How Do Trailer Hitches Handle Force in a Moving Tractor-Trailer System?

[aalsaigh]

Homework Porblem -- Forces

Homework Statement

A tractor with two trailers moves at a constant speed on a straight level road. The mass of the tractor is 5000 kg and the mass of each trailer is 25 000 kg.
a) Determine the force exerted by each of the two trailer hitches.
b) Repeat, if the tractor-trailer is acceleration at 2 m/s^2
c) Repeat, if the coefficient of friction is 0.7

Homework Equations

F friction = (coeff of friction)(F normal)
F net = ma
F g = mg

The Attempt at a Solution

a) 0
b) a = 2 m/s^2. F = ma
First one = ma = (50 000)(2) = 100 000 N
Second = ma = (25 000)(2) = 50 000 N
c) I don't know how to go about this, but I did find F g for each trailer.
last trailer: Fg = mg = 245 000
Second one: Fg = 245 000
F friction = (0.7)(245 000) = 171 500

That's as far as I got.

 

[anathema]

Thank you for your question. I can provide you with some guidance on how to approach the problem and some possible solutions.

First, let's review the concepts involved in this problem. The force exerted by an object can be calculated using the equation F = ma, where F is the force, m is the mass, and a is the acceleration. This equation is known as Newton's second law of motion. Additionally, the force of friction can be calculated using the equation F = μN, where μ is the coefficient of friction and N is the normal force.

Now, let's tackle each part of the problem individually.

a) In order to determine the force exerted by each trailer hitch, we need to first understand the forces acting on each trailer. Since the tractor and trailers are moving at a constant speed, we can assume that there is no acceleration, and therefore no net force acting on the system. This means that the forces exerted by the trailer hitches must be equal in magnitude and opposite in direction to the forces of friction acting on each trailer. In other words, the force exerted by each trailer hitch is equal to the force of friction acting on that trailer.

Using the equation F = μN, we can calculate the force of friction for each trailer. The normal force, N, is equal to the weight of the trailer, which can be found using the equation Fg = mg, where Fg is the force of gravity and m is the mass. Plugging in the given values, we get:

Fg = (25,000 kg)(9.8 m/s^2) = 245,000 N

For the first trailer, μ = 0.7, so the force of friction is:

Ff = (0.7)(245,000 N) = 171,500 N

For the second trailer, μ = 0.7, so the force of friction is also:

Ff = (0.7)(245,000 N) = 171,500 N

Therefore, the force exerted by each trailer hitch is 171,500 N.

b) In this part of the problem, the tractor-trailer system is accelerating at a rate of 2 m/s^2. This means that there is an unbalanced force acting on the system, causing it to accelerate. To find the force exerted by each trailer hitch, we can use the equation F

 

[Moetasim]

I would suggest the following response to the content provided:

Firstly, it is important to note that the force exerted by the trailer hitches is dependent on the weight of the trailers and the acceleration of the tractor-trailer system. In order to accurately determine the force exerted by each of the two trailer hitches, we must first calculate the total weight of the system. This can be done by adding the mass of the tractor (5000 kg) to the mass of each trailer (25 000 kg x 2 = 50 000 kg), giving us a total weight of 55 000 kg.

a) For a system moving at a constant speed, the net force acting on the system is zero. Therefore, the force exerted by each trailer hitch would be zero.

b) If the tractor-trailer is accelerating at 2 m/s^2, we can use the equation F = ma to calculate the force exerted by each trailer hitch. The total mass of the system is still 55 000 kg, and the acceleration is 2 m/s^2. Thus, the force exerted by the first trailer hitch would be 110 000 N (55 000 kg x 2 m/s^2), and the force exerted by the second trailer hitch would be 55 000 N (25 000 kg x 2 m/s^2).

c) In order to calculate the force exerted by each trailer hitch when the coefficient of friction is 0.7, we must consider the force of friction acting on the system. From the given information, we can calculate the weight of each trailer (25 000 kg x 9.8 m/s^2 = 245 000 N). Therefore, the total weight of the system is 495 000 N (245 000 N x 2). Using the equation F friction = μF normal, we can calculate the force of friction to be 346 500 N (0.7 x 495 000 N). This force of friction is acting in the opposite direction to the motion of the system, and thus must be subtracted from the net force. This results in a net force of 148 500 N acting on the system. Dividing this net force by the total mass of the system (55 000 kg) gives us an acceleration of 2.7 m/s^2. Using the equation F = ma