Math Problem Statement

A pushing force of F, at angle theta is applied to a box of mass m. If the box moves at a constant velocity, what is the coefficient of Friction in terms of f, m, and theta

Solution

If a box is moving at a constant velocity, the net force acting on it must be zero. This means that the pushing force and frictional force must balance each other out.

Let's break down the forces acting on the box:

  1. Pushing Force: A force FF is applied at an angle θ\theta to the horizontal. The force can be resolved into two components:

    • Horizontal component: Fx=FcosθF_x = F \cos \theta
    • Vertical component: Fy=FsinθF_y = F \sin \theta
  2. Normal Force: The normal force NN is the force exerted by the surface perpendicular to the box. It is affected by both the weight of the box and the vertical component of the pushing force. The normal force is given by: N=mgFsinθN = mg - F \sin \theta where mgmg is the gravitational force (weight) acting downward.

  3. Frictional Force: The frictional force ffrictionf_{\text{friction}} opposes the motion and is given by: ffriction=μNf_{\text{friction}} = \mu N where μ\mu is the coefficient of friction.

Since the box moves at a constant velocity, the horizontal forces must balance out. Thus: Fcosθ=ffriction=μNF \cos \theta = f_{\text{friction}} = \mu N

Substituting the expression for the normal force NN into the equation: Fcosθ=μ(mgFsinθ)F \cos \theta = \mu (mg - F \sin \theta)

Now, solving for μ\mu (the coefficient of friction): μ=FcosθmgFsinθ\mu = \frac{F \cos \theta}{mg - F \sin \theta}

So, the coefficient of friction μ\mu in terms of FF, mm, and θ\theta is: μ=FcosθmgFsinθ\boxed{\mu = \frac{F \cos \theta}{mg - F \sin \theta}}

Would you like any details on this solution or further explanation?


Here are five related questions to consider:

  1. How does increasing the angle θ\theta affect the coefficient of friction?
  2. What happens to the coefficient of friction if the pushing force FF is increased?
  3. How would the scenario change if the surface was inclined instead of horizontal?
  4. How can this equation be applied to determine the required force to move an object at a constant velocity?
  5. What role does the mass mm play in the coefficient of friction when the applied force is at an angle?

Tip: Always resolve forces into their components when dealing with forces applied at an angle. This helps in understanding how each force contributes to the overall motion.

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Math Problem Analysis

Mathematical Concepts

Force and Motion
Friction
Vectors

Formulas

Horizontal and Vertical Components of Force
Frictional Force Formula
Normal Force Calculation

Theorems

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Suitable Grade Level

High School