How Do You Calculate the Force a Tractor Exerts on a Trailer on an Incline?

[owenhigdon]

The problem states that a 3900kg trailer is pulled by a tractor up a 16 degree incline with a steady speed of 3.0m\s.

I am asked what force the tractor exerts on the trailer.

I said that the force was equal to the (W)(sin)16degrees, i get the right answer in the back of the book but i think its a coincidence...any help on this one would be much appreciated

 

[cimon9999]

me again! Your answer is right. Because you're not accelerating the trailer up the incline, there is no additional force required. All you need to do to maintain the speed of 3ms-1 is to overcome the backward force down the slope.

 

[thomate1]

Your calculation is correct. The force exerted by the tractor on the trailer can be calculated using the formula F = mgsinθ, where m is the mass of the trailer, g is the acceleration due to gravity, and θ is the angle of the incline. In this case, m = 3900 kg, g = 9.8 m/s^2, and θ = 16 degrees. Plugging these values into the formula, we get F = (3900 kg)(9.8 m/s^2)(sin 16 degrees) = 1044.6 N. This is the force exerted by the tractor on the trailer to maintain a steady speed of 3.0 m/s up the incline. It is not a coincidence that this is the same as the answer in the back of the book.