- #1

Joshuava

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## Homework Statement

A 1480(k)g car pulls a 300(k)g trailer. The car exerts a horizontal force of 3900(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.

## Homework Equations

My question really is how to solve this problem correctly, but more so, correcting the mistakes I have made in my attempt in gathering a correct solution. I will just put all my work that way anyone who looks at this can easily follow my path and be able to correct my work.

## The Attempt at a Solution

Normal force on car = 1480(kg) * 9.81(m/s^2) = 14518.8(N)

Friction on car = 14518.8(N) * 0.15 = 2177.82(N)

Normal force on trailer = 300(kg) * 9.81(m/s^2) = 2943(N)

Friction on trailer = 2943(N) * 0.15 = 441.45(N)

3900(N) in my problem will be working in the positive direction as the other forces are resisting movement due to friction:

F = 3900(N) - 2177.82(N) - 441.45(N) => 1280.73(N)

Now, using Newton's 2nd law of F = m * a, I have rearranged to solve for acceleration:

a = F / m

a = 1280.73(N) / [1480(kg) + 300(kg)] =>0.7195(m/s^2)

F(trailer)=m(trailer) * a(trailer) + any other additional forces acting on the trailer?

F(trailer) = 300(kg) * 0.7195(m/s^2) + 441.45(N) => 657.3(N)

660(N) is my solution.

Would anyone be kind enough to help correct my mistakes? I appreciate you taking the time to take a look!

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