Math Problem Statement

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

Solution

To determine if the force P=2.3lbfP = 2.3 \, \text{lbf} is sufficient to keep the 5 lbm mass in equilibrium on a 15-degree incline, we need to analyze the forces acting on the mass. Here's the step-by-step approach:

1. Convert lbm to force (weight):

Since the mass is given in pounds mass (lbm), we need to find the weight in pounds force (lbf). We use the relationship between mass and weight:

W=mgW = m \cdot g

where:

  • m=5lbmm = 5 \, \text{lbm} (mass),
  • g=32.174ft/s2g = 32.174 \, \text{ft/s}^2 (acceleration due to gravity),
  • WW is the weight in pounds-force (lbf).

Since 1 lbm corresponds to 1 lbf under Earth’s gravity, the weight of the object is:

W=5lbfW = 5 \, \text{lbf}

2. Break the weight into components:

The weight of the object can be broken into two components:

  • Parallel to the incline (down the slope): W=Wsin(θ)W_{\parallel} = W \cdot \sin(\theta)
  • Perpendicular to the incline: W=Wcos(θ)W_{\perp} = W \cdot \cos(\theta)

where θ=15\theta = 15^\circ.

W=5sin(15)W_{\parallel} = 5 \cdot \sin(15^\circ) W50.2588=1.294lbfW_{\parallel} \approx 5 \cdot 0.2588 = 1.294 \, \text{lbf}

W=5cos(15)W_{\perp} = 5 \cdot \cos(15^\circ) W50.9659=4.8295lbfW_{\perp} \approx 5 \cdot 0.9659 = 4.8295 \, \text{lbf}

3. Compare with the applied force (P):

The applied force P=2.3lbfP = 2.3 \, \text{lbf} is acting to prevent the object from sliding down the incline. To maintain equilibrium, PP must balance the component of the weight acting parallel to the incline, WW_{\parallel}.

Since W=1.294lbfW_{\parallel} = 1.294 \, \text{lbf} and the applied force P=2.3lbfP = 2.3 \, \text{lbf}, we see that:

P>WP > W_{\parallel}

Conclusion:

The force PP is greater than the component of the weight acting down the incline, so the mass will not slide down. However, because P>WP > W_{\parallel}, the force PP is more than necessary to keep the mass in equilibrium, meaning it would push the object up the incline if friction or other resistive forces are not considered.

Would you like further details or have any questions?


Here are 5 related questions for further exploration:

  1. How does friction affect equilibrium in this scenario?
  2. What happens if the angle of the incline increases?
  3. How would equilibrium change if the applied force PP decreases?
  4. What if the mass of the object was doubled?
  5. How does gravitational acceleration impact this problem on different planets?

Tip: Always break forces into components parallel and perpendicular to an inclined plane to analyze motion or equilibrium effectively.

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Math Problem Analysis

Mathematical Concepts

Statics
Inclined Plane
Force Analysis
Trigonometry

Formulas

W = m * g
W_parallel = W * sin(θ)
W_perpendicular = W * cos(θ)

Theorems

Equilibrium Condition (P > W_parallel)

Suitable Grade Level

Grades 10-12