Question

A box slides down a plank of length d that makes an angle of θ with the horizontal as shown. μ_k is the kinetic coefficient of friction and μ_s is the static coefficient of friction. Enter an expression for the minimum angle θ (in degrees) the box will begin to slide.

Answer 1

Balancing the forces,

mg sin(θ) - u_s mg cos(θ) = 0

Sin(θ) = u_s cos(θ)

u_s = tan(θ)

θ = tan^-1(μ_s)

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Answer 2

To find the minimum angle θ at which the box will begin to slide, we need to consider the forces acting on the box and set up an equation for the condition of impending motion (the point at which the box is just about to start sliding). At this point, the static friction force is at its maximum value, which is equal to the static coefficient of friction (μs) multiplied by the normal force (N).

Let's analyze the forces acting on the box:

1. Weight (W): The weight of the box acts vertically downward and is given by W = mg, where m is the mass of the box and g is the acceleration due to gravity.

2. Normal Force (N): The normal force acts perpendicular to the surface of the plank and is equal in magnitude but opposite in direction to the vertical component of the weight. Therefore, N = mg cos(θ).

3. Force along the plank (Fpar): The component of the weight parallel to the plank is given by Fpar = mg sin(θ).

Now, the condition for impending motion is given by:

Fpar = μs * N

Substitute the expressions for Fpar and N:

mg sin(θ) = μs * mg cos(θ)

Simplify and solve for θ:

sin(θ) = μs * cos(θ)

Now, divide both sides by cos(θ):

tan(θ) = μs

Finally, solve for θ:

θ = arctan(μs)

Now, to find the minimum angle θ in degrees, plug in the value of the static coefficient of friction (μs). The arctan function gives the result in radians, so convert it to degrees:

θ (degrees) = arctan(μs) * (180/π)

where π is approximately 3.14159.

So, the expression for the minimum angle θ (in degrees) at which the box will begin to slide is:

θ (degrees) = arctan(μs) * (180/π)