A 11301130-kg car is held in place by a light cable on a very smooth (frictionless) ramp (Fig. E5.85.8). The cable makes an angle of 31.0°31.0° above the surface of the ramp, and the ramp itself rises at 25.0°25.0° above the horizontal. Find the tension in the cable.

Ch 05: Applying Newton's Laws

Young & Freedman Calc - University Physics 14th Edition

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Young & Freedman Calc 14th Edition

Ch 05: Applying Newton's Laws

Problem 8b

Chapter 5, Problem 8b

A $1130$ -kg car is held in place by a light cable on a very smooth (frictionless) ramp (Fig. E $5.8$ ). The cable makes an angle of $31.0 degrees$ above the surface of the ramp, and the ramp itself rises at $25.0 degrees$ above the horizontal. Find the tension in the cable.

Verified step by step guidance

Step 1: Identify the forces acting on the car. The forces include the gravitational force (weight of the car), the tension in the cable, and the normal force exerted by the ramp. The weight of the car can be calculated using the formula:

F_{g} = m g

, where

m

is the mass of the car and

g

is the acceleration due to gravity.

Step 2: Break the forces into components along the ramp and perpendicular to the ramp. The weight of the car has a component parallel to the ramp given by

F_{g} = m g \cdot θ

, where

θ

is the angle of the ramp (25.0°). The perpendicular component is

F_{⊥} = m g \cdot θ

Step 3: Analyze the tension in the cable. The cable makes an angle of 31.0° above the ramp. The tension force has two components: one parallel to the ramp and one perpendicular to the ramp. The parallel component is

T_{∥} = T \cdot 31.0°

, and the perpendicular component is

T_{⊥} = T \cdot 31.0°

Step 4: Apply Newton's second law along the ramp. Since the car is stationary, the net force along the ramp is zero. This means the tension's parallel component balances the weight's parallel component:

T_{∥} = F_{g}

. Substitute the expressions for these components to solve for the tension.

Step 5: Solve for the tension in the cable. Rearrange the equation to isolate

T

T = F_{g} 31.0°

. Substitute the values for

m

g

, and the angles to find the tension.

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Key Concepts

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

Forces on an Inclined Plane

When an object is on an inclined plane, the forces acting on it include gravitational force, normal force, and any tension or frictional forces. The gravitational force can be resolved into two components: one parallel to the incline, which causes the object to slide down, and one perpendicular to the incline, which is balanced by the normal force. Understanding these forces is crucial for analyzing the motion and equilibrium of the object.

Tension in a Cable

Tension is the force transmitted through a cable or rope when it is pulled tight by forces acting from opposite ends. In this scenario, the tension in the cable must counteract the component of the gravitational force acting down the ramp, as well as any other forces acting on the car. Calculating the tension involves using trigonometric relationships based on the angles given in the problem.

Trigonometric Functions

Trigonometric functions, such as sine, cosine, and tangent, relate the angles of a triangle to the ratios of its sides. In physics problems involving angles, these functions are essential for resolving forces into their components. For example, the sine function can be used to find the component of the gravitational force acting parallel to the incline, while the cosine function can help determine the normal force acting perpendicular to the incline.

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Textbook Question

A man pushes on a piano with mass $180$ kg; it slides at constant velocity down a ramp that is inclined at $19.0°$ above the horizontal floor. Neglect any friction acting on the piano. Calculate the magnitude of the force applied by the man if he pushes parallel to the floor.

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Textbook Question

A $1130$ -kg car is held in place by a light cable on a very smooth (frictionless) ramp (Fig. E $5.8$ ). The cable makes an angle of $31.0°$ above the surface of the ramp, and the ramp itself rises at $25.0°$ above the horizontal. How hard does the surface of the ramp push on the car?

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