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Home / What Is the Tangential Acceleration (I.e., the Acceleration Parallel to the Road) of Car a?

What Is the Tangential Acceleration (I.e., the Acceleration Parallel to the Road) of Car a?

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Tangential Acceleration Along an Oddly Shaped Route

  • Thread starter JeYo
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Three cars are driving at 29.0m/s forth the road shown in the figure. Car B is at the lesser of the hill and automobile C is at the top. Suppose each auto suddenly brakes hard and starts to slip. What are the tangential accelerations of Cars A B and C.

Here should be the picture: http://i216.photobucket.com/albums/cc200/zucchinigrandma/knight_Figure_07_43.jpg

I am not really certain how to even approach this problem, for I accept never attempted a problem of this blazon and have never been very clear on how to find tangential acceleration.

Answers and Replies

How does the friction force compare for each car? Hint: Consider how the normal force differs for each machine.
I would assume that the frciton would be like f_k = [tex]\mu[/tex]_k * cos[tex]\theta[/tex] * north? But what would the angle be?
Kinetic friction is [itex]f_k = \mu_k Northward[/itex]. (I don't understand where you got that [itex]\cos\theta[/itex] factor.)
Apparently I was wrong in my reasoning. I realise that the normal force for car A is mg, just I honestly have no idea of what the normal forces for cars B and C would exist, if they are different, or how to notice them.
You detect the normal force by applying Newton's 2nd law. Realize that the cars B and C are centripetally accelerating. Start by indentifying the forces interim on each machine.
The forces acting on the auto should merely exist gravity, normal, and the braking, which I am not certain how to represent, correct?
Right. The three forces are gravity, normal forcefulness, and friction. Now employ Newton II to the vertical forces.
Simply how does friction account into the verticle forces when the forcefulness of friction is simply along the horizontal?
It doesn't. But to calculate friction you need the normal force, which is a vertical strength.
Ooh, well, since there is no acceleration in the y direction, i would assume that northward = westward = mg, yes? and then the friction is equal to [tex]\mu[/tex]_k * mg. Simply how does that enable me to find the acceleration?
Ooh, well, since there is no dispatch in the y direction, i would presume that n = w = mg, yes?
No. Realize that the cars are on curved tracks and that they are centripetally accelerating.
and then the friction is equal to [tex]\mu[/tex]_k * mg. But how does that enable me to discover the acceleration?
One time you correctly find the friction, y'all'll apply Newton Two to find the tangential acceleration.
I do not understand...I mean what I idea would non be correct for car A either? I mean, I don't want you to give me the respond or anything, but I just do not understand.
Automobile A is on a apartment road--no vertical acceleration there. Cars B and C are on curved sections. Big difference.
But A is accelerating in the opposite direction, considering of the braking. And I realise that there is a difference in the motions of car A and cars B and C; nevertheless, I do not sympathise how to use that divergence in a way that helps me solve the trouble.
I meant to say no vertical dispatch of Car A, thus its normal force is simply equal to mg. Only cars B & C have centripetal acceleration, which is vertical. Calculate the normal force by applying Newton II to the vertical forces.
Alright, take a expect at this:

[tex]\Sigma[/tex](F_A)_y = due north - w = m_A * a_y = 0N
[tex]\Sigma[/tex](F_A)_x = -f_k = m_A * a_x

And then, f_k = m_A * a_x, [tex]\mu[/tex]_k * thou = a_x

Right...?

Alright, accept a wait at this:

[tex]\Sigma[/tex](F_A)_y = n - w = m_A * a_y = 0N

due west = m_A*1000
(a_y = 0)
[tex]\Sigma[/tex](F_A)_x = -f_k = m_A * a_x

So, f_k = m_A * a_x, [tex]\mu[/tex]_k * g = a_x

Good. The horizontal forcefulness on machine A is [itex]F = \mu N = \mu mg[/itex]. Applying Newton 2 gives you the horizontal acceleration, [itex]a_x = \mu g[/itex].
Sugariness...at present, how exactly would I begin to classify the forces on cars B and C as in either the x- or y-management?
The iii forces are gravity, normal forcefulness, and friction. Which fashion does each human activity?
Well, gravity goes straight down. My first instinct is to say that normal goes perpindicular, but perpindicular to what? Then, maybe information technology is oposition to gravity, but I doubt it. And I am as well non sure about friction, I know information technology goes in the opposite direction of the front of the machine, only whether or not information technology has a vertical component, I am non certain.
Well, gravity goes straight down.
Correct.
My start instinct is to say that normal goes perpindicular, just perpindicular to what?
The surface. And the plane of the surface is defined past the tangent to the surface at the bespeak in question. (There's a reason why cars B and C are at the exact lesser and exact top of the hills.)
So, maybe information technology is oposition to gravity, but I doubtfulness information technology. And I am as well not sure about friction, I know information technology goes in the opposite management of the forepart of the motorcar, merely whether or not it has a vertical component, I am not sure.
Friction is parallel (tangential) to the surface.
And then the forces on cars B and C are pointing in the same directions as the forces on car A?
And then the forces on cars B and C are pointing in the same directions as the forces on automobile A?
Absolutely.
But the accelerations could not be the same for all three cars, could they?
Just the accelerations could not be the same for all three cars, could they?
No. As I suggested before, first effigy out the normal forcefulness (as that volition determine the friction and thus the tangential acceleration). Do that past analyzing the vertical forces and acceleration.
But if all of the cars have the aforementioned mass, and the mass cancels out when you exercise the y-direction of the sum of the forces, will it not come out to be the same acceleration?
Try it and see. What does Newton's 2nd law tell you about vertical forces?
It tells me that the normal force - the force of gravity = m_car * a_y = 0, aye?
I have this same problem on my homework, except with 21 thousand/southward.

How can y'all utilize these formulas if y'all take no mass or coefficient of friction?

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