) south of west (resistance due to the water), and \(\overrightarrow{\mathbf{F}}_{\mathrm{w}}\) (due to the wind). Find the magnitude and direction of the force \(\overrightarrow{\mathbf{F}}_{\mathbf{w}} .\) Express the direction as an angle with respect to due east.

A flatbed truck is carrying a crate up a hill of angle of inclination \(\theta=10.0^{\circ},\) as the figure illustrates. The coefficient of static friction between the truck bed and the crate is \(\mu_{\mathrm{s}}=0.350 .\) Find the maximum acceleration that the truck can attain before the crate begins to slip backward relative to the truck.

A bowling ball (mass \(=7.2 \mathrm{kg}\), radius \(=0.11 \mathrm{m}\) ) and a billiard ball (mass \(=0.38 \mathrm{kg}\), radius \(=0.028 \mathrm{m}\) ) may each be treated as uniform spheres. What is the magnitude of the maximum gravitational force that each can exert on the other?

The drawing shows Robin Hood (mass \(=77.0 \mathrm{kg}\) ) about to escape from a dangerous situation. With one hand, he is gripping the rope that holds up a chandelier (mass \(=195 \mathrm{kg}\) ). When he cuts the rope where it is tied to the floor, the chandelier will fall, and he will be pulled up toward a balcony above. Ignore the friction between the rope and the beams over which it slides, and find (a) the acceleration with which Robin is pulled upward and (b) the tension in the rope while Robin escapes.

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