Force exerted from car on trailer question

In summary, a 1480kg car pulling a 300kg trailer exerts a horizontal force of 3900N. Taking into account a friction coefficient of 0.15 for the trailer, the force exerted by the car on the trailer is approximately 1024N. This can be determined by setting up free-body diagrams for both the car and trailer and solving for the force of tension (T) using Newton's second and third laws.
  • #1
Joshuava
6
0

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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  • #2
It is said in the problem statement that 0.15 is for the trailer only so does that mean 3900N is a net force so kinetic friction on the car can be neglected?
 
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  • #3
This was copied directly from masteringphysics so I really can't answer that question...I just attempted to solve it as best I could, but my answer is not what it is looking for.

That would change the answer significantly.

The new F would be 3458.55(N)

acceleration would change to 1.94(m/s^2)

The new F(trailer) would be 300(kg) * 1.94(m/s^2) + 441.45 => 1024.35(N)?
 
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  • #4
Joshuava said:
This was copied directly from masteringphysics so I really can't answer that question...I just attempted to solve it as best I could, but my answer is not what it is looking for.
What is it looking for, then?
 
  • #5
lep11 said:
What is it looking for, then?

I believe the question is looking for the amount of force that is exerted onto the trailer by the car. It wants the answer in 2 significant digits, and with my experience thus far, I don't imagine an answer would be 1.0E3(N)
 
  • #6
Joshuava said:
I believe the question is looking for the amount of force that is exerted onto the trailer by the car.
Yes, but do you know what the answer should be? I got T≈1024N
 
  • #7
Draw free-body diagrams for the car and trailer

Newton second
for Car F-T=mcar*a
for Trailer T-Fμ=mtrailer*a
they have equal acceleration a and T is the force the car exerts on the trailer or vice versa(Newton III) You don't even have to calculate a, just solve for T and substitute it into car equation.
 
  • #8
Awesome, now I understand it all. Thanks so much for your help! By the way, I put in 1000 for the solution and it accepted it, even though it was asking for 2 sig figs.
 

1. What is the force exerted from a car on a trailer?

The force exerted from a car on a trailer is known as the tongue weight. It is the downward force that the trailer's tongue or hitch exerts on the car's hitch receiver. This force is important for maintaining stability and balance while towing.

2. How is the force exerted from a car on a trailer calculated?

The force exerted from a car on a trailer can be calculated by multiplying the total weight of the trailer by the tongue weight percentage. The tongue weight percentage is typically between 10-15% for conventional trailers and 20-25% for gooseneck or fifth-wheel trailers.

3. What factors can affect the force exerted from a car on a trailer?

The force exerted from a car on a trailer can be affected by several factors, including the weight of the trailer, the weight distribution of the load on the trailer, the type of trailer hitch, and the condition of the vehicle's suspension and tires.

4. Why is it important to properly distribute the force exerted from a car on a trailer?

Proper distribution of the force exerted from a car on a trailer is crucial for safe towing. An unbalanced or overloaded trailer can cause swaying, reduced steering and braking control, and potential accidents. It can also put excess strain on the vehicle's suspension and tires.

5. How can the force exerted from a car on a trailer be adjusted?

The force exerted from a car on a trailer can be adjusted by redistributing the weight of the load on the trailer, adjusting the trailer's tongue or hitch, or using a weight distribution system. It is important to consult the vehicle's manual and follow manufacturer's recommendations for proper weight distribution and adjusting the force exerted on the trailer.

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