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Ch 06: Dynamics I: Motion Along a Line
Knight Calc - Physics for Scientists and Engineers 5th Edition
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook
All textbooksKnight Calc 5th EditionCh 06: Dynamics I: Motion Along a LineProblem 28
Chapter 6, Problem 28

A 4000 kg truck is parked on a 15° slope. How big is the friction force on the truck? The coefficient of static friction between the tires and the road is 0.90.

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Identify the forces acting on the truck: The truck is subject to gravitational force, normal force, and static friction. The gravitational force can be broken into two components: one parallel to the slope (causing the truck to slide down) and one perpendicular to the slope (contributing to the normal force).
Calculate the gravitational force acting on the truck using the formula: Fg = mg, where m is the mass of the truck (4000 kg) and g is the acceleration due to gravity (9.8 m/s²).
Determine the component of the gravitational force parallel to the slope using the formula: Fparallel = Fg sin(θ), where θ is the angle of the slope (15 degrees).
Calculate the maximum static friction force using the formula: Ffriction = μs Fnormal, where μs is the coefficient of static friction (0.90) and Fnormal is the normal force, calculated as Fg cos(θ).
Compare the parallel component of the gravitational force (Fparallel) with the maximum static friction force. If Fparallel is less than or equal to the maximum static friction force, the friction force equals Fparallel, as the truck does not slide.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Static Friction

Static friction is the force that resists the initiation of sliding motion between two surfaces in contact. It acts parallel to the surfaces and is dependent on the normal force and the coefficient of static friction. The maximum static friction force can be calculated using the formula: F_friction = μ_s * N, where μ_s is the coefficient of static friction and N is the normal force.
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Normal Force

The normal force is the perpendicular force exerted by a surface to support the weight of an object resting on it. On an inclined plane, the normal force is less than the object's weight and can be calculated using the formula: N = mg * cos(θ), where m is the mass, g is the acceleration due to gravity, and θ is the angle of the incline. This force is crucial for determining the frictional force acting on the truck.
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Inclined Plane

An inclined plane is a flat surface tilted at an angle to the horizontal, which affects the forces acting on an object resting on it. The angle of the incline influences both the normal force and the gravitational force components acting on the object. Understanding how to resolve these forces is essential for calculating the friction force and analyzing the object's stability on the slope.
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Related Practice
Textbook Question

Bonnie and Clyde are sliding a 300 kg bank safe across the floor to their getaway car. The safe slides with a constant speed if Clyde pushes from behind with 385 N of force while Bonnie pulls forward on a rope with 350 N of force. What is the safe's coefficient of kinetic friction on the bank floor?

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Textbook Question

A rubber-wheeled 50 kg cart rolls down a 15° concrete incline. What is the magnitude of the cart's acceleration if rolling friction is (a) neglected and (b) included?

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Textbook Question

A 50,000 kg locomotive is traveling at 10 m/s when its engine and brakes both fail. How far will the locomotive roll before it comes to a stop? Assume the track is level.

Textbook Question

Bob is pulling a 30 kg filing cabinet with a force of 200 N, but the filing cabinet refuses to move. The coefficient of static friction between the filing cabinet and the floor is 0.80. What is the magnitude of the friction force on the filing cabinet?

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Textbook Question

A 10 kg crate is placed on a horizontal conveyor belt. The materials are such that μs\(\mu\)_{s} = 0.5 and μk\(\mu\)_{k} = 0.3. Draw a free-body diagram showing all the forces on the crate if the conveyer belt is speeding up.

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Textbook Question

A 1500 kg car skids to a halt on a wet road where μk\(\mu\)_{k} = 0.50. How fast was the car traveling if it leaves 65-m-long skid marks?

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