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Chapter 5: Problem 10

A \(1130-k g\) car is held in place by a light cable on a very smooth (frictionless) ramp, as shown in Fig. \(5.45 .\) The cable makes an angle of \(31.0^{\circ}\) above the surface of the ramp, and the ramp itself rises at \(25.0^{\circ}\) above the borizontal. (a) Draw a free-body diagram for the car. (b) Find the tension in the cable. (c) How hard does the surface of the ramp push on the car? figure can't copy

Short Answer

Expert verified
The tension in the cable is approximately 4721.5 N, and the normal force on the car is roughly 9641.6 N.

Step by step solution

01

Understand the Problem

We are dealing with a car on a frictionless ramp connected by a cable. The cable pulls at an angle to oppose gravitational force, while the ramp also exerts a force on the car.
02

Draw the Free-Body Diagram

Identify all forces acting on the car. These include the gravitational force (\(mg\), directed downward), the normal force (\(N\), perpendicular to the ramp's surface), and the tension in the cable (\(T\), making an angle with the ramp). Utilize trigonometry to resolve forces parallel and perpendicular to the ramp.
03

Resolve Forces Along the Ramp

The forces along the ramp include: component of gravitational force (\(mg\sin(25^{\circ})\)) acting down the ramp, and tension component (\(T\cos(31^{\circ})\)) acting up the ramp. Since the system is in equilibrium, these two forces must be equal: \(mg\sin(25^{\circ}) = T\cos(31^{\circ})\).
04

Solve for Tension in the Cable

Substitute the values: mass of the car \(1130 \text{ kg}\), gravitational acceleration \(9.8 \text{ m/s}^2\), \(\sin(25^{\circ})\), and \(\cos(31^{\circ})\) into the equation: \[1130 \times 9.8 \times \sin(25^{\circ}) = T\times \cos(31^{\circ})\] Rearranging and solving gives: \[T = \frac{1130 \times 9.8 \times \sin(25^{\circ})}{\cos(31^{\circ})}\] Calculate \(T\).
05

Calculate the Normal Force

Resolve forces perpendicular to the ramp: gravitational force component (\(mg\cos(25^{\circ})\)) and tension component (\(T\sin(31^{\circ})\)) act perpendicular to the ramp. The normal force (\(N\)) balances these components: \[N = mg\cos(25^{\circ}) - T\sin(31^{\circ})\] Substitute for \(T\) and calculate \(N\).

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Key Concepts

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

Free-Body Diagrams
A free-body diagram is a fundamental tool in physics that helps visualize the forces acting on an object. When tackling problems involving forces, like a car on a frictionless ramp, drawing a free-body diagram is essential.
The diagram includes:
  • The gravitational force, which always acts downward towards the Earth's center.
  • The normal force, which is perpendicular to the surface. This force is exerted by the ramp on the car.
  • The tension force from the cable, which is angled and opposes the gravitational force pulling the car down the slope.
By accurately drawing a free-body diagram, you can correctly resolve forces into components that are parallel and perpendicular to the inclined surface. This step is crucial, as it lays the groundwork for analyzing each force's role in maintaining equilibrium.
Equilibrium of Forces
Equilibrium occurs when all the forces acting on an object balance each other out, meaning the object stays in place or moves at a constant velocity. In the context of the car on a ramp:
  • The forces parallel to the ramp, such as the component of gravity pulling the car down the ramp, must balance the tension in the cable pulling it up.
  • The forces perpendicular to the ramp鈥攍ike the normal force from the ramp and any perpendicular component of the tension鈥攎ust also balance the gravitational force's contribution in that direction.
By applying these principles of equilibrium, equations can be set up. For instance, to solve for the tension in the cable, you derive the equation based on these balanced forces, ensuring that each component of force is correctly accounted for.
Trigonometry in Physics
Trigonometry plays a pivotal role in solving physics problems involving angles and inclined planes. Here鈥檚 where it comes into use for the car on the ramp:
  • Angles are used to resolve forces into components. For instance, the gravitational force can be split into two: one parallel and one perpendicular to the ramp.
  • Trigonometric functions like sine and cosine help determine these components. The sine function corresponds to the opposite side, while the cosine function corresponds to the adjacent side of an angle in a right triangle.
In this problem:
  • Using \(\sin(25^\circ)\) to find the component of gravitational force pulling the car down the ramp.
  • And \(\cos(31^\circ)\) for the tension component acting upwards along the ramp.
With this understanding, you can confidently resolve forces and analyze their effects based on their direction, ensuring all vector components are accurately treated in equilibrium scenarios.

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Most popular questions from this chapter

A man pushes on a piano with mass 180 \(\mathrm{kg}\) so that it slides at constant velocity down a ramp that is inclined at \(11.0^{\circ}\) above the horizontal floor. Neglect any friction acting on the piano. Calculate the magnitude of the force applied by the man if he pushes (a) parallel to the incline and (b) parallel to the floor.

Stopping Distance. (a) If the coefficient of kinctic friction between tires and dry pavement is 0.80 , what is the shortest distance in which you can stop an automobile by locking the brakes when traveling at 28.7 \(\mathrm{m} / \mathrm{s}\) (about 65 \(\mathrm{mi} / \mathrm{h} ) ?\) (b) On wet pavement the coefficient of kinetic friction may be only \(0.25 .\) How fast should you drive on wet pavement in order to be able to stop in the same distance as in part (a)? (Note: Locking the brakes is not the safest way to stop.)

Maximum Safe Speed. As you travel every day to campus, the road makes a large turn that is approximately an are of a circle. You notice the warning sign at the start of the turm, asking for a maximum speed of 55 \(\mathrm{mi} / \mathrm{h}\) . You also notice that in the curved portion the road is level - that is, not banked at all. On a dry day with very little traffic, you enter the turn at a constant speed of 80 \(\mathrm{mi} / \mathrm{h}\) and feel that the car may skid if you do not slow down quickly. You conclude that your speed is at the limit of safety for this curve and you slow down. However, you remember reading that on dry pavement new tires have an average coefficient of static friction of about 0.76 .while under the worst winter driving conditions, you may encounter wet ice for which the coefficient of static friction can be as low as \(0.20 .\) Wet ice is not unheard of on this road, so you ask yourself whether the speed limit for the turn on the roadside warning sign is for the worst-case scenario. (a) Estimate the radius of the curve from your \(80-\) mi/h experience in the dry turn. (b) Use this estimate to find the maximum speed limit in the turn under the worst wet-ice conditions. How does this compare with the speed limit on the sign? Is the sign misleading drivers?(c) On a rainy day, the coefficient of static friction would be about \(0.37 .\) What is the maximum safe speed of for the turn when the road is wet? Does your answer help you understand the maximum-speed sign?

A small button placed on a horizontal rotating platform with diameter 0.320 \(\mathrm{m}\) will revolve with the platform when it is brought up to a speed of 40.0 rev/min, provided the button is no more than 0.150 \(\mathrm{m}\) from the axis. (a) What is the coefficient of station between the button and the platform? (b) How far from the axis can the button be placed, without slipping, if the platform rotates at 60.0 rev/min?

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 \(\mathrm{g} ?\) (b) What is the apparent weight of the pilot at the lowest point of the pullout?
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