91Ó°ÊÓ

In physics, especially when dealing with vector quantities like force, it's important to decompose these vectors into their components. This makes it easier to work with them since we can handle each component separately.
Force components refer to breaking down a vector force into perpendicular components. If you think of directions on a map:

• The x-component might represent east-west directions
• The y-component might represent north-south directions

In this exercise, Paul's force of 48.0 N is directed 61.0° south of west. To analyze this, we break it into x and y components.To find the x-component (west-east direction), we use:\[ F_{P,x} = F \, \cos(\text{{angle}}) \]Therefore, for Paul, \( F_{P,x} = 48.0 \, \cos(61.0°) \).For the y-component (north-south direction), we have:\[ F_{P,y} = F \, \sin(\text{{angle}}) \]So, \( F_{P,y} = 48.0 \, \sin(61.0°) \).Since the direction is south of west:

• The x-component (F_{P,x}) is negative because it's towards the west
• The y-component (F_{P,y}) is negative as it's towards the south

Understanding these components is crucial for calculating the interaction of forces in multiple directions.

A displacement vector defines how far and in what direction an object moves from its initial position. Just like forces, displacement vectors can be broken into components to simplify calculations.In this scenario, Paul's crate moves 12.0 m at an angle of 22.0° east of north.
We break this displacement into:

• The x-component (east-west direction): \( d_{x} = 12.0 \, \sin(22.0°) \).
• The y-component (north-south direction): \( d_{y} = 12.0 \, \cos(22.0°) \).

Since the displacement is east of north:

• The x-component (\(d_{x}\)) is positive, moving towards the east.
• The y-component (\(d_{y}\)) is also positive, indicating motion towards the north.

Displacement vectors help us understand precise motion by showing not just how much something moves, but also where it moves.

The angle between vectors is a vital aspect when calculating work done, as the direction between the force and displacement directly affects the magnitude of work. In mathematical terms, the work done by a force is the dot product of the force and displacement vectors.The formula for work is:\[ W = F \, d \, \cos(\theta) \]where \( \theta \) is the angle between the force and displacement vectors.To find the angle between the vectors, you can use their components in:\[ \vec{F_{P}} \cdot \vec{d} = F_{P,x} \, d_{x} + F_{P,y} \, d_{y} \]The resulting formula helps derive how much of the force contributes to the displacement in its given direction.A positive result implies that the force aids the displacement, while a negative result shows it does the opposite, which is evident with Paul's force acting against the movement direction of the crate.
When we calculated this, we found:\[ \vec{F_{P}} \cdot \vec{d} \approx -626.6 \text{ J} \]This negative sign confirms that the force had an angle greater than 90° from the displacement, showing force against the opportunity.