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 lb) is sufficient to keep the mass in equilibrium on the incline, we need to analyze the forces acting on the mass and check if the net force is zero. If the net force is not zero, the mass will either slide up or down the incline.

### Step 1: Determine the forces acting on the mass
The forces acting on the mass are:

1. **Weight of the mass (W)**: The weight of the mass can be calculated as:
   \[
   W = m \cdot g
   \]
   where \( m = 5 \text{ lbm} \) and \( g = 32.2 \text{ ft/s}^2 \) (standard acceleration due to gravity in imperial units).
   
   In this case, the weight force \( W \) is:
   \[
   W = 5 \cdot 32.2 = 161 \text{ lbm-ft/s}^2 = 5 \text{ lbf} \quad (\text{as lbf and lbm cancel out with gravity constant})
   \]

2. **Components of the weight along the incline**:
   - The component of the weight **parallel to the incline**:
     \[
     W_{\parallel} = W \cdot \sin(\theta) = 5 \cdot \sin(15^\circ) = 5 \cdot 0.2588 \approx 1.294 \text{ lbf}
     \]
   - The component of the weight **perpendicular to the incline** (this does not affect sliding but creates normal force):
     \[
     W_{\perp} = W \cdot \cos(\theta) = 5 \cdot \cos(15^\circ) = 5 \cdot 0.9659 \approx 4.829 \text{ lbf}
     \]

### Step 2: Analyze the equilibrium
To keep the mass in equilibrium, the applied force \( P \) (2.3 lbf) must counterbalance the weight component acting parallel to the incline, which is \( W_{\parallel} = 1.294 \text{ lbf} \).

- If \( P = W_{\parallel} \), the mass will stay in equilibrium.
- If \( P > W_{\parallel} \), the mass will move **up** the incline.
- If \( P < W_{\parallel} \), the mass will move **down** the incline.

### Step 3: Compare the forces
Here, \( P = 2.3 \text{ lbf} \) and \( W_{\parallel} = 1.294 \text{ lbf} \).

Since \( P > W_{\parallel} \), the force is more than enough to overcome the weight component. Therefore, the mass will not stay in equilibrium, and it will **slide up** the incline due to the excess force applied.

### Conclusion:
The 2.3 lbf force is sufficient to overcome the weight component along the incline, and it will cause the mass to slide up the incline.

Would you like more details or have any questions?

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### 5 Follow-up Questions:
1. How would the situation change if the incline angle were increased?
2. What if friction were introduced to the incline? How would that affect the equilibrium?
3. How would the force required change if the mass were doubled?
4. How would the calculation differ if this problem were in metric units?
5. What would happen if the applied force were less than the parallel weight component?

**Tip:** Always decompose forces into parallel and perpendicular components when working with inclined planes to simplify analysis.

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 how to calculate whether a 5 lbm mass on a 15° incline remains in equilibrium when a 2.3 lb force is applied. This problem involves force decomposition, weight calculation, and determining if the mass will slide up or down. Suitable for high school physics students in Grades 10-12.

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