# Homework Help: Basic Mechanics Confusion

1. May 11, 2015

### FaroukYasser

1. The problem statement, all variables and given/known data
The question is made of a smooth inclined ramp at an angle of 30 degrees. There is a particle on the ramp with mass m. The question asks to find R (The reaction force on the ball) in terms of mg

2. Relevant equations
Resolving the forces with the ramp and equating upward and downward forces to find the reaction.

3. The attempt at a solution
I know the question is really basic. Most people resolve the weight force so that one is opposite the Reaction and on is down the ramp or in other words:
R = mgcos(30) and mgsin(30) down the ramp.

My question is why don't we resolve the Reaction force so that we have our system in an x-y coordinate like for example:
Rcos(30) = mg and Rsin(30) to the side, and since Rcos(30) = mg then R = mgsec(30)
I know what I wrote is wrong but I just don't understand why. why do we have to resolve the forces along the ramp?

Thanks :)

2. May 11, 2015

### TSny

The basic starting point is always Newton's second law: $\sum{F_x} = ma_x$ and $\sum{F_y} = ma_y$.

If you take your x and y axes to be horizontal and vertical, respectively, then note that neither $a_x$ nor $a_y$ is zero. So, your equation Rcos(30) = mg does not take into account the acceleration in the vertical direction.

3. May 11, 2015

### FaroukYasser

Thanks for your reply. So do we take our y perpendicular to the ramp because we are sure the object is not moving off of it or into so the forces on it must be 0 and so the acceleration must be 0 and in which case we can use Newton's second?

4. May 11, 2015

### TSny

Yes, there is no component of acceleration in the direction perpendicular to the ramp. So, the sum of the components of the forces acting on the object in this direction must equal zero.

5. May 11, 2015

### FaroukYasser

Thanks a lot!

6. May 11, 2015

### TSny

Let me add a couple of more comments. You said, "So do we take our y perpendicular to the ramp because we are sure the object is not moving off of it or into so the forces on it must be 0 and so the acceleration must be 0 and in which case we can use Newton's second?"

It seems that you might not quite have the logic correct here. Let's take your statement in parts. First, "So do we take our y perpendicular to the ramp because we are sure the object is not moving off of it or into..". Yes, that statement is good.

Then, "...so the forces on it must be 0 and so the acceleration must be 0". Here, I think you have it backwards. Since the object moves in a straight line down the slope, it can only have acceleration in the direction parallel to the slope. From that you can conclude that the component of acceleration perpendicular to the slope is 0. Then using Newton's second law, you can conclude that the sum of the components of force perpendicular to the slope is 0. So, you see the order of the logic is to first deduce that there is no acceleration perpendicular to the slope and then use the second law to deduce that the sum of the forces perpendicular to the slope must be zero.

Finally, "...and in which case we can use Newton's second?". You can use Newton's law for any choice of orientation of the axes. So, if you wanted to take your x and y axes horizontal and vertical, then you can still apply Newton's second law to the x and y directions and get the correct result for the reaction force and the acceleration. It just won't be as convenient as orienting the axes parallel and perpendicular to the slope.

7. May 11, 2015

### FaroukYasser

Thanks for taking the time to write all of this! Definitely cleared any misconception I had :)