A crate is placed on an inclined board as shown in the figure below. One end of the board is hinged so that the angle θ is adjustable. If the coefficient of static friction between the crate and the board is μS = 0.37, what is the value of θ at which the crate just begins to slip?
The equation of motion (in vector form)
according to the 2nd Newton’s Law is
vector(m•g) +vector F(fr)+ vector N =0.
Projections on the x- and y-axes
x: m•g•sinα - F(fr) = 0,
y: - m•g•cosα + N = 0.
F(fr) = μ•N = μ •m•g•cosα,
m•g•sinα - μ •m•g•cosα =0,
tanα = μ,
α =arctan μ =20.3º
To find the value of θ at which the crate just begins to slip, we need to consider the forces acting on the crate.
Let's label the forces:
- The weight of the crate acts straight downward and has a magnitude of mg, where m is the mass of the crate and g is the acceleration due to gravity.
- The normal force, FN, acts perpendicular to the incline.
- The frictional force, Ff, acts parallel to the incline and opposes the motion of the crate.
The force of static friction can be calculated as: Ff = μS * FN, where μS is the coefficient of static friction.
Since the crate is just about to slip, the maximum possible static friction force is equal to the force pushing it downhill, which is the component of the weight parallel to the incline: Ff = m * g * sin(θ).
Setting these two expressions for Ff equal to each other, we have: μS * FN = m * g * sin(θ).
The normal force is equal to the component of the weight perpendicular to the incline: FN = m * g * cos(θ).
Substituting this expression for FN into the equation above, we get:
μS * (m * g * cos(θ)) = m * g * sin(θ).
Dividing both sides of the equation by m * g, we have:
μS * cos(θ) = sin(θ).
Rearranging the equation, we get:
μS / sin(θ) = 1 / cos(θ).
Using the trigonometric identity sin(θ) = 1 / csc(θ) and cos(θ) = 1 / sec(θ), the equation becomes:
μS / (1 / csc(θ)) = 1 / (1 / sec(θ)).
Simplifying, we get:
μS * csc(θ) = sec(θ).
Recall that csc(θ) = 1 / sin(θ) and sec(θ) = 1 / cos(θ), so the equation becomes:
μS / sin(θ) = 1 / cos(θ).
This is the same equation we started with, so we can conclude that:
μS * FN = m * g * sin(θ).
Since μS = 0.37, we can substitute this value into the equation:
0.37 * (m * g * cos(θ)) = m * g * sin(θ).
Canceling out the mass and acceleration due to gravity:
0.37 * cos(θ) = sin(θ).
To solve for θ, we can take the inverse tangent of both sides of the equation:
θ = tan^(-1)(0.37).
Using a calculator, we find that θ is approximately 21.8 degrees.
Therefore, the value of θ at which the crate just begins to slip is approximately 21.8 degrees.
To determine the value of θ at which the crate just begins to slip, we can analyze the forces acting on the crate. Let's break it down step-by-step:
1. Identify the forces acting on the crate:
- The weight of the crate (mg), which acts vertically downward.
- The normal force (N) exerted by the board on the crate, which acts perpendicular to the surface of the board.
- The force of static friction (fS) between the crate and the board, which acts parallel to the surface of the board.
2. Determine the components of the weight force:
- The weight force can be broken down into two components: one parallel to the board (mg*sinθ) and one perpendicular to the board (mg*cosθ).
- In this case, the component parallel to the board (mg*sinθ) will be pertinent to our analysis.
3. Set up an equation for the condition of impending slip:
- For the crate to begin slipping, the force of static friction must reach its maximum value, which is given by:
fS_max = μS * N, where μS is the coefficient of static friction between the crate and the board, and N is the normal force.
4. Express the normal force in terms of the weight force component perpendicular to the board:
- Since the normal force (N) counteracts the perpendicular component of the weight, it can be expressed as N = mg*cosθ.
5. Substitute the expression for the normal force into the equation for the maximum static friction:
- fS_max = μS * mg*cosθ
6. Equate the maximum static friction to the weight force component parallel to the board:
- fS_max = mg*sinθ
7. Set up the equation for the condition of impending slip:
- Equating the two expressions for fS_max, we have: μS * mg*cosθ = mg*sinθ
8. Solve for θ:
- Dividing both sides of the equation by mg: μS * cosθ = sinθ
- Rearranging the equation: μS * cosθ - sinθ = 0
At this point, we can solve for θ either graphically or numerically.
- Graphical method: Plot the left-hand side of the equation as a function of θ and find the value(s) of θ where it crosses zero.
- Numerical method: Use an iterative approach like the Newton-Raphson method to find the root(s) of the equation.
Using either method, you can find the value of θ at which the crate just begins to slip.