Time to slide down a ramp w/ no friction

[TrpnBils]

Homework Statement

A wet bar of soap (m=150g) slides down a 2m long ramp at 7.3° relative to the horizontal. How long does it take to reach the bottom?

Homework Equations

d = 1/2at²

The Attempt at a Solution

The answer is apparently 1.81 seconds, although I can't figure out how.

I can figure out the height of the ramp to be 0.25m, but using the equation above tells me that it should only take 0.22 seconds if you use gravity as the acceleration. What am I doing wrong?

 

[azizlwl]

You have the equation just substitute.

 

[TrpnBils]

How do you figure? If I substitute 9.8 for acceleration and 2m for distance I get 0.63 seconds, and if I substitute 0.25m for distance I get 0.22 seconds for time...and neither of which are right.

 

[azizlwl]

How do you figure? If I substitute 9.8 for acceleration and 2m for distance I get 0.63 seconds, and if I substitute 0.25m for distance I get 0.22 seconds for time...and neither of which are right.

You should be able to differentiate between g and a.

 

[TrpnBils]

isn't gravity the only acceleration here?

 

[Yukoel]

Hello TrpnBils,
Resolve the acceleration due to gravity parallel to the incline because the actual path the soap travels is along the inclined plane.This should give you the the "a" in your equation and the "d" is already given.
regards
Yukoel

 

[azizlwl]

isn't gravity the only acceleration here?

As you see the body is accelerating not towards the Earth but slanting.
If you drop a mass it will accelerate at the rate of g.
In your case it is moving to the left or right as well as downward.

 

[TrpnBils]

okay, so if I assume 9.8 is pointing down (the Y axis) and I use sin(7.3) = 9.8/L then that gives me a diagonal component of 77m/s^2? That doesn't make sense...

 

[azizlwl]

okay, so if I assume 9.8 is pointing down (the Y axis) and I use sin(7.3) = 9.8/L then that gives me a diagonal component of 77m/s^2? That doesn't make sense...

Analogy 6=4+2

Since acceleration is vector, we can also make 2 components.
BUT the total of these components should be equal to original vector.
Evaluate you calculation.

 

[TrpnBils]

Like the other one I threw on here tonight, I see now that the acceleration is less than that of gravity...I just can't figure out why you do the calculation the way that you do. In this problem, if I take sin(7.3) * 9.8, I get an acceleration of 1.24m/s^2 which works out to be right. I just don't know WHY I multiply the sin(7.3) by gravity when the laws we all learned in trig were Sin(theta) = Opposite/Hypotenuse...and in this case, the 9.8 is the "opposite" while the acceleration along the length of the ramp is the unknown (hypotenuse). In my mind, if you plug them in that way, you should be DIVIDING 9.8 by Sin(7.3).

Why is it the other way?

 

[azizlwl]

As you see from my analogy, 6 is bigger than any of its components.
So in vector addition with right angle triangle, you should know where is the highest value and its 2 components.

 

[TrpnBils]

Yeah, the hypotenuse is always going to be the largest...but that still doesn't answer my question about why I'm multiplying by sine of the angle rather than dividing by it.

 

[azizlwl]

Yeah, the hypotenuse is always going to be the largest...but that still doesn't answer my question about why I'm multiplying by sine of the angle rather than dividing by it.

Ok you get that right.
Then how do you calculate the value of opposite and adjacent with the given hypotenuse?

 

[TrpnBils]

Isn't the value of gravity (9.8) the value for opposite since it's the vertical acceleration?

 

[Saitama]

Like the other one I threw on here tonight, I see now that the acceleration is less than that of gravity...I just can't figure out why you do the calculation the way that you do. In this problem, if I take sin(7.3) * 9.8, I get an acceleration of 1.24m/s^2 which works out to be right. I just don't know WHY I multiply the sin(7.3) by gravity when the laws we all learned in trig were Sin(theta) = Opposite/Hypotenuse...and in this case, the 9.8 is the "opposite" while the acceleration along the length of the ramp is the unknown (hypotenuse). In my mind, if you plug them in that way, you should be DIVIDING 9.8 by Sin(7.3).

Why is it the other way?

No, the diagonal is acceleration due to gravity. A vector can be resolved into two components. When you find the resultant of the resolved vectors, you get the same vector back. See the figure below:

(Sorry for an inaccurate figure, i know its dirty. :P)
We have to find the component of g along the ramp.
Do you now see why you use 9.8sin(7.3)?

 

[TrpnBils]

The diagram makes sense, yes, but why would the diagonal vector be 9.8 considering gravity isn't accelerating in the diagonal direction? This seems like it should be a pretty straightforward problem and I don't know why I'm having problems with this or the other one I asked about tonight...

 

[azizlwl]

The diagram makes sense, yes, but why would the diagonal vector be 9.8 considering gravity isn't accelerating in the diagonal direction? This seems like it should be a pretty straightforward problem and I don't know why I'm having problems with this or the other one I asked about tonight...

You should rotate the diagram anticlockwise 45°.
Then you see the g is pointing down.
And the opposite side becomes parallel to the inclined plane.

 

[TrpnBils]

I sketched that out on my paper, but logically I have absolutely no idea why you would do that. If I am standing in at the base of a ramp, gravity is not accelerating at the angle of that ramp down towards my feet.

 

[azizlwl]

I sketched that out on my paper, but logically I have absolutely no idea why you would do that. If I am standing in at the base of a ramp, gravity is not accelerating at the angle of that ramp down towards my feet.

True.
Now try standing on a slippery ramp .
Do you feel something pulling you forward parallel to the inclined.

 

[TrpnBils]

Maybe it would help me if you all saw what I'm picturing in my head...

In my mind, I'm picturing this ramp at 7.3 degrees from the horizontal. Gravity acts vertically downward, not along the length of the ramp...

I'm wrong, but I don't see the logic behind doing it so that gravity acts down the length of the ramp. To me, doing it that way just seems to arbitrarily rotate the ramp to point gravity in a new direction...why not just turn the whole thing upside down? It seems just as logical that way to me...

 

[Saitama]

The diagram makes sense, yes, but why would the diagonal vector be 9.8 considering gravity isn't accelerating in the diagonal direction? This seems like it should be a pretty straightforward problem and I don't know why I'm having problems with this or the other one I asked about tonight...

You are looking at it from the wrong direction, the incline is making you a little confused.
You should not consider the incline while resolving the vector into its components.

 

[TrpnBils]

True.
Now try standing on the ramp.
Do you feel something pulling you forward.

Yes... okay, that's starting to make sense now. So gravity is acting downward, but because of the slope of the ramp and the fact that the ramp is in the way of me and the floor, the only way I CAN be moved by gravity is to move at an angle down the ramp until I hit the floor...is that right?

 

[azizlwl]

Yes... okay, that's starting to make sense now. So gravity is acting downward, but because of the slope of the ramp and the fact that the ramp is in the way of me and the floor, the only way I CAN be moved by gravity is to move at an angle down the ramp until I hit the floor...is that right?

Yes that's right
Remember another component but this is balanced by normal force which prevent you from passing through the inclined plane.

 

[TrpnBils]

ok - got it... just needed to walk away for a few minutes... thanks everyone!