Forces and Particles Equilibrium
Moment of a Force about a Line
Scalars, vectors and forces
A scalar is a quantity that is fully described by a single numerical value and its associated unit, devoid of any directional component. Scalars are used to represent measurements that have magnitude but no direction, such as temperature, mass, and time. Unlike vectors, which possess both magnitude and direction, scalars simplify the analysis of physical systems by focusing solely on the size of a quantity. Mathematical operations involving scalars, such as addition, subtraction, multiplication, and division, follow straightforward rules that facilitate the manipulation of physical quantities in a range of scientific and engineering applications.
A scalar is represented in standard text without boldface. For example, a scalar s can be written as:
s = 5
where s denotes the scalar quantity with a numerical value of 5.
A vector is a quantity characterized by both magnitude and direction, represented as an ordered pair or triplet in space. Vectors are essential for describing physical phenomena that involve directionality, such as velocity, force, and acceleration. Unlike scalars, which only convey size, vectors incorporate both the amount and the orientation of a quantity. Mathematical operations with vectors include addition, subtraction, and scalar multiplication, which adhere to specific rules and often involve geometric interpretations such as dot products and cross products. These operations are crucial for accurately modeling and analyzing the behavior of physical systems where direction plays a significant role.
Vectors are represented in boldface to distinguish them from scalars. For example, a vector \mathbf{v} is written as:
\mathbf{v} = \begin{bmatrix} v_x \\ v_y \\ v_z \end{bmatrix}
where v_x, v_y, and v_z are the components of the vector in a three-dimensional space.
A force is a fundamental interaction that causes an object to accelerate or change its state of motion. It is a vector quantity, meaning it has both magnitude and direction. In the International System of Units (SI), force is measured in newtons (\text {N}). In the Imperial system, force is measured in pounds-force (\text{lbf}). The conversion between these units is given by 1 \text{ lbf} = 4.448 \text{ N}.
Equilibrium of forces
Newton’s First Law of Motion states:
An object at rest will remain at rest, and an object in motion will continue in motion with a constant velocity, unless acted upon by a net external force.
if \mathbf{F}_{\text{net}} = 0, then \mathbf{v} = \text{constant}, where \mathbf{F}_{\text{net}} is the net external force and \mathbf{v} is the velocity of the object.
Therefore, for be in equilibrium the sum of all forces acting on a body must be zero:
\sum \mathbf{F}_i = 0
In two dimensions and Cartesian coordinates, this is expressed as:
\begin{aligned} \sum F_x & = 0 \\ \sum F_y & = 0 \end{aligned}
where F_x and F_y are the components of the forces along the x and y axes, respectively.
In three dimensions, it is expressed as:
\begin{aligned} \sum F_x & = 0 \\ \sum F_y & = 0 \\ \sum F_z & = 0 \end{aligned}
where F_x, F_y, and F_z are the components of the forces along the x, y, and z axes, respectively.
The equations for static equilibrium are valid for axes that are mutually orthogonal. For example, in polar coordinates, for 2D systems, the equilibrium conditions are:
\begin{aligned} \sum F_r & = 0 \\ \sum F_\theta &= 0 \end{aligned}
where F_r is the force component in the radial direction and F_\theta is the force component in the tangential direction.
So there are 2 or 3 independent equations balancing the forces in each orthogonal direction.
Free Body Diagram
The Free Body Diagram (FBD) method is a technique in static engineering used to analyze forces acting on a system. In this approach, we isolate a body or a system, removing it from its surroundings, and represent all the external forces, moments, and reactions acting on it. The FBD includes all known forces, such as gravitational, normal, frictional, and any applied loads, as well as unknown reaction forces at points of contact or support. Each force is represented as a vector with a specified direction and magnitude, either known or to be determined.
In static analysis, the FBD method is primarily used to ensure that the system is in equilibrium, meaning the sum of forces and acting on the body must be zero. The mathematical formulation of this is expressed as:
\sum \mathbf{F} = 0
where \mathbf{F} represents forces in two or three dimensions. These equations can be resolved to determine unknown forces or moments, making FBD an important for solving structural and mechanical problems.
The method is versatile and applicable across various fields and simplifies complex interactions by breaking them into manageable components, enabling engineers to solve for unknowns in static equilibrium problems.
I created some Python classes and function to assist in the drawing of these diagrams.
The first step is to identify the body of interest.
In the example we choose the point where the weight is applied and the cable are attached and we have:
- the weight M;
- the tension of the cable 1 acting with a \theta_1 angle on the left;
- the tension of the cable 2 acting with a \theta_2 angle on the right.
Once the body is identified it is possible to draw the particle as a point free of constraints, and then apply the external forces acting on it; in this case we have:
- the weight Mg;
- the tension T_1 acting on the left;
- the tension T_2 acting on the right.
The last step is to add the dimensions.
Moment of a force
The moment of a force (or torque) is the measure of the tendency of a force to rotate an object about a point or axis.
Mathematically, it is defined as the cross product of the position vector (\mathbf{r}) from the point of rotation to the point where the force is applied, and the force vector (\mathbf{F}):
\mathbf{M} = \mathbf{r} \times \mathbf{F}
The magnitude of the moment is given by:
M = r F \sin(\theta)
where: - r is the distance from the point or axis to the point of force application, - F is the magnitude of the applied force, - \theta is the angle between the position vector and the force vector.
In a 2D problem where the force lies in a plane, such as the xy-plane, the moment (or torque) generated by the force will always be perpendicular to that plane. Moments will always be along the axis perpendicular to this plane (e.g., the z-axis for the xy-plane).
This means that in 2D problems, there is only one possible direction for the moment, which simplifies the problem to a scalar value indicating the magnitude of the moment along the perpendicular axis (e.g., the z-axis).
Calculation with the vector method
To find the moment of a force using the vector method, it is possible to use the cross product between the position vector \mathbf{r} and the force vector \mathbf{F}.
The moment (or torque) \mathbf{M} is given by:
\mathbf{M} = \mathbf{r} \times \mathbf{F}
where:
- \mathbf{r} = \begin{bmatrix} x_r & y_r & z_r \end{bmatrix}^T is the position vector from the point of rotation to where the force is applied.
- \mathbf{F} = \begin{bmatrix} F_x & F_y & F_z \end{bmatrix}^T is the force vector.
The cross product in matrix form is:
\mathbf{M} = \begin{vmatrix} \mathbf i & \mathbf j & \mathbf k \\ x_r & y_r & z_r \\ F_x & F_y & F_z \end{vmatrix}
Expanding this determinant gives:
\mathbf{M} = \left(y_r F_z - z_r F_y \right) \mathbf i - \left(x_r F_z - z_r F_x \right) \mathbf j + \left(x_r F_y - y_r F_x \right) \mathbf k
For a 3D case, assuming:
\begin{aligned} & \mathbf{r} = \begin{bmatrix} 2 & 3 & 4 \end{bmatrix}^T \\ & \mathbf{F} = \begin{bmatrix} 5 & -2 & 1 \end{bmatrix}^T \end{aligned}
Then, the moment \mathbf{M} is:
\begin{aligned} \mathbf{M} & = \left( 3 \times 1 - 4 \times (-2) \right) \mathbf i - \left( 2 \times 1 - 4 \times 5 \right) \mathbf j + \hat{k} \left( 2 \times (-2) - 3 \times 5 \right) \mathbf k \\ & = (3 + 8) \mathbf i - (2 - 20) \mathbf j + (-4 - 15) \mathbf k \\ & = (11) \mathbf i - (-18) \mathbf j + (-19) \mathbf k \end{aligned}
Thus, the moment vector is:
\mathbf{M} = \begin{bmatrix} 11 & 18 & -19 \end{bmatrix}^T
In 2D (e.g., the xy-plane), z_r = F_z = 0, and the moment will only have a z-component:
Assume:
\begin{aligned} & \mathbf{r} = \begin{bmatrix} 3 & 4 \end{bmatrix}^T \\ &\mathbf{F} = \begin{bmatrix} 6 & -1 \end{bmatrix}^T \end{aligned}
Then the moment \mathbf{M} is:
M_z = 3 \times (-1) - 4 \times 6 = -3 - 24 = -27
The moment is along the z-axis:
\mathbf{M} = \begin{bmatrix} 0 & 0 & -27 \end{bmatrix}^T
This is the vector form of the moment for both 2D and 3D cases using the cross product method.
Varignon theorem
Varignon’s Theorem states that the moment of a force about any point is equal to the sum of the moments of the components of the force about the same point.
Mathematically:
\mathbf M_O = \mathbf r \times \mathbf F = \mathbf r \times \mathbf F_x + \mathbf r \times \mathbf F_y + \mathbf z \times \mathbf k
Where: - \mathbf M_O is the moment of the force about point O - \mathbf r is the position vector from point O to the point of application of the force - \mathbf F is the applied force - \mathbf F_x, \mathbf F_y and \mathbf F_z are the components of the force in the x, y and and z directions.
In vector form:
\mathbf{M}_O = \mathbf{r} \times (F_x \mathbf{i}) + \mathbf{r} \times (F_y \mathbf{j}) + \mathbf{r} \times (F_z \mathbf{k})
To prove Varignon’s theorem in 3D, we start by considering a force \mathbf{F} applied at a point with position vector \mathbf{r} relative to point O, and we want to calculate the moment of the force \mathbf{M}_O about point O.
The components of the moment vector \mathbf{M}_O was computed previously:
\mathbf{M}_O = \left( (r_y F_z - r_z F_y) \right) \mathbf{i} - \left( (r_x F_z - r_z F_x) \right) \mathbf{j} + \left( (r_x F_y - r_y F_x) \right) \mathbf{k} Now, express \mathbf{F} as the sum of its components along the x, y, and z axes:
\mathbf{F} = F_x \mathbf{i} + F_y \mathbf{j} + F_z \mathbf{k}
The moment \mathbf{M}_O can be written as the sum of the moments of each component of \mathbf{F}:
\mathbf{M}_O = \mathbf{r} \times (F_x \mathbf{i}) + \mathbf{r} \times (F_y \mathbf{j}) + \mathbf{r} \times (F_z \mathbf{k})
By calculating each term individually, we see that the total moment is the sum of the moments of the components. This verifies Varignon’s theorem in 3D.
Moment of a force about a line
To compute a moment of a force about a line, the first step is to calculate the moment about a point O:
\mathbf{M}_O = \mathbf{r} \times \mathbf{F}
Let \mathbf{u} be the unit vector along the axis (line) of interest. The scalar projection of \mathbf{M}_O onto the axis (direction of \mathbf{u}) is:
M_{\text{axis}} = \mathbf{u} \cdot \mathbf{M}_O = \mathbf{u} \cdot (\mathbf{r} \times \mathbf{F})
This gives the scalar magnitude of the moment about the axis, but we want the vector component of the moment along the axis. To obtain the vector component of the moment along the axis, we need to project the scalar magnitude M_{\text{axis}} onto the unit vector \mathbf{u}:
\mathbf{M}_{\text{axis}} = M_{\text{axis}} \mathbf{u} = \left[ \mathbf{u} \cdot (\mathbf{r} \times \mathbf{F}) \right] \mathbf{u}
Moment due to a couple
A couple consists of two equal and opposite forces, \mathbf{F}_1 and \mathbf{F}_2, separated by a distance. The moment of a couple is the torque generated by these two forces, and it is independent of the reference point.
Let:
- \mathbf{F}_1 = \mathbf{F} be one of the forces.
- \mathbf{F}_2 = -\mathbf{F} be the opposite force.
- \mathbf{r}_1 and \mathbf{r}_2 be the position vectors of the points of application of \mathbf{F}_1 and \mathbf{F}_2, respectively.
The moment of a couple, \mathbf{M}_{\text{couple}}, is defined as the sum of the moments produced by each force about a point:
\mathbf{M}_{\text{couple}} = \mathbf{r}_1 \times \mathbf{F}_1 + \mathbf{r}_2 \times \mathbf{F}_2
Since \mathbf{F}_2 = -\mathbf{F}_1 = -\mathbf{F}, the forces are equal in magnitude but opposite in direction, the moment becomes:
\mathbf{M}_{\text{couple}} = \mathbf{r}_1 \times \mathbf{F} + \mathbf{r}_2 \times (-\mathbf{F}) = (\mathbf{r}_1 - \mathbf{r}_2) \times \mathbf{F} = \mathbf{d} \times \mathbf{F}
The moment of the couple is the cross product of the separation vector \mathbf{d} and the force \mathbf{F}, and it produces a pure rotational effect independent of the reference point.
Lets consider an example:
\begin{aligned} & \mathbf A = (-1,0) \\ & \mathbf B = (4,0) \end{aligned}
And \mathbf F is on a {}^3{\mskip -5mu/\mskip -3mu}_4 slope.
Computing the moments around the origin \mathbf O:
\begin{aligned} & \mathbf M_{OA} = \mathbf OA \times \mathbf F = -\mathbf i \times \left(\frac{3}{5} \mathbf F \mathbf i + \frac{4}{5} \mathbf F \mathbf M_{OA} = \mathbf j \right) = -\frac{3}{5}\mathbf F \mathbf k \\ & \mathbf OB \times \mathbf F = 4 \mathbf i \times \left(-\frac{3}{5} \mathbf F \mathbf i - \frac{4}{5} \mathbf F \mathbf j \right) = -\frac{12}{5} \mathbf F \mathbf k \\ & \mathbf M = \mathbf M_{OA} + \mathbf M_{OA} = -5\mathbf F \mathbf k \end{aligned}
This is consistent with the formula:
\mathbf{M}_{\text{couple}} = \mathbf{d} \times \mathbf{F} = -5\mathbf F \mathbf k
And does not depends from the reference point.
Static equilibrium equations
To account for the moments in static equilibrium, the sum of the moments about any point must also be zero in addition to the sum of forces derived above. The moment of a force is the cross product of the position vector and the force vector. In two and three dimensions, the conditions for static equilibrium are:
In two dimensions, the forces and moments yield three equations of equilibrium:
\begin{array}{lll} \sum F_x = 0 & \sum F_y = 0 & \sum M_z = 0 \end{array}
where M_z is the moment about the z-axis (perpendicular to the plane).
In three dimensions, both forces and moments yield six equations of equilibrium:
\begin{array}{lll} \sum F_x = 0 & \sum F_y = 0 & \sum F_z = 0 \\ \sum M_x = 0 & \sum M_y = 0 & \sum M_z = 0 \end{array}
where M_x, M_y, and M_z are the moments about the x-, y-, and z-axes, respectively.
These six equations are necessary and sufficient to ensure that the object is in complete static equilibrium.
It is not necessary to use Cartesian coordinates. It is sufficient to use any set of orthogonal directions. The equations of equilibrium can be written in terms of forces and moments along these orthogonal directions. Additionally, the six equilibrium conditions do not need to be strictly separated into three force and three moment equations. They can be a combination of forces and moments, as long as the chosen directions are independent and cover all degrees of freedom.
Principle of transmissibility
The principle of transmissibility states that the effect of a force on a rigid body is unchanged if the force is moved along its line of action. This means that as long as a force is applied in the same direction and along the same line, the point of application can be shifted without affecting the external motion (translation and rotation) of the rigid body.
In essence, the force produces the same net effect (both in terms of forces and moments) on the body, regardless of where it is applied along its line of action. However, this principle holds true only for rigid bodies, as it does not account for internal stresses or deformations within the object.
Equivalent systems
Equivalent Systems refer to two or more force systems that produce the same external effect on a rigid body. This means they create the same resultant force and the same resultant moment about any point on the body. Even if the forces are applied at different points or have different magnitudes and directions, they can still be equivalent if their combined effect (resultant force and moment) is identical.
For example, a force acting at a certain point can be replaced by an equivalent force and a couple (moment) at a different point, as long as the net effect on the body remains the same.
Consider two systems, \text{System A} and \text{System B}, that act on a rigid body. Let \mathbf{F}_A and \mathbf{F}_B represent the resultant forces of systems A and B, respectively, and \mathbf{M}_A and \mathbf{M}_B represent the resultant moments about a point O.
For the systems to be equivalent, they must satisfy two conditions.
The resultant forces of both systems must be equal:
\mathbf{F}_A = \mathbf{F}_B
The resultant moments about point O must also be equal:
\mathbf{M}_A = \mathbf{M}_B
Alternatively, if system A has a force \mathbf{F}_A applied at point P_A and system B has a force \mathbf{F}_B applied at point P_B, the moments generated by these forces about point O can be related by:
\begin{aligned} \mathbf{M}_A & = \mathbf{r}_{OA} \times \mathbf{F}_A \\ \mathbf{M}_B & = \mathbf{r}_{OB} \times \mathbf{F}_B \end{aligned}
where \mathbf{r}_{OA} and \mathbf{r}_{OB} are the position vectors from point O to points P_A and P_B, respectively.
For equivalence, the forces and moments must satisfy:
\begin{aligned} & \mathbf{F}_A = \mathbf{F}_B \\ & \mathbf{r}_{OA} \times \mathbf{F}_A = \mathbf{r}_{OB} \times \mathbf{F}_B \end{aligned}
It is possible to do an example, considering a cantilever beam to which there is a force \mathbf F = 200\mathbf i + 350\mathbf j acting at point A and a moment \mathbf M = -100 \mathbf k acting on point B and try to find the equivalent system at point C. This is System A, and we want to compute the equivalent force and moment for system B.
We apply the balance of forces:
\begin{aligned} \Sigma F_x & = 200 \\ \Sigma F_y & = 350 \end{aligned}
Since the forces are the same in both systems, we have:
\begin{aligned} \Sigma F_x |_A & = \Sigma F_x |_B = 200 \\ \Sigma F_y |_A & = \Sigma F_y |_B = 350 \end{aligned}
Now, we apply the balance of moments about point C. The moment at C is composed of the moment at point A plus the contribution of the force \mathbf{F} acting at a distance from C.
The moment at point A is:
M_A = -100
The position vectors are:
\begin{aligned} \mathbf{r_A} & = (10, 2) \\ \mathbf{r_C} & = (3, 0) \end{aligned}
The distance between A and C is:
\mathbf{r_{CA}} = \mathbf{r_A} - \mathbf{r_C} = (7, 2)
Now, compute the moment contribution from the force:
M_{C, A} = r_{CA_x} F_y - r_{CA_y} F_x = 7 \cdot 350 - 2 \cdot 200 = 2450 - 400 = 2050
Thus, the total moment at C is:
M_C = M_A + M_{C, A} = -100 + 2050 = 1950
So, the equivalent system at point C has the same forces and a moment of 1950.
It is possible to combine the two forces into a single resultant force, we will compute the magnitude (modulus) of the force and the angle it makes with the x-axis.
The magnitude F_R of the resultant force is given by:
F_R = \sqrt{F_x^2 + F_y^2} = \sqrt{200^2 + 350^2} = \sqrt{162500} \approx 403
The angle \theta that the resultant force makes with the x-axis is given by:
\theta = \tan^{-1}\left(\frac{F_y}{F_x}\right) = \tan^{-1}\left(\frac{350}{200}\right) = \tan^{-1}(1.75) \approx 60.26^\circ
Coplanar forces
Coplanar forces are forces that lie in the same plane. When dealing with coplanar force systems, the forces are restricted to two dimensions (usually in the xy-plane with an opportune choice of axis), and the equations of equilibrium simplify because there are no components in the z direction.
Coplanar forces can be reduced to a single resultant force, acting at an appropriate point, which produces the same effect as the original system of forces. This is possible due to the fact that all the forces lie in the same plane, allowing their vector sum and moment to fully describe the system.
The system of coplanar forces can be reduced to a single resultant force \mathbf{F}_R, where:
\mathbf{F}_R = \sum \mathbf{F}_i
This resultant force has components:
\begin{aligned} & F_{Rx} = \sum F_x \\ & F_{Ry} = \sum F_y \end{aligned}
The total moment about any point O must account for the sum of the moments of all the forces:
M_O = \sum (\mathbf{r}_i \times \mathbf{F}_i)
where \mathbf{r}_i is the position vector of the point of application of the i-th force relative to O.
The original system can be replaced by:
- A single resultant force \mathbf{F}_R
- Acting at an appropriately chosen point such that the resultant moment is equal to the sum of the moments of the original forces.
The point where the resultant force should act can be found by ensuring that the moment produced by \mathbf{F}_R about the reference point O is the same as the original system’s total moment:
\mathbf{r}_{\text{eq}} \times \mathbf{F}_R = M_O
where \mathbf{r}_{\text{eq}} is the position vector of the point where the resultant force acts.
Therefore, the coplanar force system is reduced to a single resultant force acting at a specific point, providing the same overall effect on the body.
We can take the previous example and compute the coplanar force. We already calculated:
- Resultant force F_R = 403.11 acting at an angle \theta = 60.26^\circ with the x-axis.
- Moment about point C (from previous calculation): M_C = 1950.
The first step is to calculate the perpendicular distance d the distance from point C to the line of action of the resultant force F_R is given by:
M_C = F_R \cdot d
Solving for d:
d = \frac{M_C}{F_R} = \frac{1950}{403.11} \approx 4.84
The system can be reduced to a single resultant force F_R = 403.11 acting at a distance d \approx 4.84 units from point C, along the direction defined by the angle \theta = 60.26^\circ with the x-axis.
Let’s compute the equation of a line that is parallel to the line passing through point C(3, 0) and has a slope of \tan(60.26^\circ). We want this line to be at a perpendicular distance of d = 4.84 from point C(3, 0).
The slope m of the line is given by:
m = \tan(60.26^\circ) \approx 1.75
The equation of the line passing through point C(3, 0) with slope m = 1.75 is:
y = 1.75(x - 3) = 1.75x - 5.25
We need to find a parallel line to this one that is at a perpendicular distance of d = 4.84 from point C(3, 0).
The general formula for the perpendicular distance d from a point (x_1, y_1) to a line ax + by + c = 0 is:
d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}
For the line y = 1.75x - 5.25, we can write it in the form:
1.75x - y - 5.25 = 0
So, a = 1.75, b = -1, and c = -5.25.
Using the point (3, 0) and the perpendicular distance d = 4.84, we substitute into the distance formula:
4.84 = \frac{|1.75 \times 3 - 1 \times 0 - 5.25|}{\sqrt{1.75^2 + (-1)^2}} = \frac{|5.25 - 5.25|}{\sqrt{3.0625 + 1}} = \frac{|0 + c|}{2.015}
Thus, to find c (the offset from the original line):
4.84 \times 2.015 = |c| \approx 9.75
So, the possible values of c are:
c = 9.75 \quad \text{or} \quad c = -9.75
Given the original line equation y = 1.75x - 5.25, the parallel lines at the specified distance can be written as:
- y = 1.75x - 5.25 + 9.75 = 1.75x + 4.5
- y = 1.75x - 5.25 - 9.75 = 1.75x - 15
These are the two parallel lines at a distance d = 4.84 from point C(3, 0) with a slope of 1.75. The correct line will depend on the direction of the moment that needs to be generated by the force.
For the first line let’s choose x = 0 to simplify the calculation, which gives y = 4.5. The position vector relative to point C(3, 0): \mathbf{r}_{CA} = (-3, 4.5).
The moment about C for this point is:
M_C = (-3 \times 350) - (4.5 \times 200) = -1050 - 900 = -1950 \, \text{Nm}
For the second line let’s choose again x = 0, which gives y = -15. the position vector relative to point C(3, 0): \mathbf{r}_{CA} = (-3, -15).
The moment about C for this point is:
M_C = (-3 \times 350) - (-15 \times 200) = -1050 + 3000 = 1950 \, \text{Nm} Therefore:
- Line 1: y = 1.75x + 4.5 generates a moment of -1950 \, \text{Nm} (clockwise).
- Line 2: y = 1.75x - 15 generates a moment of +1950 \, \text{Nm} (counterclockwise).
Since the desired moment is +1950 \, \text{Nm} (counterclockwise), the correct line of action is:
y = 1.75x - 15
This line intersects the x-axis at approximately x = \frac{15}{1.75} \approx 8.57.
As check, the moment on this point is given by:
M_C = \mathbf{r}_{CA} \times \mathbf{F} = (5.57 \mathbf{i}) \times (200 \mathbf{i} + 350 \mathbf{j})
Since \mathbf{i} \times \mathbf{i} = 0 and \mathbf{i} \times \mathbf{j} = \mathbf{k}, the moment is:
M_C = 5.57 \times 350 \mathbf{k} = 1950 \mathbf{k} \, \text{Nm}
Which confirm the result.
Coulomb friction
When a force P is applied to a body resting on a surface, the resistance to motion arises from friction between the body and the surface, commonly described by Coulomb friction.
Initially, as the force P increases, the body does not move due to static friction, which adjusts to counterbalance P up to a maximum value given by F_{max} = \mu_s N, where \mu_s is the coefficient of static friction, and N is the normal force. As long as P is less than F_{max}, the body remains stationary in the “No Slip” condition.
Once P exceeds F_{max}, the body begins to slide, transitioning into kinetic friction, where the frictional force is constant and given by F_k = \mu_k N, with \mu_k being the coefficient of kinetic friction, typically lower than \mu_s.
In this regime, the force opposing motion is F_k, and the body experiences a “Slip” condition. The forces mg, representing gravity, N, the normal force, and f, the frictional force, are key components in determining the behavior of the system under both static and kinetic friction.
There will be cases where, instead of slipping, a body under the influence of an applied force P may experience an impending tip or rotation about an edge of contact. This occurs when the applied force P, in combination with the weight of the body mg, creates a moment large enough to lift one side of the object off the surface.
In this scenario, the frictional force f is still resisting motion, but rather than the body sliding, the normal force N begins to shift towards one edge of the contact area. As P increases further, the normal force on the opposite side of the tipping point reduces to zero. At this critical moment, the body is on the verge of tipping, and the system reaches what is called an impending tip condition.
Here, the frictional force no longer has a uniform distribution across the base of the object, and the entire reaction force is concentrated at the tipping edge. This situation is distinct from sliding, as the body may rotate about the contact point without significant lateral movement.