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

  • Thread starter Thread starter owenhigdon
  • Start date Start date
  • Tags Tags
    Incline
AI Thread Summary
To calculate the force a tractor exerts on a trailer on a 16-degree incline, the weight of the trailer (3900 kg) must be considered along with the incline angle. The force required to maintain a steady speed of 3.0 m/s is equal to the component of the trailer's weight acting down the slope, calculated using the formula (W)(sin 16 degrees). Since the trailer is not accelerating, no additional force is needed beyond overcoming this gravitational component. Therefore, the calculation aligns with the correct answer provided in the textbook. Understanding the balance of forces is crucial in this scenario.
owenhigdon
Messages
4
Reaction score
0
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
 
Physics news on Phys.org
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.
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanged mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top