Problem 13 - Physics for Scientists and Engineers with Modern Physics

Question

(II) A 1280 -kg car pulls a 350 -kg trailer. The car exerts a horizontal force of \(3.6 \times 10^{3} \mathrm{~N}\) against the ground in order to accelerate. What force does the car exert on the trailer? Assume an effective friction coefficient of 0.15 for the trailer.

Step-by-Step Solution

Answer

The car exerts approximately 1180.05 N on the trailer.

Step 1: Identify the Systems

First, we need to identify the systems involved in the problem. There are two main systems: the car and the trailer. The car and trailer are connected, and the force exerted by the car helps to accelerate both.

Step 2: Calculate the Total Mass

The total mass of the system is the sum of the mass of the car and the trailer. Therefore, \( \text{Total Mass} = 1280 \text{ kg} + 350 \text{ kg} = 1630 \text{ kg} \).

Step 3: Determine Total Force Required to Overcome Friction

For the trailer, the force of friction can be calculated using the formula \( F_{friction} = \mu \times m_{trailer} \times g \), where \( \mu = 0.15 \) is the coefficient of friction, \( m_{trailer} = 350 \text{ kg} \) is the mass of the trailer, and \( g = 9.8 \text{ m/s}^2 \) is the acceleration due to gravity.\[ F_{friction} = 0.15 \times 350 \text{ kg} \times 9.8 \text{ m/s}^2 = 514.5 \text{ N} \]

Step 4: Calculate the Acceleration of the System

The force exerted by the car on the ground is used for accelerating the whole system. The net force is the applied force minus the force of friction. Therefore, \( F_{net} = 3.6 \times 10^3 \text{ N} - 514.5 \text{ N} = 3085.5 \text{ N} \). Using Newton's Second Law, \( F_{net} = (m_{car} + m_{trailer}) \times a \), we solve for acceleration \( a \).\[ 3085.5 \text{ N} = 1630 \text{ kg} \times a \] \[ a = \frac{3085.5 \text{ N}}{1630 \text{ kg}} \approx 1.893 \text{ m/s}^2 \]

Step 5: Calculate the Force on the Trailer

Now, we find the force that the car exerts on the trailer using the calculated acceleration. For the trailer, the force exerted on it by the car must overcome the friction and provide the acceleration. Using \( F = m \times a \):\[ F_{trailer} = m_{trailer} \times a + F_{friction} \] \[ F_{trailer} = 350 \text{ kg} \times 1.893 \text{ m/s}^2 + 514.5 \text{ N} = 665.55 \text{ N} + 514.5 \text{ N} \approx 1180.05 \text{ N} \]

Step 6: Finalize the Result

The final force exerted by the car on the trailer is approximately \( 1180.05 \text{ N} \).

Key Concepts

Force of Friction

When an object moves across a surface, friction acts as a force opposing this movement. This isn’t just inconvenient, it’s an important factor in calculating movements and forces in physics problems.
The force of friction is the product of the friction coefficient ( \( \mu \) ) and the normal force ( \( F_{normal} \) ).
In this scenario with the car and trailer, it is the force of friction that the trailer's wheels experience from the ground. The force of friction is calculated as \( F_{friction} = \mu \times m \times g \).
Here:

• \( \mu = 0.15 \), which represents the friction coefficient for the trailer.
• \( m = 350 \text{ kg} \) is the mass of the trailer.
• \( g = 9.8 \text{ m/s}^2 \) is the acceleration due to Earth's gravity.

After calculation, it results in \( 514.5 \text{ N} \), acting against the direction of movement. This force must be overcome by the car’s force to initiate or sustain movement.

Net Force

Newton’s second law gives insight into motion, where net force is responsible for any acceleration. The net force is the sum of all forces acting on an object.
In our exercise, the car exerts a force of \( 3.6 \times 10^3 \text{ N} \) to move itself and the trailer.
To find the net force acting on the system, we subtract the force of friction from the car's force:

• \( F_{net} = 3.6 \times 10^3 \text{ N} - 514.5 \text{ N} \)

This yields \( 3085.5 \text{ N} \) of net force.
The net force is what moves the whole system (car + trailer) and allows them to accelerate. The concept of net force is crucial here, as it establishes the relationship between force, mass, and acceleration according to Newton's Second Law \( F = m \times a \).
Understanding net force helps explain how much force remains available to propel the system forward after accounting for friction losses.

Acceleration

Acceleration is a measure of how quickly an object changes its speed. According to Newton's Second Law, the acceleration of an object is determined by the net force acting on it and its mass.
Our problem involves calculating the system's acceleration using the equation \( a = \frac{F_{net}}{m_{total}} \).
We now have:

• The net force: \( 3085.5 \text{ N} \).
• The total mass: \( 1630 \text{ kg} \), combining both car and trailer.

When applying these values, the acceleration is calculated as:\[a = \frac{3085.5 \text{ N}}{1630 \text{ kg}} \approx 1.893 \text{ m/s}^2\]This acceleration signifies how quickly the car-trailer system gains speed. It is important in predicting how long it will take to reach certain speeds or how forceful the vehicle’s movements will feel. This fundamental principle not only applies to cars but to any scenario involving moving connected masses under various forces.