Find the tension in the cable and the force the ramp exerts on the tires

In summary, the conversation discusses the use of a rotated coordinate system to solve for the tension in a cable and the force exerted by a ramp on a car. The weight of the car is resolved into its x and y components, and the sum of forces in each direction must be zero for equilibrium. The normal force is equal to the weight of the car, and the tension force is the horizontal component of the weight. The conversation also considers the scenario of no tension and the role of friction in keeping the car on the incline.
  • #1
mirs08

Homework Statement


A 5000 kg car rests on a 32° slanted ramp attached to a trailer. Only a cable running from the trailer to the car prevents the car from rolling off the ramp (the car's transmission is in neutral and its brakes are off). Find the tension in the cable and the force that the ramp exerts on the tires

**please keep in mind that I have been using Ft instead of T as indicated in the picture as our prof encourages us to do so, so we don't get confused between T for temperature and T for tension in future problems

upload_2017-10-20_12-17-19.png

Homework Equations


Fgx = mgsinΘ
Fgy = -mgcosΘ
Θ+β+90° = 180°
β = 90°-Θ (1)

α + β= 90° ∴ 90°- β (2)
α = 90° - (90°- θ) = Θ

The Attempt at a Solution


Our prof used a rotated coordinate system, putting +y in the n direction, and +x downwards and parallel to the surface of the incline (i.e. directly opposite/antiparallel to FT). He also used a bunch of different angles/triangle rules, letting Θ = the angle of the incline, and α = the angle that the -y-axis makes with w, creating a triangle. β = the angle opposite to α.

I don't really understand how to use the rotated coordinate system, so I started off by separating the forces into their x and y components based on how they are oriented in the picture, assuming that Fg has no horizontal component and both the tension force and the normal force have both horizontal and vertical components. I also assumed that the net force of the system is zero since the car is still, so forces should be in equilibrium.

x-direction:
Fg = 0
FTx
nx

y-direction:
Fg = -mg = -49000 N
FTy
ny

ΣFx + ΣFy = 0I'm assuming I will need to know the triangle rules to find the x and y components of the normal force, and I think this is where I'm struggling, so this might just be basic geometry skills that I am lacking. I'm guessing that the normal force is slightly reduced (not exactly equal and opposite to the weight of the car) since there is a pulling force from the cable, and the weight is shared between FT and n. Is there a different way to do this without having to break it up into two different triangles, or could someone please explain how I should start breaking it down?
 

Attachments

  • upload_2017-10-20_12-17-10.png
    upload_2017-10-20_12-17-10.png
    7.8 KB · Views: 633
  • upload_2017-10-20_12-17-19.png
    upload_2017-10-20_12-17-19.png
    8.1 KB · Views: 1,398
Physics news on Phys.org
  • #2
mirs08 said:
Is there a different way to do this without having to break it up into two different triangles, ...
There is no different way. You can choose the coordinate axes any way you please, but some ways make writing vector components more complicated than others. The axes chosen by your professor are the simplest because two vectors are already along the principal axes. All you have to do is resolve the weight into two components. If you don't know how to do that, please read and understand the relevant section(s) fron your textbook or notes. It is a very important skill that you need to acquire as soon as you can.
 
Last edited:
  • #3
In addition to what the previous poster said, this
mirs08 said:
ΣFx + ΣFy = 0
while technically correct is not the whole story. The sum of the forces in each direction need to be zero individually, i.e., ΣFx = 0 and ΣFy = 0,
 
  • #4
kuruman said:
There is no different way. You can choose the coordinate axes any way you please, but some ways make writing vector components more complicated than others. The axes chose by your professor are the simplest because two vectors are already along the principal axes. All you have to do is resolve the weight into two components. If you don't know how to do that, please read and understand the relevant section(s) fron your textbook or notes. It is a very important skill that you need to acquire as soon as you can.

Ok, my geometry skills are quite rusty but I think I finally figured out why the angle between the -y-axis and weight is also 32 degrees. So then Fgy = mgcos32 and Fgy = mgsin32? So if that is the weight of the car, then the weight that the ramp exerts on the tire would just be the normal force, and this would be mgcos32 = 41554 N? Then would the tension force just be the horizontal component?
 
  • #5
What if there was no tension? What if it was friction/traction from the tires that is keeping the car on the incline -- would this friction/traction force then be the horizontal component of the gravity? And the normal force is always perpendicular, therefore it would always only be the vertical component of the car's weight?
 
  • #6
mirs08 said:
Ok, my geometry skills are quite rusty but I think I finally figured out why the angle between the -y-axis and weight is also 32 degrees. So then Fgy = mgcos32 and Fgy = mgsin32? So if that is the weight of the car, then the weight that the ramp exerts on the tire would just be the normal force, and this would be mgcos32 = 41554 N? Then would the tension force just be the horizontal component?
Are you working in the tilted coordinate system now? "Horizontal" would usually to imply in the horizontal plane parallel to the ground. Is this what you intend or is it the x-axis as introduced by your teacher?

mirs08 said:
What if there was no tension? What if it was friction/traction from the tires that is keeping the car on the incline -- would this friction/traction force then be the horizontal component of the gravity? And the normal force is always perpendicular, therefore it would always only be the vertical component of the car's weight?
Same problem, specify what you mean by "vertical" here.
 
  • #7
Orodruin said:
Are you working in the tilted coordinate system now? "Horizontal" would usually to imply in the horizontal plane parallel to the ground. Is this what you intend or is it the x-axis as introduced by your teacher?Same problem, specify what you mean by "vertical" here.
Sorry yes, I am working in the tilted coordinate system; by horizontal I mean parallel to the surface of the plane.
 
  • #8
mirs08 said:
Sorry yes, I am working in the tilted coordinate system; by horizontal I mean parallel to the surface of the plane.
Then yes, but I suggest that you do not use that terminology. Vertical and horizontal are usually taken to refer to directions parallel to and orthogonal to the gravitational field by definition. These would not be your coordinate directions so it is better to use "x-direction" and "y-direction".
 
  • #9
Orodruin said:
Then yes, but I suggest that you do not use that terminology. Vertical and horizontal are usually taken to refer to directions parallel to and orthogonal to the gravitational field by definition. These would not be your coordinate directions so it is better to use "x-direction" and "y-direction".

Will keep that in mind!
 

Related to Find the tension in the cable and the force the ramp exerts on the tires

What is the tension in the cable?

The tension in the cable is the pulling force that is transmitted through the cable. In this scenario, the tension is the force that is pulling the ramp upward. It can be calculated using the formula T = mg + ma, where T is the tension, m is the mass of the object, g is the acceleration due to gravity, and a is the acceleration of the object.

How do you calculate the tension in the cable?

To calculate the tension in the cable, you will need to know the mass of the object on the ramp, the acceleration due to gravity (9.8 m/s²), and the acceleration of the object. Once you have these values, you can plug them into the formula T = mg + ma and solve for T.

What is the force that the ramp exerts on the tires?

The force that the ramp exerts on the tires is the reaction force to the tension in the cable. As the tension pulls the ramp upward, the ramp exerts an equal and opposite force on the tires. This force is necessary to keep the tires from sliding down the ramp.

How can the force exerted by the ramp on the tires be measured?

The force exerted by the ramp on the tires can be measured using a force meter or a dynamometer. These tools can accurately measure the force being applied to an object. It is important to make sure the measurement is taken at the point where the ramp and tires make contact.

What factors can affect the tension in the cable and the force exerted by the ramp on the tires?

The tension in the cable and the force exerted by the ramp on the tires can be affected by a variety of factors, including the mass of the object on the ramp, the angle of the ramp, the coefficient of friction between the ramp and the tires, and any external forces acting on the object. It is important to consider these factors when calculating or measuring these forces.

Similar threads

  • Introductory Physics Homework Help
Replies
31
Views
3K
  • Introductory Physics Homework Help
Replies
4
Views
4K
  • Introductory Physics Homework Help
Replies
10
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
2K
  • Introductory Physics Homework Help
Replies
5
Views
4K
  • Introductory Physics Homework Help
Replies
31
Views
16K
Replies
9
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
6
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
3K
  • Introductory Physics Homework Help
Replies
3
Views
1K
Back
Top