A 1.30 Kg Glass Is On A Tray That Is Inclined At $18.0^{\circ}$. What Is The Y-component Of The Weight Of The Glass? $w_{y} = [?] \, \text{N}$

Introduction

In this problem, we are tasked with finding the y-component of the weight of a 1.30 kg glass that is placed on a tray inclined at an angle of $18.0^{\circ}$. To solve this problem, we need to understand the concept of vectors and how to resolve them into their components.

The Concept of Vectors

A vector is a quantity that has both magnitude and direction. In this case, the weight of the glass is a vector that has a magnitude (the amount of force exerted on the glass) and a direction (the direction in which the force is exerted). The weight of the glass can be resolved into its x and y components using trigonometry.

Resolving the Weight into Components

To resolve the weight into its components, we need to use the following formulas:

$$w_{x} = w \sin(\theta)$$

$$w_{y} = w \cos(\theta)$$

where $w_{x}$ and $w_{y}$ are the x and y components of the weight, $w$ is the magnitude of the weight, and $\theta$ is the angle of inclination.

Calculating the Magnitude of the Weight

The magnitude of the weight of the glass can be calculated using the following formula:

$$w = mg$$

where $m$ is the mass of the glass and $g$ is the acceleration due to gravity.

Given Values

• Mass of the glass: $m = 1.30 , \text{kg}$
• Angle of inclination: $\theta = 18.0^{\circ}$
• Acceleration due to gravity: $g = 9.80 , \text{m/s}^2$

Calculating the Magnitude of the Weight

Using the formula $w = mg$, we can calculate the magnitude of the weight as follows:

$$w = 1.30 \, \text{kg} \times 9.80 \, \text{m/s}^2 = 12.74 \, \text{N}$$

Calculating the Y-Component of the Weight

Now that we have the magnitude of the weight, we can calculate the y-component of the weight using the formula:

$$w_{y} = w \cos(\theta)$$

Substituting the values, we get:

$$w_{y} = 12.74 \, \text{N} \times \cos(18.0^{\circ}) = 12.74 \, \text{N} \times 0.951 = 12.14 \, \text{N}$$

Conclusion

In this problem, we have calculated the y-component of the weight of a 1.30 kg glass that is placed on a tray inclined at an angle of $18.0^{\circ}$. The y-component of the weight is $12.14 , \text{N}$.

Final Answer

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Introduction

In our previous article, we explored the problem of a 1.30 kg glass on a tray inclined at an angle of $18.0^{\circ}$. We calculated the y-component of the weight of the glass using trigonometry. In this article, we will answer some frequently asked questions related to this problem.

Q: What is the significance of the angle of inclination?

A: The angle of inclination is crucial in determining the components of the weight of the glass. The angle of inclination affects the magnitude and direction of the weight components.

Q: How do you resolve the weight into its components?

A: To resolve the weight into its components, we use the following formulas:

$$w_{x} = w \sin(\theta)$$

$$w_{y} = w \cos(\theta)$$

where $w_{x}$ and $w_{y}$ are the x and y components of the weight, $w$ is the magnitude of the weight, and $\theta$ is the angle of inclination.

Q: What is the difference between the x and y components of the weight?

A: The x and y components of the weight are different because they are perpendicular to each other. The x component of the weight is the force exerted on the glass in the horizontal direction, while the y component of the weight is the force exerted on the glass in the vertical direction.

Q: How do you calculate the magnitude of the weight?

A: The magnitude of the weight can be calculated using the following formula:

$$w = mg$$

where $m$ is the mass of the glass and $g$ is the acceleration due to gravity.

Q: What is the acceleration due to gravity?

A: The acceleration due to gravity is the rate at which an object falls towards the ground. On Earth, the acceleration due to gravity is approximately $9.80 , \text{m/s}^2$.

Q: How do you calculate the y-component of the weight?

A: To calculate the y-component of the weight, we use the formula:

$$w_{y} = w \cos(\theta)$$

Substituting the values, we get:

$$w_{y} = 12.74 \, \text{N} \times \cos(18.0^{\circ}) = 12.74 \, \text{N} \times 0.951 = 12.14 \, \text{N}$$

Q: What is the final answer?

A: The final answer is $12.14 , \text{N}$.

Conclusion

In this article, we have answered some frequently asked questions related to the problem of a 1.30 kg glass on a tray inclined at an angle of $18.0^{\circ}$. We hope that this article has provided a better understanding of the concepts involved in this problem.

Final Answer

The final answer is: $\boxed{12.14}$