Math Problem Statement

A crate of mass M = 100kg is on a rocking boat, tilted at an angle θ. Part (a) asks for a free-body diagram, and part (b) asks for the angle at which the box begins to slide, given the coefficient of static friction μ_s = 0.55.

Solution

The problem in the image is focused on friction force and involves a crate on a tilted boat due to a wave. The setup is a wooden crate on a surface inclined at an angle θ\theta, and the crate has a mass of M=100kgM = 100 \, \text{kg}. There are two parts to this question:

Part (a):

Draw a free-body diagram of the crate.

We need to represent all forces acting on the crate. These forces are:

  1. Weight (WW): The gravitational force pulling the crate downwards, which is equal to MgMg, where gg is the gravitational acceleration (9.8m/s29.8 \, \text{m/s}^2).

    • The weight can be broken into two components:
      • W=Mgcos(θ)W_{\perp} = Mg \cos(\theta): Perpendicular to the inclined plane.
      • W=Mgsin(θ)W_{\parallel} = Mg \sin(\theta): Parallel to the inclined plane, pulling the crate down the slope.
  2. Normal Force (NN): The reaction force exerted by the surface, which is perpendicular to the inclined plane and equal in magnitude to WW_{\perp}, so N=Mgcos(θ)N = Mg \cos(\theta).

  3. Frictional Force (FfF_f): The force resisting the motion, acting parallel to the plane in the opposite direction of the crate's motion. The maximum frictional force is given by: Ff=μsN=μsMgcos(θ)F_f = \mu_s N = \mu_s Mg \cos(\theta) where μs=0.55\mu_s = 0.55 is the coefficient of static friction.

Part (b):

At what angle θ\theta will the box begin to slide?

To determine the angle at which the box begins to slide, we need to consider the balance of forces parallel to the incline.

  • The force that tends to move the box down the slope is Mgsin(θ)Mg \sin(\theta).
  • The maximum static friction force is Ff=μsMgcos(θ)F_f = \mu_s Mg \cos(\theta).

The box will begin to slide when the downslope force exceeds the maximum static friction: Mgsin(θ)=μsMgcos(θ)Mg \sin(\theta) = \mu_s Mg \cos(\theta)

We can cancel MgMg from both sides and rearrange: tan(θ)=μs\tan(\theta) = \mu_s

Now, solve for θ\theta: θ=tan1(μs)=tan1(0.55)28.81\theta = \tan^{-1}(\mu_s) = \tan^{-1}(0.55) \approx 28.81^\circ

Thus, the box will begin to slide when the angle θ\theta exceeds approximately 28.8128.81^\circ.


Would you like more details on any part of the explanation? Here are a few follow-up questions to consider:

  1. What is the significance of the normal force in this problem?
  2. How does the coefficient of friction affect the angle at which the box will slide?
  3. What would happen if the coefficient of static friction were increased or decreased?
  4. How would the free-body diagram change if the box was moving instead of stationary?
  5. Can we extend this problem to account for dynamic (kinetic) friction after the box starts sliding?

Tip: When working on problems involving inclined planes, always break the forces into components parallel and perpendicular to the incline for easier analysis.

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Math Problem Analysis

Mathematical Concepts

Physics
Statics
Friction
Inclined Plane

Formulas

Weight (W) = Mg
Normal Force (N) = Mg * cos(θ)
Frictional Force (F_f) = μ_s * N = μ_s * Mg * cos(θ)
tan(θ) = μ_s

Theorems

Newton's Laws of Motion
Static Friction Force

Suitable Grade Level

Grades 11-12 or Introductory University Physics