# Homework Help: Gravity's effect on a cart sliding down an incline

1. Oct 4, 2004

### deanchhsw

Hello, I've just joined this forum in hopes of finding some help to my physics problems. I usually can find out what to do by myself, but this problem seems to be requiring some advanced concepts...or it maybe is so that I'm just not using my noodle.

A car is sliding down an inclined plane, say, of 20 degrees to the horizontal.
And the distance traveled, s, is 2.0818 meters, and the time elapsed is
2.33 seconds.

therefore, the calculated value of acceleration of the cart is .767m/s^2.
It is very easy up to this point, however, the teacher asks us to solve for acceleration due to gravity, otherwise known as 9.8 m/s^2, to compare the error margin to the accepted value. However, it doesn't come to me how you can solve for acceleration due to gravity just with the cart's acceleration.

Essentially, the only force of acceleration applied to the cart is that of the gravity, but the incline is reducing the accelertion, as it is not the "free fall".
which would be the case if the incline was 90(or 270) degrees. So my assupmtion is that there must be a certain ratio to solve this...but I just can't hold a grasp of it. I tried to get a brain blast(ha), didn't work for 20 minutes and I just ended up with a headache.

Of course, since one is solving for Ag, it cannot be incorporated in solving process- .. I have not been this frustrated with a problem for a long time..

for your convenience, the data again is summed up here:

displacement = 2.0818 m
time = 2.33 s
acceleration cart = 0.767m/s^2
angle of the incline = 20

solve for Acc. due to gravity.

Once again, thank you, Dean.
* p.s. air resistance and friction is considered negligible in this lab.

Last edited: Oct 4, 2004
2. Oct 4, 2004

### Staff: Mentor

Consider that the only force accelerating the car down the incline is the component of its weight parallel to the incline. Find that component. (You'll need a little trig.) Then apply Newton's 2nd law.

3. Oct 4, 2004

### deanchhsw

I suppose it'd work this way, but the weight of the car cannot be measured;

Although the answer is slightly ambiguous to me(my fault), at least I have a glimpse of how that'd work- the only problem is, we were not allowed to measure the mass of the cart, so the weight idea is gone.

Only equipments allowed for us were an incline plane, a spark timer(from which we could determine the t value), and a cart. Thus, only assets given in this problem is time elapsed, displacement the car traveled, and the height in which incline started from, which gives us the angle of inclination...

How can this problem be solved?...

4. Oct 4, 2004

### Staff: Mentor

Don't give up so easily. You don't need to measure the mass--it will cancel out. Just call the mass "m". What's the weight? (Do it with symbols, not numbers.) Then what's the component of that weight down the incline? Try it.

If that's too abstract, try this way: If the acceleration due to gravity is a vector of magnitude "g" pointing down, what is its component parallel to the incline? (That parallel component is what your experiment measured. Now work backwards to find "g".)

5. Oct 4, 2004

### deanchhsw

I'm really, really sorry, I still don't understand completely...

I understand that acceleration(or force, for that matter) is a vector quantity, therefore those measurements can combine to do trig, etc...

Force equals mass x acceleration, therefore
the component of the weight going down that incline would be

mass*(0.767m/s^2)..

and Newton's second law of motion suggests

net force = mass x acceleration

thus, since Net force is the component of the weight going down that incline(since it's weight which includes gravity), I come to a conclusion that

mass * (0.767m/s^2) = mass * acc.

...and it doesnt' do anything but to reconfirm that .767 is the acceleration of the cart. It's the conceptual part I don't understand, that how the INCLINE affects the force of gravity going down.

I've tried drawing that component method and trig you mentioned, and I've been so accustomed to over few months now..

so there's a vector straight down, 9.8 m/s^2.
and provided that the angle of incline is x, the angle theta between the inclination vector and Ag vector would be 90-x. I understand up to that point.

But the downward vector 9.8 m/s^2 is an acceleration.
if I am to place the inclination vector(which has to be in a form of acceleration, a 0.767m/s^2)....the resultant vector tells nothing.
Also, if I am to do trig and solve for angle theta, the angle the incline makes with the horizontal plane, I can't have 0.767 as a hypotenuse and Ag as the vertical vector..it doesn't make sense mathematically that way.

I suppose you've gone down to specifics already, but could you sheperd this little lamb and provide more details?...

Note that we are not allowed to use Ag in the calculation, since that's what I'm solving for. Although, I think what you were implying in "If that's too abstract, try this way:..." here was to just get me to understand the concept.

*p.s. I understand that you're not giving straightforward answer- I really do respect that way of guidance, I think people really need to think before they give up and reach out for answers.

6. Oct 5, 2004

### phrontist

FBDs

Has your teacher introduced you to Free Body Diagrams? If not, the idea is simple. Just sketch out all the forces involved before beginning the problem. Then break up any vectors into more workable component vectors. For easy problems it seems like overkill, but being practiced at it will help when you run into difficult problems.

7. Oct 5, 2004

### Pyrrhus

Also, it's better to put your coordinate system in the easiest way possible, in this problem i will put it rotated 20 degrees counter clockwise from its original position.

8. Oct 5, 2004

### Staff: Mentor

In terms of mass (m) and g, what is the weight? And what is the component of the weight that acts down the incline? You will need to use a little trig. It's easy... do it!

Once you find the force down the incline, then apply Newton's 2nd law. The component of the weight down the incline must equal the mass X the acceleration you measured.

That's because you're going in a bit of a circle. Do what I say above.

Good. So what's the problem? Find the component of the acceleration along the incline.

Since you are finding the components of Ag, Ag is the hypontenuse of a triangle.
You must use Ag in your thinking and equations, else you'll never be able to calculate it.

So we've talked about two ways to relate the acceleration due to gravity to the acceleration down the incline, which you measured:
(1) Using Newton's 2nd law. Big hint: the weight = mg. So what's the component down the incline?
(2) Finding the component of g directly. What's the component of g down the incline?

Both methods require you to take a vector pointing down (either weight or acceleration) and find the component parallel to the incline. That's the problem you're having. You need to understand better how to use trig to find that component.

9. Oct 6, 2004

### deanchhsw

I just wanted to post that thank you for being patient with me.
In fact, I just realized how to solve this while I was coming back to
clarify more things...sigh.

I know, I know, I've never been more astounded by my stupidity..
teacher looked at me looking over your things and said
"he's giving you the answer, and you can't grasp it"...