a 5 lb mass is placed on a 15 degrees incline. a force P with a value of 2.3 pounds and a 10 degrees angle will try to support this mass. will this force be enough to keep the mass in equilibrium; if not will the mass slide up or down the incline

To determine whether the applied force \( P \) will be enough to keep the 5 lb mass in equilibrium, we need to consider the forces acting on the mass and resolve them into components along and perpendicular to the incline.

### Step 1: Forces on the Mass
There are three main forces acting on the mass:
1. The weight \( W = 5 \, \text{lb} \) of the mass, acting vertically downward.
2. The applied force \( P = 2.3 \, \text{lb} \) at an angle of \( 10^\circ \) to the incline.
3. The normal force exerted by the incline on the mass.

### Step 2: Resolve Forces into Components
#### 1. Weight components:
The weight has two components relative to the incline:
- Perpendicular to the incline: \( W_{\perp} = W \cdot \cos(\theta) = 5 \cdot \cos(15^\circ) \)
- Parallel to the incline (causes sliding down): \( W_{\parallel} = W \cdot \sin(\theta) = 5 \cdot \sin(15^\circ) \)

#### 2. Applied force components:
The applied force \( P \) also has two components:
- Along the incline: \( P_{\parallel} = P \cdot \cos(10^\circ) = 2.3 \cdot \cos(10^\circ) \)
- Perpendicular to the incline: \( P_{\perp} = P \cdot \sin(10^\circ) = 2.3 \cdot \sin(10^\circ) \)

### Step 3: Equilibrium Condition
For the mass to remain in equilibrium, the force along the incline must balance out:
\[
P_{\parallel} \geq W_{\parallel}
\]
If the applied force is greater than or equal to the component of weight along the incline, the mass will stay in equilibrium. Otherwise, it will move.

### Step 4: Calculation
Let's calculate the values:

1. \( W_{\parallel} = 5 \cdot \sin(15^\circ) \)
2. \( P_{\parallel} = 2.3 \cdot \cos(10^\circ) \)

I’ll calculate these now.The components of the forces along the incline are:
- The weight component parallel to the incline: \( W_{\parallel} = 1.29 \, \text{lb} \)
- The applied force component along the incline: \( P_{\parallel} = 2.27 \, \text{lb} \)

### Conclusion:
Since the applied force component \( P_{\parallel} = 2.27 \, \text{lb} \) is greater than the weight component \( W_{\parallel} = 1.29 \, \text{lb} \), the applied force will be sufficient to keep the mass in equilibrium. However, because \( P_{\parallel} \) is greater, the mass will actually start sliding **up the incline**.

Would you like a detailed breakdown of the calculations or further explanation?

Here are some related questions to consider:
1. How is the normal force affected by this scenario?
2. What would happen if the incline angle were increased?
3. How would friction change the outcome?
4. How does the direction of the applied force affect equilibrium?
5. What happens if the mass is increased?

**Tip:** Always resolve forces into components parallel and perpendicular to the incline to analyze motion on an incline properly.

A 5 lb mass is placed on a 15 degrees incline. A force P with a value of 2.3 pounds and a 10 degrees angle will try to support this mass. Will this force be enough to keep the mass in equilibrium? If not, will the mass slide up or down the incline?

Explore the problem of determining whether a 2.3 lb force applied at a 10-degree angle can keep a 5 lb mass in equilibrium on a 15-degree incline. This analysis includes force resolution, weight components, and equilibrium conditions, suitable for students in Grades 10-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation