Math Problem Statement

A flatbed truck is carrying a crate up a hill of angle of inclination θ = 7.00°, as the figure illustrates. The coefficient of static friction between the truck bed and the crate is μs = 0.310. Find the maximum acceleration that the truck can attain before the crate begins to slip backward relative to the truck.

Solution

We need to find the maximum acceleration amaxa_{\text{max}} that the truck can have before the crate starts slipping backward. This involves analyzing the forces acting on the crate and the role of friction. The key concepts are:

  1. The force due to static friction, fsf_s, which prevents the crate from slipping.
  2. The gravitational force components acting on the crate, specifically parallel (mgsinθmg \sin \theta) and perpendicular (mgcosθmg \cos \theta) to the incline.
  3. The maximum static friction is given by fsmax=μsFNf_s^{\text{max}} = \mu_s F_N, where μs\mu_s is the coefficient of static friction and FNF_N is the normal force.

Step-by-Step Analysis:

  1. Forces in the yy-direction (perpendicular to the incline): The normal force FNF_N balances the perpendicular component of the gravitational force: FN=mgcosθF_N = mg \cos \theta where:

    • mm is the mass of the crate,
    • gg is the acceleration due to gravity,
    • θ=7\theta = 7^\circ.
  2. Maximum static friction: The maximum static friction force is: fsmax=μsFN=μsmgcosθf_s^{\text{max}} = \mu_s F_N = \mu_s mg \cos \theta where μs=0.310\mu_s = 0.310.

  3. Forces in the xx-direction (parallel to the incline): The static friction force opposes the relative motion (slipping), while the gravitational force component mgsinθmg \sin \theta pulls the crate down the incline. To find the maximum acceleration amaxa_{\text{max}}, we set up the following equation of motion along the incline: ma=fsmaxmgsinθma = f_s^{\text{max}} - mg \sin \theta Substituting fsmaxf_s^{\text{max}} into this equation: ma=μsmgcosθmgsinθma = \mu_s mg \cos \theta - mg \sin \theta Simplifying, the mass mm cancels out: a=g(μscosθsinθ)a = g (\mu_s \cos \theta - \sin \theta)

Substituting the values:

  • g=9.81m/s2g = 9.81 \, \text{m/s}^2,
  • μs=0.310\mu_s = 0.310,
  • θ=7\theta = 7^\circ.

I'll now calculate the maximum acceleration amaxa_{\text{max}}.The maximum acceleration amaxa_{\text{max}} that the truck can attain before the crate begins to slip backward is approximately 1.82m/s21.82 \, \text{m/s}^2.

Would you like further clarification or details on any part of the process?

Here are some related questions you might find helpful:

  1. How does the angle of inclination affect the maximum acceleration?
  2. What would happen if the coefficient of friction were higher?
  3. How does the mass of the crate impact the maximum acceleration?
  4. How is the frictional force related to the normal force?
  5. What if the truck were moving downhill instead of uphill?

Tip: When solving problems involving inclined planes, always break forces into components parallel and perpendicular to the incline for simpler calculations.

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Math Problem Analysis

Mathematical Concepts

Statics
Inclined Plane
Newton's Laws
Friction

Formulas

Normal force: FN = mg cos θ
Maximum static friction: f_s_max = μ_s FN
Newton's second law: ma = f_s_max - mg sin θ
Final formula for maximum acceleration: a = g(μ_s cos θ - sin θ)

Theorems

Newton's Second Law of Motion
Static Friction Theorem

Suitable Grade Level

Grades 11-12 (High School Physics)