/*! 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 66 An inclined plane makes an angle... [FREE SOLUTION] | 91Ó°ÊÓ

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An inclined plane makes an angle of \(30^{\circ}\) with the horizontal. Find the constant force, applied parallel to the plane, required to cause a \(15-\mathrm{kg}\) box to slide \((a)\) up the plane with acceleration \(1.2 \mathrm{~m} / \mathrm{s}^{2}\) and \((b)\) down the incline with acceleration \(1.2 \mathrm{~m} / \mathrm{s}^{2} .\) Neglect friction forces.

Short Answer

Expert verified
Upward force needed: 91.5 N; downward: -55.5 N.

Step by step solution

01

Identify Forces Acting on the Box

An inclined plane problem typically involves gravitational force, normal force, and the applied force. For this box, its weight can be decomposed into components parallel and perpendicular to the plane. The parallel component of the gravitational force is \( mg \sin\theta \). Here, \( m = 15 \, \text{kg} \), \( g = 9.8 \, \text{m/s}^2 \), and \( \theta = 30^\circ \).
02

Calculate Gravitational Force Component

Calculate the component of the gravitational force acting down the slope of the incline: \[F_{\text{gravity, parallel}} = mg \sin\theta = 15 \times 9.8 \times \sin(30^\circ)\]\[F_{\text{gravity, parallel}} = 73.5 \, \text{N}\]
03

Part A: Calculate Force for Upward Acceleration

For the box to slide upward with acceleration, the applied force must overcome the gravitational component and provide additional force for the acceleration:\[F_{\text{up}} - F_{\text{gravity, parallel}} = ma\]Substituting the known values:\[F_{\text{up}} - 73.5 = 15 \times 1.2\]\[F_{\text{up}} = 91.5 \, \text{N}\]
04

Part B: Calculate Force for Downward Acceleration

To find the force that causes the box to slide down the incline with acceleration, reduce the gravitational component by the acceleration force:\[F_{\text{down}} + F_{\text{gravity, parallel}} = ma\]Substituting the known values:\[F_{\text{down}} + 73.5 = 15 \times 1.2\]\[F_{\text{down}} = -55.5 \, \text{N}\]The negative sign indicates that the applied force is in the opposite direction as the box moves down.

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Key Concepts

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

Forces on an Incline
Understanding the forces acting on an incline is crucial when dealing with inclined plane problems. When a box is placed on an inclined plane, several forces come into play. Firstly, there is the gravitational force, which acts vertically downward. This force can be split into two components: one parallel to the slope and one perpendicular to it.

The normal force acts perpendicular to the plane surface, counterbalancing the perpendicular component of weight. The key player in inclined plane scenarios is the component of gravity that acts parallel to the slope. This is the force that causes the box to potentially slide down the incline. In many problems, an additional force is applied to either push or pull the box in the desired direction along the slope.

To summarize, the forces typically include:
  • Gravitational force (acting vertically)
  • Normal force (acting perpendicular to the slope)
  • Applied force (pushing or pulling parallel to the slope)
  • Frictional force (often neglected or simplified in beginners' problems)
Understanding how these forces interact is essential to solving mechanics problems involving inclined planes.
Gravitational Force Components
The gravitational force acting on an object on an inclined plane can be decomposed into two components using trigonometric functions. This is a key technique in physics for solving inclined plane problems.

The component acting parallel to the incline is given by:\[ F_{\text{gravity, parallel}} = mg \sin \theta \]This component is responsible for the object sliding down the plane.
This term is crucial when calculating the net force along the plane, especially in problems where you aim to either resist or enhance the natural sliding motion.

The perpendicular component of gravity does not cause motion along the plane but determines the normal force. It's given by:\[ F_{\text{gravity, perpendicular}} = mg \cos \theta \]This component is balanced by the normal force, which ensures the object doesn't accelerate vertically into the plane.

By understanding these components, you can break down complex forces into simpler, more manageable parts. This greatly helps in using Newton's equations to determine other variables like acceleration or the required applied force.
Newton's Second Law
Newton's Second Law of motion states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. This fundamental principle is expressed as:\[ F = ma \]Where \( F \) is the net force on the object, \( m \) is the mass, and \( a \) is the acceleration.

In the context of inclined plane problems, Newton's Second Law helps us determine the forces required to achieve a specific motion. When calculating forces on a slope:
  • We need to consider the direction and magnitude of both the gravitational component and any external forces, like an applied push.
  • For upward acceleration, the applied force needs to counteract gravity's pull down the slope and provide enough force for the desired acceleration.
  • For downward acceleration, the gravitational force is the primary motive force, so the applied force, if present, will assist or resist this motion.
Using this law, we can frame and solve equations to find unknown forces or accelerations. It’s a powerful tool because it helps translate physical observations into mathematical expressions.
Mechanics Problems
Mechanics problems, particularly involving inclined planes, are essential for understanding the application of physics laws to real-world scenarios. These problems help break down complex systems into simpler parts, making them easier to analyze.

Common types of such problems include calculating the force required to move an object up or down an incline, finding the acceleration of an object given a particular force, and determining the impact of angles and mass on motion.

Key steps to solve mechanics problems effectively involve:
  • Drawing a clear free-body diagram to visualize all forces and their directions.
  • Using known equations, such as those from Newton's laws, to set up relationships between forces, mass, and acceleration.
  • Breaking forces into components where necessary, especially on inclined planes.
  • Ignoring or including friction based on the problem’s requirements.
This systematic approach ensures that no force is overlooked and helps identify the crucial factors affecting motion. Practice with these problems enhances understanding and prepares students to tackle more complex real-world physics applications.

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Most popular questions from this chapter

The coefficient of static friction between a box and the flat bed of a truck is \(0.60\). What is the maximum acceleration the truck can have along level ground if the box is not to slide? The box experiences only one \(x\) -directed force, the friction force. When the box is on the verge of slipping, \(F_{\mathrm{f}}=\mu_{s} F_{W}\), where \(F_{W}\) is the weight of the box. As the truck accelerates, the friction force must cause the box to have the same acceleration as the truck: otherwise, the box will slip. When the box is not slipping, \(\sum F_{x}=m a_{x}\) applied to the box gives \(F_{\mathrm{f}}=m a_{x}\) However, if the box is on the verge of slipping, \(F_{\mathrm{f}}=\mu_{s} F_{W}\) so that \(\mu_{s} F_{W}=m a_{x} .\) Because \(F_{W}=m g\), $$ a_{x}=\frac{\mu_{s} m g}{m}=\mu_{s} g=(0.60)\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)=5.9 \mathrm{~m} / \mathrm{s}^{2} $$ as the maximum acceleration without slipping.

A \(5.0-\mathrm{kg}\) block rests on a \(30^{\circ}\) incline. The coefficient of static friction between the block and the incline is \(0.20\). How large a horizontal force must push on the block if the block is to be on the verge of sliding \((a)\) up the incline and \((b)\) down the incline?

A 700-N man stands on a scale on the floor of an elevator. The scale records the force it exerts on whatever is on it. What is the scale reading if the elevator has an acceleration of ( \(a\) ) \(1.8 \mathrm{~m} / \mathrm{s}^{2}\) up? (b) \(1.8 \mathrm{~m} / \mathrm{s}^{2}\) down? (c) \(9.8 \mathrm{~m} / \mathrm{s}^{2}\) down?

A cord passing over a light frictionless pulley has a \(7.0\) -kg mass hanging from one end and a \(9.0-\mathrm{kg}\) mass hanging from the other, as seen in Fig. 3-20. (This arrangement is called Atwood's machine.) Find the acceleration of the masses and the tension in the cord. Because the pulley is easily turned, the tension in the cord will be the same on each side. The forces acting on each of the two masses are drawn in Fig. 3-20. Recall that the weight of an object is \(m g\). It is convenient in situations involving objects connected by cords to take the overall direction of motion of the system as the positive direction. That's often indicated by the direction of motion of the pulley when the system is let free to move. In the present case, the pulley would turn clockwise, and so we take \(u p\) positive for the \(7.0\) -kg mass, and down positive for the \(9.0\) -kg mass. (If we do this, the acceleration will be positive for each mass. Because the cord doesn't stretch, the accelerations are numerically equal.) Writing \(\sum F_{y}=m a_{y}\) for each mass in turn, $$ +\uparrow \sum F_{y A}=F_{T}-(7.0)(9.81) \mathrm{N}=(7.0 \mathrm{~kg})(a) \text { and }+\downarrow \sum F_{y B}=(9.0)(9.81) \mathrm{N}-F_{T}=(9.0 \mathrm{~kg})(a) $$ Add these two equations and the unknown \(F_{T}\) drops out, giving $$ (9.0-7.0)(9.81) \mathrm{N}=(16 \mathrm{~kg})(a) $$ for which \(a=1.23 \mathrm{~m} / \mathrm{s}^{2}\) or \(1.2 \mathrm{~m} / \mathrm{s}^{2}\). Now substitute \(1.23 \mathrm{~m} / \mathrm{s}^{2}\) for \(a\) in either equation and obtain \(F_{T}=77 \mathrm{~N}\).

A car coasting at \(20 \mathrm{~m} / \mathrm{s}\) along a horizontal road has its brakes suddenly applied and eventually comes to rest. What is the shortest distance in which it can be stopped if the friction coefficient between tires and road is \(0.90 ?\) Assume that all four wheels brake identically. If the brakes don't lock, the car stops via static friction. The friction force at one wheel, call it wheel 1 , is $$ F_{\mathrm{fl}}=\mu_{5} F_{N 1}=\mu_{s} F_{W 1} $$ where \(F_{W}\) is the weight carried by wheel 1 . We obtain the total friction force \(F_{\mathrm{f}}\) by adding such terms for all four wheels: $$ F_{\mathrm{f}}=\mu_{s} F_{W 1}+\mu_{s} F_{W_{2}}+\mu_{s} F_{W 3}+\mu_{s} F_{W 4}=\mu_{s}\left(F_{W_{1}}+F_{W 2}+F_{W 3}+F_{W_{4}}\right)=\mu_{s} F_{w} $$ where \(F_{W}\) is the total weight of the car. (Notice that we are assuming optimal braking at each wheel.) This friction force is the only unbalanced force on the car (we neglect air friction). Writing \(F=m a\) for the car with \(F\) replaced by \(-\mu_{s} F_{W}\) gives \(-\mu_{s} F_{W}=m a\), where \(m\) is the car's mass and the positive direction is taken as the direction of motion. However, \(F_{W}=m g\); so the car's acceleration is $$ a=-\frac{\mu_{s} F_{W}}{m}=-\frac{\mu_{s} m g}{m}=-\mu_{s} g=(-0.90)\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)=-8.829 \mathrm{~m} / \mathrm{s}^{2} $$ We can determine how far the car went before stopping by solving the constant- \(a\) motion problem. Knowing that \(v_{i}=20 \mathrm{~m} / \mathrm{s}, v_{f}=0\), and \(a=-8.829 \mathrm{~m} / \mathrm{s}^{2}\), we find from \(v_{f}^{2}-v_{i}^{2}=2 a x\) that $$ x=\frac{(0-400) \mathrm{m}^{2} / \mathrm{s}^{2}}{-17.66 \mathrm{~m} / \mathrm{s}^{2}}=22.65 \mathrm{~m} \quad \text { or } \quad 23 \mathrm{~m} $$ If the four wheels had not all been braking optimally, the stopping distance would have been longer.

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