Sledding

Catherine Medeiros
Period C

I have a 10 year old sister who thinks any amount of snow is fun. Needless to say, when there is about 1 inch of snow on the ground, she wants to go sledding. We went down two different hills, one in our front yard and one at our local park. The one at our local park is substantially steeper than the one at our house. But, how does this affect the speed at which my sister travels down the hill? With some common sense, a person would think that the steeper hill would cause her to travel faster. Is this true and how much faster?

My sister weighs about 95 pounds or 43 kilograms. Although my sister does not start sledding from rest, I figured that by solving for her acceleration, one can see how fast she is going by substituting into the equation

acceleration = velocity / time

First the small hill in my front yard.

That is my sister going down the hill. Now, she doesn't go very fast. 

Free body diagram:

Rotate the axis so that Friction is the Y Axis and Normal is the X Axis. The angle of the hill (37 degrees) will go in between the Force of Gravity and the Negative Y Axis. 

Net Force Equations:

sum of Forces x = Normal - mg(sin37) = 0

sum of Forces y = Friction - mg(cos37) = ma

When you substitute numbers in you get:

Fx = N - 43(9.8)sin37 = 0

Fy = f - 43(9.8)cos37 = 43a 

We want to solve for a. We must change the second equation. Remember that 

f = µN

Therefore Fy = µN - 43(9.8)cos37 = 43a as well. 

Now, solve the Fx for N and substitute into the new Fy equation.  

Fx = N = 43(9.8)sin37

Fy = µN - 43(9.8)cos37 = 43a

Fy = µ(43(9.8)sin37)- 43(9.8)cos37 = 43a

Now, you must solve for a.

-271.19µ - 322.55 = 43a

-6.31µ - 7.5 = a 

Acceleration will be a negative number. This is because we set the positive direction of the Y Axis to be opposite the acceleration. (It just happened like this and I did not realize I did that until this point.)

Now for the larger hill. (I began recording but then my phone died because it was so cold out)

Free Body Diagram:

 

The free body diagram is the same for both situations because they are the same situation. Here, the angle will be larger (68 degrees). Rotate the axis so that Friction is the Y Axis and Normal is the X Axis. The angle of the hill (68 degrees) will go in between the Force of Gravity and the Negative Y Axis. 

Net Force Equations:

sum of Forces x = Normal - mg(sin68) = 0

sum of Forces y = Friction - mg(cos68) = ma

When you substitute numbers in you get:

Fx = N - 43(9.8)sin68 = 0

Fy = f - 43(9.8)cos68 = 43a 

We want to solve for a. We must change the second equation. Remember that 

f = µN

Therefore Fy = µN - 43(9.8)cos68 = 43a as well. 

Now, solve the Fx for N and substitute into the new Fy equation.  

Fx = N = 43(9.8)sin68

Fy = µN - 43(9.8)cos68 = 43a

Fy = µ(43(9.8)sin68)- 43(9.8)cos68 = 43a

Now, you must solve for a.

-378.39µ - 185.48 = 43a

-8.799µ - 4.31 = a 

Again the acceleration is negative due to how we set the Axis. 

The acceleration for the smaller hill is -6.31µ - 7.5 = a

The acceleration for the larger hill is -8.799µ - 4.31 = a 

To change them to be positive, you get:

The acceleration for the smaller hill is 6.31µ + 7.5 = a

The acceleration for the larger hill is 8.799µ + 4.31 = a 

In both cases, µ remains the same. This is because both cases had the same conditions (same kid, same sled, same amount of snow etc). This makes it easy to compare. Using logic, we know that the larger hill will have the higher acceleration. Comparing the two accelerations we see that the larger hill has a larger number before µ. However, it does not have a larger number added to it. 

Because the question we are solving for is how much faster and µ remains the same for both, we can use a number for µ and solve for a. Lets use 2. 

Smaller Hill: 

6.31µ + 7.5 = a

6.31(2) + 7.5 = a

20.12 = a

Larger Hill

8.799µ + 4.31 = a

8.799(2) + 4.31 = a

21.908 = a

We can see that the acceleration in the larger hill is greater than the acceleration in the smaller hill (this is true for all values of µ greater than one). If µ is less than one, the smaller hill has a higher acceleration. Because of this change, you would need to know the exact value of µ to solve. We do not know the value of µ; therefore, solving this problem out completely becomes very challenging.