Wally Goes Skiing on a Windy Day
Wally (45 kg), is going skiing. He skis down a frictionless slope that is at a 15° incline from horizontal. A modest wind pushes him from behind with a force of 50 N.
Question 1: What is the net force acting upon Wally in his direction of motion?
There are three forces acting on Wally that we can deduce from the problem statement:
- The wind force pushes Wally from behind, parallel to the slope and his direction of motion.
- The normal force acts perpendicularly to the slope, preventing him from sinking into the snow.
- The gravitational force pulls Wally downward, at an oblique angle to the slope and directly towards the center of the Earth.
We will sketch our problem and orient our coordinate system so that the positive x-axis is in line with the direction of motion. This will make our coming calculations easier:
Figure 1: Our sketch of the forces acting on Wally
Since Wally is moving in a direction parallel to the slope of the hill, we are only interested in the components of the forces that work in that direction, along our x-axis.
\[ F_\text{net} = F_{\text{wind}_\text{x}} + F_{\text{normal}_\text{x}} + F_{\text{gravity}_\text{x}} \]
Since the force of the wind is acting in the same direction as Wally's direction of motion, we know the full magnitude contributes to his acceleration:
\[ F_{\text{wind}_\text{x}} = 50 \, \text{N} \]
The normal force acts perpendicularly to Wally's direction of motion, so we know it does not contribute to Wally's acceleration:
\[ F_{\text{normal}_\text{x}} = 0 \, \text{N} \]
The force of gravity can be calculated using Newton's Second Law of Motion, the mass of Wally, and the gravitational acceleration of Earth.
\[ F_{\text{gravity}} = m_\text{wally} \times g \]
However, the force of gravity acts at an angle to Wally's direction of motion, so we must calculate its contribution using trigonometric ratios.
\[ F_{\text{gravity}_\text{x}} = F_{\text{gravity}} \times \cos(90^\circ - \theta) \]
\[ F_{\text{gravity}_\text{x}} = ( m_\text{wally} \times g ) \times \cos(90^\circ - \theta) \]
\[ F_{\text{gravity}_\text{x}} = ( 45 \, \text{kg} \times 9.8 \, \text{m/s}^2 ) \times \cos(90^\circ - 15^\circ) \]
\[ F_{\text{gravity}_\text{x}} = ( 45 \, \text{kg} \times 9.8 \, \text{m/s}^2 ) \times \cos(75^\circ) \]
Therefore, our net force is:
\[ F_\text{net} = F_{\text{wind}_\text{x}} + F_{\text{normal}_\text{x}} + F_{\text{gravity}_\text{x}} \]
\[ F_\text{net} = ( 50 \, \text{N} ) + ( 0 \, \text{N} ) + ( ( 45 \, \text{kg} ) \times ( 9.8 \, \text{m/s}^2 ) ) \times \cos(75^\circ) \]
\[ F_\text{net} \approx 164.14 \, \text{N} \]
Question 2: What is the magnitude of Wally's acceleration?
Again, we will apply Newton's Second Law of Motion to find the acceleration due to the net force:
\[ F_\text{net} = m_\text{wally} \times a \]
Solving our equation for \(a\):
\[ a = F_\text{net} / m_\text{wally} \]
\[ a \approx ( 164.14 \, \text{N} ) / ( 45 \, \text{kg} ) \]
We arrive at our final aceleration:
\[ a \approx 3.65 \, \text{m/s}^2 \]