Math Problem Statement

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

Solution

To determine whether the force PP (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=mgW = m \cdot g where m=5 lbmm = 5 \text{ lbm} and g=32.2 ft/s2g = 32.2 \text{ ft/s}^2 (standard acceleration due to gravity in imperial units).

    In this case, the weight force WW is: W=532.2=161 lbm-ft/s2=5 lbf(as lbf and lbm cancel out with gravity constant)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=Wsin(θ)=5sin(15)=50.25881.294 lbfW_{\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=Wcos(θ)=5cos(15)=50.96594.829 lbfW_{\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 PP (2.3 lbf) must counterbalance the weight component acting parallel to the incline, which is W=1.294 lbfW_{\parallel} = 1.294 \text{ lbf}.

  • If P=WP = W_{\parallel}, the mass will stay in equilibrium.
  • If P>WP > W_{\parallel}, the mass will move up the incline.
  • If P<WP < W_{\parallel}, the mass will move down the incline.

Step 3: Compare the forces

Here, P=2.3 lbfP = 2.3 \text{ lbf} and W=1.294 lbfW_{\parallel} = 1.294 \text{ lbf}.

Since P>WP > 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?


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.

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Math Problem Analysis

Mathematical Concepts

Forces on an incline
Decomposition of forces
Equilibrium

Formulas

Weight (W) = m * g
Parallel force component: W_parallel = W * sin(θ)
Perpendicular force component: W_perpendicular = W * cos(θ)

Theorems

Newton's First Law of Motion

Suitable Grade Level

Grades 10-12 (High School Physics)