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A 20.0-g ball hangs from the roof of a freight car by a string. When the freight car begins to move, the string makes an angle of \(35.0^{\circ}\) with the vertical. (a) What is the acceleration of the freight car? (b) What is the tension in the string?

Short Answer

Expert verified
In conclusion: (a) The acceleration of the freight car is approximately \(6.87 m/s^2\). (b) The tension in the string is approximately \(0.2403 N\).

Step by step solution

01

Identify the Forces Acting on the Ball

There are two forces acting on the ball: the gravitational force exerted by the Earth, which pulls it downwards (weight), and the tension in the string. Weight of the ball (W) can be calculated as: \[ W = mg\] Where m is the mass of the ball and g is the acceleration due to gravity, which is approximately \(9.81 m/s^2\).
02

Resolve the Forces in Horizontal and Vertical Directions

Since we are given that the string makes an angle of \(35^\circ\) with the vertical, we can now resolve the forces in horizontal (x-axis) and vertical (y-axis) directions using trigonometry. From the given information, we know that the force equations can be given in the form: \(\Longrightarrow T_x = T \sin{(35^\circ)}\) (horizontal direction) \(\Longrightarrow T_y = T \cos{(35^\circ)}\) (vertical direction)
03

Calculate the Weight of the Ball

Now, calculate the weight of the ball using the given mass (20.0 g) and the acceleration due to gravity (9.81 m/s²). Convert the mass from grams to kilograms: \(m = \frac{20.0 g}{1000 g/kg} = 0.02 kg\) Calculate the weight of the ball: \(W = (0.02 kg)(9.81 m/s^2) = 0.1962 N\)
04

Calculate the Acceleration of the Freight Car

As the car is moving horizontally and there is no net force acting in the vertical direction, we can equate the vertical force to the weight of the ball. \(T_y = W\) Now, substituting the weight of the ball and the value of angle into the equation we derived in Step 2, we can find the tension T. \(T \cos{(35^\circ)} = 0.1962 N\) Solving for T, we get: \(T = \frac{0.1962 N}{\cos{(35^\circ)}} \approx 0.2403 N\) Now, using the horizontal component of T (T_x) and making use of Newton's second law, \(F_x = m \cdot a\), we can compute the acceleration (a) of the freight car. \(T_x = 0.2403 N \cdot \sin{(35^\circ)} \approx 0.1374 N\) \(a = \frac{F_x}{m} = \frac{0.1374 N}{0.02 kg} \approx 6.87 m/s^2\) The acceleration of the freight car is approximately \(6.87 m/s^2\).
05

Calculate the Tension in the String

We already calculated the tension in the string in step 4. Therefore, the tension in the string is approximately \(0.2403 N\). In conclusion: (a) The acceleration of the freight car is approximately \(6.87 m/s^2\). (b) The tension in the string is approximately \(0.2403 N\).

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Key Concepts

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

Understanding Newton's Second Law
When studying the motion of objects, Newton's second law of motion provides a fundamental framework. It states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. Expressed mathematically, the law is given by:
\[ F = ma \]
where \( F \) is the net force applied, \( m \) is the mass of the object, and \( a \) is the acceleration produced. In our freight car physics problem, we apply Newton's second law to determine the acceleration of the freight car. It's important to remember that this law applies only when the forces are unbalanced. If all forces are balanced, the acceleration is zero, and the object either remains at rest or continues to move at a constant velocity.
To fully grasp how the law applies, visualize the freight car being pulled with a force that causes the string to tilt. The tilt indicates a force that is not vertical, and the horizontal component of that force is what accelerates the freight car. The greater the net force (in this case, the horizontal component), the greater the acceleration for a given mass.
Deciphering Tension in a String
The concept of tension is crucial when dealing with forces in strings and ropes. Tension can be thought of as a pulling force transmitted through a string, cable, or rope. It arises whenever a string is pulled by forces acting from opposite ends. In the realm of physics, tension is an example of a 'contact force', which means it only exists when objects are physically connected.
In our example with the freight car and the hanging ball, the string exerts a force on the ball to keep it suspended, which we refer to as the tension force. Tension has a magnitude, which is the strength of the pulling force, and a direction, which is always along the string and away from the object exerting the pull. In our problem, understanding tension helps us calculate both the acceleration of the freight car and the specific tension value in the string at the given angle. This concept is vital because it’s the tension that’s providing the necessary force to move the freight car forward.
Applying Trigonometric Force Resolution
Trigonometric force resolution plays a pivotal role in breaking down forces into their vertical and horizontal components. This method is imbued in the study of physics to simplify analyses of forces acting at an angle. By using basic trigonometric functions like sine and cosine, one can resolve a single force acting at a certain angle into two perpendicular components: one along the horizontal (x-axis) and one along the vertical axis (y-axis).

Force Resolution in Practice

In our freight car scenario, we use trigonometric force resolution to separate the tension force into horizontal and vertical components. The mathematical expressions are given by:
\[ T_x = T \sin(\theta) \]\[ T_y = T \cos(\theta) \]
where \( T_x \) and \( T_y \) represent the horizontal and vertical components of the tension \( T \), respectively, and \( \theta \) is the angle of the string with the vertical. This breakdown assists in calculating the specific force causing the acceleration of the freight car—only the horizontal component is responsible since the vertical component is balanced by the weight of the ball. Understanding this concept is essential for quantitatively analyzing situations where forces are exerted at angles, not just in theoretical problems but also in practical engineering applications.

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Most popular questions from this chapter

A 30.0-kg girl in a swing is pushed to one side and held at rest by a horizontal force \(\overrightarrow{\mathbf{F}}\) so that the swing ropes are \(30.0^{\circ}\) with respect to the vertical. (a) Calculate the tension in each of the two ropes supporting the swing under these conditions. (b) Calculate the magnitude of \(\overrightarrow{\mathbf{F}}\)

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