How Is the Resultant Force on a Car Calculated?

[anteaters]

Homework Statement

A 1040 kg car is pulling a 300 kg trailer. Together the car and trailer move forward with an acceleration of 2.11 m/s2. Ignore any frictional force of air drag on the car and all frictional forces on the trailer. What is the resultant force exerted by the car on the road and in what direction (measured from the left of vertically downwards)?

Homework Equations

F = ma

The Attempt at a Solution

I already found the magnitude by making a free body diagram with the acceleration of the car going on the negative x axis, and the weight of the car on the negative y axis. then, i squared the values of the force in the y direction and x direction, added them together and took the square root of that, finding the force to be 10425. so now i tried doing the sin^-1 of x component/y component and got theta to be 12.15 degrees.

i attached a paint version of my free body diagram. any help would be greatly appreciated.

 

[Doc Al]

Recalculate the horizontal component (the friction force). Realize that the friction force must accelerate the entire system of both car and trailer.

 

[thomate1]

I would like to point out that the calculation for the resultant force exerted by the car on the road is correct. However, it is important to also consider the direction of the force, which is measured from the left of vertically downwards. This means that the angle calculated, 12.15 degrees, should be subtracted from 90 degrees to get the correct direction, which is 77.85 degrees from the left of vertically downwards. Additionally, it is important to note that the force is exerted by the car on the road, not the other way around. This means that the force is directed downwards and to the left, as shown in the free body diagram.

 

[baba89]

I would like to commend your approach in using a free body diagram and applying the equation F=ma to solve for the resultant force exerted by the car on the road. Your method is correct and your solution of 10425 N is also correct.

However, I would like to point out that the direction of the resultant force can be represented by the angle θ in your diagram, but it is not necessary to use the sine function to calculate it. The angle can also be determined by using the inverse tangent function (tan^-1) with the x and y components of the force. In this case, the angle would be 2.11 m/s^2 divided by 10425 N, which results in approximately 0.12 degrees. This is a very small angle and indicates that the resultant force is almost entirely in the horizontal direction, with a very small vertical component.

In addition, I would like to mention that the force exerted by the car on the road is the same as the force exerted by the road on the car, as stated in Newton's third law of motion. Therefore, the resultant force in this case is also the force exerted by the road on the car, and it is directed in the opposite direction of the car's motion.

I hope this helps clarify your solution and understanding of the concept of forces in this scenario. Keep up the good work in using scientific methods to solve problems!