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

To determine if the 2.3-pound force \( P \) is enough to keep the 5-pound mass in equilibrium on the 15-degree incline, we need to analyze the forces acting on the object.

### Step 1: Break Down the Forces
There are three main forces acting on the mass:
1. **Gravitational force**: The weight of the object, \( W = 5 \, \text{lbm} \).
2. **Force applied (P)**: A 2.3-pound force trying to support the mass.
3. **Normal force**: The perpendicular reaction force from the surface of the incline.

### Step 2: Components of the Gravitational Force
The gravitational force can be split into two components relative to the incline:
- **Parallel to the incline**: \( W_{\parallel} = W \sin(\theta) \)
- **Perpendicular to the incline**: \( W_{\perp} = W \cos(\theta) \)

Where \( \theta = 15^\circ \).

Given:
- \( W = 5 \, \text{lbm} \)
- \( \sin(15^\circ) = 0.2588 \)
- \( \cos(15^\circ) = 0.9659 \)

### Step 3: Calculate the Components
- **Parallel component**: \( W_{\parallel} = 5 \times 0.2588 = 1.294 \, \text{lbm} \)
- **Perpendicular component**: \( W_{\perp} = 5 \times 0.9659 = 4.829 \, \text{lbm} \)

The force \( W_{\parallel} = 1.294 \, \text{lbm} \) is the force that tries to pull the mass down the incline due to gravity.

### Step 4: Analyze the Force \( P \)
If the force \( P = 2.3 \, \text{lb} \) is applied **up the incline**, it will counteract the downward component of the gravitational force, which is \( 1.294 \, \text{lb} \).

### Step 5: Compare Forces
- The gravitational force pulling the mass down the incline is \( 1.294 \, \text{lb} \).
- The applied force \( P = 2.3 \, \text{lb} \).

Since \( 2.3 \, \text{lb} > 1.294 \, \text{lb} \), the applied force is **greater** than the gravitational force acting parallel to the incline.

### Conclusion:
- The 2.3-pound force is **more than sufficient** to keep the mass in equilibrium.
- In fact, the mass will tend to **move up the incline** because the applied force is stronger than the force trying to pull it down the incline.

Would you like further details on the calculations or have any questions?

Here are some related questions to deepen your understanding:
1. How would friction affect the motion of the mass on the incline?
2. What happens if the angle of the incline is increased to 30 degrees?
3. How do we calculate the normal force on the incline?
4. What happens if the applied force is less than the parallel component of the weight?
5. What is the role of the perpendicular component of the weight on an inclined plane?

**Tip**: Always resolve forces into components parallel and perpendicular to the incline when dealing with problems on an inclined plane.

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?

Learn how to analyze forces acting on a 5 lb mass placed on a 15-degree incline. This problem involves calculating gravitational force components and determining if a 2.3 lb applied force is enough to maintain equilibrium. Suitable for students in Grades 9-12.

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