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

[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

Homework Equations

F_gx = mgsinΘ
F_gy = -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 F_T). 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 F_g 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
F_Tx
n_x

y-direction:
Fg = -mg = -49000 N
F_Ty
n_y

ΣF_x + ΣF_y = 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 F_T 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?

 

[kuruman]

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.

 

[Orodruin]

In addition to what the previous poster said, this

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,

 

[mirs08]

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?

 

[mirs08]

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?

 

[Orodruin]

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?

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.

 

[mirs08]

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.

 

[Orodruin]

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".

 

[mirs08]

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!