A 5 lbm mass is placed on a 15 degrees incline. A force (P) with a value of 2.3 pound 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 force \( P = 2.3 \) pounds is enough to keep the 5 lbm mass in equilibrium on a 15° incline, we need to analyze the forces acting on the mass.

### Forces Acting on the Mass:

1. **Weight of the Mass (W)**:  
   The weight \( W \) of the mass is the force due to gravity. Since the mass is 5 lbm, and assuming standard gravity (\( g = 32.2 \, \text{ft/s}^2 \)), we can use the following approximation for force due to weight (where lbm and lbf are numerically the same in English units):
   \[
   W = 5 \, \text{lbf}
   \]
   
2. **Force Components Along the Incline**:  
   We can break the weight \( W \) into two components:
   - A component **parallel to the incline** that tries to pull the mass down: 
     \[
     W_{\parallel} = W \sin \theta = 5 \sin(15^\circ)
     \]
   - A component **perpendicular to the incline**, which doesn't affect the sliding but contributes to the normal force:
     \[
     W_{\perp} = W \cos \theta = 5 \cos(15^\circ)
     \]
   
3. **Applied Force (P)**:  
   The applied force \( P = 2.3 \, \text{lbf} \) is acting up the incline to prevent the mass from sliding down.

### Step 1: Calculate \( W_{\parallel} \)
\[
W_{\parallel} = 5 \sin(15^\circ) \approx 5 \times 0.2588 \approx 1.29 \, \text{lbf}
\]

### Step 2: Compare the Applied Force \( P \) and the Parallel Component \( W_{\parallel} \)
To keep the mass in equilibrium, the applied force \( P \) must be equal to or greater than \( W_{\parallel} \).

Given that:
- Applied force \( P = 2.3 \, \text{lbf} \)
- Parallel weight component \( W_{\parallel} = 1.29 \, \text{lbf} \)

The applied force \( P \) is greater than the weight component pulling the mass down. Therefore, the mass **will not slide down** the incline. 

### Conclusion:
Since \( P > W_{\parallel} \), the applied force is strong enough to prevent the mass from sliding down. However, the applied force is **more than sufficient**, meaning that the mass will likely slide **up** the incline because the applied force is greater than the force required to keep it in equilibrium.

### Summary:
- The force \( P = 2.3 \, \text{lbf} \) is **more than enough** to keep the mass in equilibrium.
- The mass will **slide up the incline** due to the excess applied force.

Would you like more details on this analysis or help with a related concept?

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### 5 Follow-up Questions:
1. How does the angle of the incline affect the forces acting on a mass?
2. What is the significance of the friction in problems involving inclined planes?
3. How would the required force change if the angle of the incline were steeper (e.g., 30 degrees)?
4. What happens to the mass if the applied force equals the parallel component exactly?
5. How can we include friction in the calculation if it were present?

### Tip:
Always break forces into components along the incline and perpendicular to it when dealing with inclined plane problems. This simplifies your calculations significantly.

A 5 lbm mass is placed on a 15 degrees incline. A force (P) with a value of 2.3 pounds 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 whether a 2.3 lb force is enough to keep a 5 lb mass in equilibrium on a 15° incline. This detailed analysis uses statics and force components along an inclined plane to determine if the mass will slide. Suitable for Grades 10-12 students.

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