Problem 42 - College Physics

Question

A \(3000-\mathrm{kg}\) truck tows a \(1500-\mathrm{kg}\) car by a chain. If the net forward force on the truck by the ground is \(3200 \mathrm{~N}\), (a) what is the acceleration of the car, and (b) what is the tension in the connecting chain?

Step-by-Step Solution

Answer

a) The acceleration is approximately 0.711 m/s². b) The tension in the chain is 1066.5 N.

Step 1: Calculate Total Mass

To find the acceleration, we need to calculate the total mass of the system, which includes both the truck and the car.The total mass is:\[ M = m_{ ext{truck}} + m_{ ext{car}} = 3000 ext{ kg} + 1500 ext{ kg} = 4500 ext{ kg} \]

Step 2: Find the Acceleration

We use Newton's second law to find the acceleration of the entire system. The equation is given by:\[ F_{ ext{net}} = M imes a \]where \( F_{ ext{net}} = 3200 ext{ N} \) and \( M = 4500 ext{ kg} \). Solving for \( a \):\[ a = \frac{F_{ ext{net}}}{M} = \frac{3200 ext{ N}}{4500 ext{ kg}} \approx 0.711 ext{ m/s}^2 \]This acceleration applies to both the truck and the car since they are connected.

Step 3: Determine the Tension in the Chain

To find the tension in the chain, consider only the car, which is being accelerated by the tension in the chain. Apply Newton's second law to the car:\[ F_{ ext{tension}} = m_{ ext{car}} imes a \]Here, \( m_{ ext{car}} = 1500 ext{ kg} \) and \( a = 0.711 ext{ m/s}^2 \):\[ F_{ ext{tension}} = 1500 ext{ kg} imes 0.711 ext{ m/s}^2 = 1066.5 ext{ N} \]

Key Concepts

Net Force

Understanding the concept of net force is crucial in solving problems involving motion. The term "net force" refers to the total force acting on an object when all the individual forces are combined. In the context of Newton's Second Law, the net force is what causes an object to accelerate.

For our truck and car scenario, the net force is the force exerted by the ground on the truck, which is primarily responsible for moving both the truck and car forward. This force is given as 3200 N in the exercise.

It’s important to remember:

• Net force is a vector, so it has both magnitude and direction.
• If multiple forces are acting on an object, they should be added vectorially to find the net force.
• An object will accelerate in the direction of the net force.

In our exercise, since the net force is responsible for the entire system's acceleration, it defines how the truck and car move together. This force must overcome any resistive forces acting against the truck and car, such as friction.

Acceleration

Acceleration is the rate at which an object changes its velocity. Newton's Second Law allows us to calculate acceleration if the net force and mass are known. The law is expressed in the formula:\[ F_{\text{net}} = M \times a \]
Here, \( a \) is the acceleration, \( F_{\text{net}} \) is the net force, and \( M \) is the total mass being accelerated.

In the exercise, the net force applied is 3200 N, and the total mass of the truck and car is 4500 kg. To find the acceleration, rearrange Newton's Second Law to solve for \( a \):\[ a = \frac{F_{\text{net}}}{M} \]
Substitute the values:\[ a = \frac{3200 \text{ N}}{4500 \text{ kg}} \approx 0.711 \text{ m/s}^2 \]

This means both the truck and car accelerate at \( 0.711 \text{ m/s}^2 \). This value is uniform for the whole system as they are connected. Understanding acceleration helps us see how forces change an object's motion over time.

Tension in Chain

The concept of tension is essential in mechanics, especially when dealing with ropes or chains connecting objects. In this context, the tension in the chain is the force transmitted through the chain that accelerates the car.

To find the tension, apply Newton’s Second Law specifically to the car. The car is accelerated by both the truck's movement and the tension in the chain. When considering only the car:\[ F_{\text{tension}} = m_{\text{car}} \times a \]
Where \( m_{\text{car}} = 1500 \text{ kg} \) and the acceleration \( a = 0.711 \text{ m/s}^2 \).
Plug in the values to find:\[ F_{\text{tension}} = 1500 \text{ kg} \times 0.711 \text{ m/s}^2 = 1066.5 \text{ N} \]

This tension is the force responsible for pulling the car forward. It's important to remember that tension is uniform along the chain's length, assuming the chain is massless and there's no friction.