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Chapter 5: Problem 60

A horizontal wire holds a solid uniform ball of mass \(m\) in place on a tilted ramp that rises \(35.0^{\circ}\) above the horizontal. The surface of this ramp is perfectly smooth, and the wire is directed away from the center of the ball (Fig. P5.60). (a) Draw a free-body diagram for the ball. (b) How hard does the surface of the ramp push on the ball? (c) What is the tension in the wire?

Short Answer

Expert verified
The surface pushes with a force of \(mg\cos(35.0^{\circ})\), and the tension is \(mg\sin(35.0^{\circ})\).

Step by step solution

01

Analyze Forces and Free-Body Diagram

Consider the forces acting on the ball. The gravitational force acts vertically downward, the normal force acts perpendicular to the ramp's surface, and the tension in the wire acts horizontally away from the ball. The angle of inclination is given as \(35.0^{\circ}\). In the free-body diagram, label the gravitational force \(mg\), normal force \(N\), and tension \(T\). These forces must balance out since the ball is stationary.
02

Resolve Gravitational Force Components

The gravitational force \(mg\) can be resolved into two components: one parallel to the ramp (\(mg\sin\theta\)) and one perpendicular to the ramp (\(mg\cos\theta\)). Here, \(\theta = 35.0^{\circ}\). The parallel component is balanced by the tension in the wire, and the perpendicular component is balanced by the normal force.
03

Calculate Normal Force

The normal force \(N\) is equal to the perpendicular component of the gravitational force. Therefore, \(N = mg\cos\theta\). Substitute the known values to find \(N\): \[ N = mg\cos(35.0^{\circ}) \]
04

Calculate Tension in the Wire

The tension \(T\) in the wire balances the parallel component of the gravitational force. Hence, \(T = mg\sin\theta\). Using the values given, calculate \(T\):\[ T = mg\sin(35.0^{\circ}) \]

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Key Concepts

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

Free-body diagram
A free-body diagram is an essential tool in physics for visualizing the forces acting on an object. In the scenario of a ball held in place on a ramp, it helps identify how different forces interact. Start by imagining the ball as a single point to simplify the process.
  • The gravitational force, depicted as an arrow pointing downward, represents the weight of the ball, which is denoted by \( mg \).
  • The normal force is perpendicular to the ramp and is illustrated by an arrow pointing away from the ramp's surface.
  • Tension in the wire appears as a horizontal arrow pointing away from the ball, showing how the wire pulls the ball.
Taking these steps to do a free-body diagram ensures clarity, especially when analyzing forces on objects in complex situations like inclined planes.
Normal force calculation
Normal force is a key concept in physics, especially when dealing with objects on inclines. It is the force exerted by a surface to support the weight of an object resting on it. In this exercise, the ramp provides the normal force to the ball.To calculate the normal force, we focus on the component of gravitational force acting perpendicular to the ramp's surface. This component is given by \( mg\cos\theta \), where \( \theta \) is the angle of the incline (\( 35.0^{\circ} \) in this case).Using this equation, the normal force \( N \) can be calculated as:\[ N = mg\cos(35.0^{\circ}) \]This shows how the force from the ramp balances out the perpendicular gravitational component and helps keep the ball stationary.
Tension in the wire
Tension in a wire is a force that pulls an object and is directed along the wire's length. In this scenario, the tension maintains the ball's position on the ramp.To compute the tension, consider the gravitational force component parallel to the ramp's surface. This component is \( mg\sin\theta \), where \( \theta \) is the ramp's angle.The tension in the wire must balance this parallel gravitational component to keep the ball from sliding down. Thus, tension \( T \) is calculated as:\[ T = mg\sin(35.0^{\circ}) \]By understanding how tension relates to other forces, you can solve similar problems involving inclined planes and cables.
Forces on an inclined plane
When an object rests or moves on an inclined plane, several forces come into play, determining its behavior. These forces can be broken down and analyzed to understand their effects.
  • Gravitational Force: This always acts downwards but can be resolved into two components relative to the incline: parallel \( mg\sin\theta \) and perpendicular \( mg\cos\theta \). This helps in calculating other forces like tension and normal force.
  • Normal Force: Perpendicular to the inclined surface, counteracting the perpendicular gravitational force.
  • Friction (if applicable): Although friction is absent in this problem due to the smooth surface, it often acts parallel to the surface, opposing motion.
  • Tension: If a wire or rope is involved, it acts along the wire's length, opposing motion parallel to the plane.
Understanding these forces allows for predicting an object's motion and solving related physics problems efficiently.

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Most popular questions from this chapter

You are standing on a bathroom scale in an elevator in a tall building. Your mass is 64 \(\mathrm{kg}\) . The elevator starts from rest and travels upward with a speed that varies with time according to \(v(t)=\left(3.0 \mathrm{m} / \mathrm{s}^{2}\right) t+\left(0.20 \mathrm{m} / \mathrm{s}^{3}\right) t^{2} .\) When \(t=4.0 \mathrm{s},\) what is the reading of the bathroom scale?

A 50.0 -kg stunt pilot who has been diving her airplane vertically pulls out of the dive by changing her course to a circle in a vertical plane. (a) If the plane's speed at the lowest point of the circle is \(95.0 \mathrm{m} / \mathrm{s},\) what is the minimum radius of the circle for the acceleration at this point not to exceed 4.00\(g ?\) (b) What is the apparent weight of the pilot at the lowest point of the pullout?

A machine part consists of a thin 40.0 -cm-long bar with small 1.15 -kg masses fastened by screws to its ends. The screws can support a maximum force of 75.0 \(\mathrm{N}\) without pulling out. This bar rotates about an axis perpendicular to it at its center. (a) As the bar is turning at a constant rate on a horizontal, frictionless surface, what is the maximum speed the masses can have without pulling out the screws? (b) Suppose the machine is redesigned so that the bar turns at a constant rate in a vertical circle. Will one of the screws be more likely to pull out when the mass is at the top of the circle or at the bottom? Use a free-body diagram to see why. (c) Using the result of part (b), what is the greatest speed the masses can have without pulling a screw?

You are riding in an elevator on the way to the 18 th floor of your dormitory. The elevator is accelerating upward with \(a=1.90 \mathrm{m} / \mathrm{s}^{2} .\) Beside you is the box containing your new computer; the box and its contents have a total mass of 28.0 \(\mathrm{kg} .\) While the elevator is accelerating upward, you push horizontally on the box to slide it at constant speed toward the elevator door. If the coefficient of kinetic friction between the box and the elevator floor is \(\mu_{\mathrm{k}}=0.32,\) what magnitude of force must you apply?

A block with mass \(M\) is attached to the lower end of a vertical, uniform rope with mass \(m\) and length \(L\) A constant upward force \(\vec{\boldsymbol{F}}\) is applied to the top of the rope, causing the rope and block to accelerate upward. Find the tension in the rope at a distance \(x\) from the top end of the rope, where \(x\) can have any value from 0 to \(L .\)
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