Math Problem Statement

Solve this friction problem

Determine the maximum force P that can be applied to the 250 lb crate with a center of gravity at G, considering the coefficient of static friction is 0.4.

Solution

To solve this problem, let’s break down the forces and analyze the static equilibrium conditions. Here’s a step-by-step guide.

Problem Information:

  1. Weight of crate, W=250W = 250 lbs
  2. Distance to center of gravity GG from base AA: 2.52.5 ft vertically.
  3. Dimensions of crate: 4.54.5 ft tall and 3.03.0 ft wide.
  4. Coefficient of static friction μs=0.4\mu_s = 0.4.
  5. Goal: Find the maximum horizontal force PP that can be applied without causing movement.

Steps for Solution

  1. Identify forces and moments:

    • The crate is in static equilibrium, so the sum of all forces and moments must equal zero.
    • There are three forces acting on the crate:
      • The applied force PP horizontally at the upper left corner.
      • The weight W=250W = 250 lbs acting downwards at point GG.
      • The normal force NN acting upwards at point AA.
    • There’s also a frictional force ff at point AA, which resists the horizontal motion and is given by f=μsNf = \mu_s N.
  2. Set up equilibrium equations:

    Sum of Forces in the Vertical Direction ( ΣFy=0\Sigma F_y = 0 ):

    N=W=250 lbsN = W = 250 \text{ lbs}

    Sum of Forces in the Horizontal Direction ( ΣFx=0\Sigma F_x = 0 ):

    f=PμsN=Pf = P \Rightarrow \mu_s N = P P=μs×W=0.4×250=100 lbsP = \mu_s \times W = 0.4 \times 250 = 100 \text{ lbs}

  3. Moment about Point AA: To ensure the crate doesn’t tip over, take moments about point AA.

    Moment Equation:

    • The clockwise moment caused by PP is: MP=P×4.5 ftM_P = P \times 4.5 \text{ ft}
    • The counterclockwise moment due to the weight WW at the center of gravity GG is: MW=W×1.5 ftM_W = W \times 1.5 \text{ ft} Setting the moments equal for static equilibrium: P×4.5=250×1.5P \times 4.5 = 250 \times 1.5 P=250×1.54.5=3754.5=83.33 lbsP = \frac{250 \times 1.5}{4.5} = \frac{375}{4.5} = 83.33 \text{ lbs}
  4. Conclusion: The limiting factor for the applied force PP is the tipping moment, giving us: P=83.33 lbsP = 83.33 \text{ lbs}

So, the maximum force PP that can be applied to the crate without causing movement is 83.33 lbs.

Would you like further explanation or have any questions?


Related Questions:

  1. How does the coefficient of friction affect the maximum force PP in this scenario?
  2. What would happen if the crate's weight increased? How would that change the maximum force PP?
  3. How would the result change if the force PP were applied lower on the crate?
  4. What role does the center of gravity play in determining tipping risk in static equilibrium problems?
  5. How does the width of the crate affect the tipping point calculation?

Tip:

When solving static equilibrium problems, always start by identifying all forces, establishing coordinate axes, and writing down equilibrium equations for both forces and moments.

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Math Problem Analysis

Mathematical Concepts

Statics
Friction
Equilibrium
Moments

Formulas

f = μ_s * N (frictional force)
ΣF_x = 0 (horizontal force balance)
ΣF_y = 0 (vertical force balance)
ΣM_A = 0 (moment equilibrium at point A)

Theorems

Static Equilibrium: ΣF = 0 and ΣM = 0
Moment about a point

Suitable Grade Level

Undergraduate Engineering (Statics or Physics)