A 175-N box is placed on an inclined plane that makes a 30.0° angle with the horizontal. Find the component of the weight force parallel to the plane's surface.
weight=mgSin30 down the plane
To find the component of the weight force parallel to the inclined plane's surface, you need to find the weight force acting on the box first.
The weight force can be found using the formula:
Weight = mass × gravitational acceleration
Where:
- Mass is the mass of the box
- Gravitational acceleration is the acceleration due to gravity, which is approximately 9.8 m/s² on Earth
The formula for weight can also be expressed as:
Weight = force × gravitational acceleration
Given that the weight force acting on the box is 175 N, we can use the formula to find the mass:
Mass = Weight / gravitational acceleration
Mass = 175 N / 9.8 m/s²
Now, to find the component of the weight force parallel to the inclined plane's surface, we need to consider the angle the inclined plane makes with the horizontal.
Let's call the component of the weight force parallel to the inclined plane's surface "F_parallel."
F_parallel = Weight × sin(angle)
The angle between the inclined plane and the horizontal is given as 30°. Therefore:
F_parallel = Weight × sin(30°)
Finally, substitute the known values:
F_parallel = 175 N × sin(30°)
To find the component of the weight force parallel to the inclined plane's surface, we can use trigonometry.
Let's break down the weight force into its components. The weight force can be represented by the vector W, which has two components: one perpendicular to the plane (W⊥) and one parallel to the plane (W//).
The component of the weight force parallel to the plane's surface (W//) can be found using the formula: W// = W * sin(θ)
Where:
W is the magnitude of the weight force (175 N in this case)
θ is the angle between the weight force vector and the inclined plane (30.0° in this case)
Let's substitute the given values into the formula:
W// = 175 N * sin(30.0°)
To get the answer, we need to calculate the sine of 30.0° and then multiply it by the weight force magnitude:
W// = 175 N * 0.5
W// = 87.5 N
Therefore, the component of the weight force parallel to the inclined plane's surface is 87.5 N.