Engineering Mechanics – Problem Set Geological Engineering

Fundamentals of Force Systems in Engineering Mechanics

• Vector Representation of Forces:

  • A force is a vector quantity defined by its magnitude, direction, line of action, and point of application.
  • Concurrent forces are forces whose lines of action intersect at a single common point.

• The Parallelogram Law of Forces:

  • If two concurrent forces acting at a point are represented in magnitude and direction by two adjacent sides of a parallelogram drawn from that point, their resultant force is represented in magnitude and direction by the diagonal of the parallelogram passing through that point.
  • For two forces $P$ and $Q$ acting at an angle $\theta$ relative to each other, the magnitude of the resultant force $R$ is given by:     $R = \sqrt{P^2 + Q^2 + 2PQ\cos(\theta)}$
  • The directional angle $\alpha$ of the resultant force $R$ relative to the force $P$ is calculated using:     $\tan(\alpha) = \frac{Q\sin(\theta)}{P + Q\cos(\theta)}$$\alpha = \arctan\left(\frac{Q\sin(\theta)}{P + Q\cos(\theta)}\right)$

• Component Method for Systems of Concurrent Forces:

  • For a system of multiple concurrent forces $F_1, F_2, \dots, F_n$ acting at angles $\theta_1, \theta_2, \dots, \theta_n$ relative to a positive reference axis:
  • Algebraic sum of horizontal components:     $R_x = \sum_{i=1}^{n} F_{i,x} = \sum_{i=1}^{n} F_i \cos(\theta_i)$
  • Algebraic sum of vertical components:     $R_y = \sum_{i=1}^{n} F_{i,y} = \sum_{i=1}^{n} F_i \sin(\theta_i)$
  • Magnitude of the resultant force $R$:     $R = \sqrt{R_x^2 + R_y^2}$
  • Directional angle $\theta$ of the resultant with respect to the horizontal reference line:     $\theta = \arctan\left(\frac{R_y}{R_x}\right)$

The $\lambda, \mu$ Theorem in Vector Mechanics

• Mathematical Statement of the Theorem:

  • Let two concurrent forces acting at a point $O$ be represented in magnitude and direction by vector multiples $\lambda \mathbf{\vec{OA}}$ and $\mu \mathbf{\vec{OB}}$, where $\lambda$ and $\mu$ are real scalar coefficients.
  • The resultant force $\mathbf{\vec{R}}$ of these two forces is represented in magnitude and direction by:     $\mathbf{\vec{R}} = (\lambda + \mu)\mathbf{\vec{OC}}$
  • The point $C$ lies on the straight line segment $AB$ and divides $AB$ internally such that:     $AC : CB = \mu : \lambda$$\frac{AC}{CB} = \frac{\mu}{\lambda}$

• Geometric and Physical Implications:

  • Division Ratio: Point $C$ divides segment $AB$ inversely proportional to the respective scalar multipliers $\lambda$ and $\mu$.
  • Position of Point $C$:
  • If $\lambda > \mu$, then $AC < CB$, meaning $C$ lies closer to point $A$.
  • If $\mu > \lambda$, then $CB < AC$, meaning $C$ lies closer to point $B$.
  • If $\lambda = \mu$, then $AC = CB$, meaning $C$ is the midpoint of segment $AB$.
  • Direction of Resultant: The line passing through point $O$ and point $C$ defines the exact line of action of the resultant force $\mathbf{\vec{R}}$.

Comprehensive Solutions to Geological Engineering Problems

• Problem 1: Resultant of Forces Acting on a Rock Block

  • Problem Parameters:
  • Force 1: $P = 10\,\text{kN}$
  • Force 2: $Q = 6\,\text{kN}$
  • Angle between forces: $\theta = 60^\circ$
  • Part (a): Magnitude of Resultant Force
  • Formula application:       $R = \sqrt{P^2 + Q^2 + 2PQ\cos(\theta)}$
  • Substitution:       $R = \sqrt{(10)^2 + (6)^2 + 2(10)(6)\cos(60^\circ)}$$R = \sqrt{100 + 36 + 120(0.5)}$$R = \sqrt{136 + 60} = \sqrt{196} = 14\,\text{kN}$
  • Part (b): Direction of Resultant with Respect to the $10\,\text{kN}$ Force
  • Formula application:       $\tan(\alpha) = \frac{Q\sin(\theta)}{P + Q\cos(\theta)}$
  • Substitution:       $\tan(\alpha) = \frac{6\sin(60^\circ)}{10 + 6\cos(60^\circ)}$$\tan(\alpha) = \frac{6\left(\frac{\sqrt{3}}{2}\right)}{10 + 6(0.5)} = \frac{3\sqrt{3}}{10 + 3} = \frac{3\sqrt{3}}{13}$$\tan(\alpha) \approx \frac{5.19615}{13} \approx 0.3997$$\alpha = \arctan(0.3997) \approx 21.79^\circ$

• Problem 2: Forces Acting on a Rock Mass Near a Tunnel

  • Problem Parameters:
  • Force 1: $Q = 15\,\text{kN}$
  • Force 2: $P = 20\,\text{kN}$
  • Angle between forces: $\theta = 45^\circ$
  • Part (a): Magnitude of Resultant Force
  • Formula application:       $R = \sqrt{P^2 + Q^2 + 2PQ\cos(\theta)}$
  • Substitution:       $R = \sqrt{(20)^2 + (15)^2 + 2(20)(15)\cos(45^\circ)}$$R = \sqrt{400 + 225 + 600\left(\frac{\sqrt{2}}{2}\right)}$$R = \sqrt{625 + 300\sqrt{2}} \approx \sqrt{625 + 424.264} = \sqrt{1049.264} \approx 32.39\,\text{kN}$
  • Part (b): Direction of Resultant with Respect to the $20\,\text{kN}$ Force
  • Formula application:       $\tan(\alpha) = \frac{15\sin(45^\circ)}{20 + 15\cos(45^\circ)}$
  • Substitution:       $\tan(\alpha) = \frac{15\left(\frac{\sqrt{2}}{2}\right)}{20 + 15\left(\frac{\sqrt{2}}{2}\right)} \approx \frac{10.6066}{20 + 10.6066} = \frac{10.6066}{30.6066} \approx 0.3465$$\alpha = \arctan(0.3465) \approx 19.11^\circ$

• Problem 3: Equal Forces Acting on a Geological Block

  • Problem Parameters:
  • Two equal forces: $P = 10\,\text{kN}$ and $Q = 10\,\text{kN}$
  • General formula for equal forces:       $R = \sqrt{P^2 + P^2 + 2P^2\cos(\theta)} = \sqrt{2P^2(1 + \cos(\theta))} = 2P\cos\left(\frac{\theta}{2}\right) = 20\cos\left(\frac{\theta}{2}\right)\,\text{kN}$
  • Part (a): Angle $\theta = 0^\circ$$R = 20\cos(0^\circ) = 20(1) = 20\,\text{kN}$
  • Part (b): Angle $\theta = 60^\circ$$R = 20\cos(30^\circ) = 20\left(\frac{\sqrt{3}}{2}\right) = 10\sqrt{3} \approx 17.32\,\text{kN}$
  • Part (c): Angle $\theta = 90^\circ$$R = 20\cos(45^\circ) = 20\left(\frac{1}{\sqrt{2}}\right) = 10\sqrt{2} \approx 14.14\,\text{kN}$
  • Part (d): Angle $\theta = 120^\circ$$R = 20\cos(60^\circ) = 20(0.5) = 10\,\text{kN}$
  • Part (e): Angle $\theta = 180^\circ$$R = 20\cos(90^\circ) = 20(0) = 0\,\text{kN}$
  • Extreme Values and Corresponding Angles:
  • Greatest possible resultant: $20\,\text{kN}$ occurring at an angle of $\theta = 0^\circ$ (forces act in the same direction).
  • Least possible resultant: $0\,\text{kN}$ occurring at an angle of $\theta = 180^\circ$ (forces act in directly opposite directions).

• Problem 4: Maximum and Minimum Resultant in a Slope System

  • Problem Parameters:
  • Force 1: $P = 25\,\text{kN}$
  • Force 2: $Q = 15\,\text{kN}$
  • Part (a): Greatest Possible Resultant     $R_{\text{max}} = P + Q = 25\,\text{kN} + 15\,\text{kN} = 40\,\text{kN}$
  • Part (b): Least Possible Resultant     $R_{\text{min}} = |P - Q| = |25\,\text{kN} - 15\,\text{kN}| = 10\,\text{kN}$
  • Part (c): Corresponding Angles
  • Angle for greatest resultant $R_{\text{max}} = 40\,\text{kN}$: $\theta = 0^\circ$
  • Angle for least resultant $R_{\text{min}} = 10\,\text{kN}$: $\theta = 180^\circ$

• Problem 5: Direction of Resultant Force on a Slope Block

  • Problem Parameters:
  • Force 1: $P = 30\,\text{kN}$
  • Force 2: $Q = 20\,\text{kN}$
  • Angle between forces: $\theta = 70^\circ$
  • Part (a): Magnitude of Resultant Force     $R = \sqrt{(30)^2 + (20)^2 + 2(30)(20)\cos(70^\circ)}$$R = \sqrt{900 + 400 + 1200(0.34202)}$$R = \sqrt{1300 + 410.424} = \sqrt{1710.424} \approx 41.36\,\text{kN}$
  • Part (b): Angle Made by Resultant with the $30\,\text{kN}$ Force     $\tan(\alpha) = \frac{20\sin(70^\circ)}{30 + 20\cos(70^\circ)}$$\tan(\alpha) = \frac{20(0.93969)}{30 + 20(0.34202)} = \frac{18.7938}{30 + 6.8404} = \frac{18.7938}{36.8404} \approx 0.51014$$\alpha = \arctan(0.51014) \approx 27.03^\circ$

• Problem 6: Resultant of a System of Concurrent Forces at a Tunnel Support

  • Problem Parameters:
  • Force 1: $F_1 = 12\,\text{kN}$ at $\theta_1 = 0^\circ$
  • Force 2: $F_2 = 8\,\text{kN}$ at $\theta_2 = 90^\circ$
  • Force 3: $F_3 = 10\,\text{kN}$ at $\theta_3 = 180^\circ$
  • Part (a): Horizontal and Vertical Components of Resultant
  • Horizontal component $R_x$:       $R_x = \sum F_x = F_1\cos(0^\circ) + F_2\cos(90^\circ) + F_3\cos(180^\circ)$$R_x = 12(1) + 8(0) + 10(-1) = 12 + 0 - 10 = 2\,\text{kN}$
  • Vertical component $R_y$:       $R_y = \sum F_y = F_1\sin(0^\circ) + F_2\sin(90^\circ) + F_3\sin(180^\circ)$$R_y = 12(0) + 8(1) + 10(0) = 0 + 8 + 0 = 8\,\text{kN}$
  • Part (b): Magnitude of Resultant Force     $R = \sqrt{R_x^2 + R_y^2} = \sqrt{(2)^2 + (8)^2} = \sqrt{4 + 64} = \sqrt{68} \approx 8.25\,\text{kN}$
  • Part (c): Direction of Resultant Force     $\theta = \arctan\left(\frac{R_y}{R_x}\right) = \arctan\left(\frac{8}{2}\right) = \arctan(4) \approx 75.96^\circ$ relative to the reference line ($0^\circ$).

• Problem 7: Application of the $\lambda, \mu$ Theorem to a Rock-Support System

  • Problem Parameters:
  • Forces: $\lambda \mathbf{\vec{OA}}$ and $\mu \mathbf{\vec{OB}}$ with $\lambda = 3$ and $\mu = 2$
  • Proof and Evaluation:
  • Using the $\lambda, \mu$ theorem, the resultant vector $\mathbf{\vec{R}}$ is given by:       $\mathbf{\vec{R}} = \lambda \mathbf{\vec{OA}} + \mu \mathbf{\vec{OB}}$
  • The ratio in which point $C$ divides $AB$ internally is:       $AC : CB = \mu : \lambda = 2 : 3$
  • Substitute values into resultant vector equation:       $\mathbf{\vec{R}} = (\lambda + \mu)\mathbf{\vec{OC}} = (3 + 2)\mathbf{\vec{OC}} = 5\mathbf{\vec{OC}}$
  • Role of Position $C$ in Determining Resultant Direction:
  • The position of point $C$ along segment $AB$ defines the exact orientation of vector $\mathbf{\vec{OC}}$.
  • Because the resultant vector $5\mathbf{\vec{OC}}$ is a scalar multiple of $\mathbf{\vec{OC}}$, the line of action of the resultant force passes through origin $O$ and point $C$.
  • Shifts in the ratio $AC : CB$ rotate the vector $\mathbf{\vec{OC}}$, thereby directing the resultant force closer to the larger force component.

• Problem 8: Resultant of Geological Forces Using the $\lambda, \mu$ Theorem

  • Problem Parameters:
  • Forces: $4\mathbf{\vec{OA}}$ and $3\mathbf{\vec{OB}}$ ($\lambda = 4$, $\mu = 3$)
  • Part (a): Division Ratio of Point $C$$AC : CB = \mu : \lambda = 3 : 4$
  • Part (b): Coefficient of $\mathbf{\vec{OC}}$ in the Resultant     $\text{Coefficient} = \lambda + \mu = 4 + 3 = 7$
  • Part (c): Vector Representation of the Resultant Force     $\mathbf{\vec{R}} = 7\mathbf{\vec{OC}}$

• Problem 9: Application of the $\lambda, \mu$ Theorem to a Rock Block

  • Problem Parameters:
  • Forces: $5\mathbf{\vec{OA}}$ and $2\mathbf{\vec{OB}}$ ($\lambda = 5$, $\mu = 2$)
  • Part (a): Division Ratio $AC : CB$$AC : CB = \mu : \lambda = 2 : 5$
  • Part (b): Position of $C$ Relative to Points $A$ and $B$
  • Point $C$ divides $AB$ such that $AC = \frac{2}{7}AB$ and $CB = \frac{5}{7}AB$.
  • Therefore, point $C$ lies closer to point A$.\n - Part (c): Vector Representation of the Resultant Force\n    \mathbf{\vec{R}} = (\lambda + \mu)\mathbf{\vec{OC}} = (5 + 2)\mathbf{\vec{OC}} = 7\mathbf{\vec{OC}}\n - Explanation of Directional Representation:\n - The relative position of point C$closer to point$A$reflects the dominant influence of the force component$5\mathbf{\vec{OA}}$over$2\mathbf{\vec{OB}}.\n - The resultant line of action along \mathbf{\vec{OC}}$naturally aligns closer to the larger component force direction$\mathbf{\vec{OA}}.\n\n- Problem 10: Combined Application of Resultant and the \lambda, \mu Theorem\n - Problem Parameters:\n - Concurrent forces of magnitudes 6\,\text{kN}$and$4\,\text{kN}$along directions$\mathbf{\vec{OA}}$and$\mathbf{\vec{OB}}\n - Angle between directions: \theta = 60^\circ\n - Part (a): Magnitude of Resultant Using Parallelogram Law\n    R = \sqrt{(6)^2 + (4)^2 + 2(6)(4)\cos(60^\circ)}\n    R = \sqrt{36 + 16 + 48(0.5)} = \sqrt{52 + 24} = \sqrt{76} \approx 8.72\,\text{kN}\n - Part (b): Direction of Resultant\n    \tan(\alpha) = \frac{4\sin(60^\circ)}{6 + 4\cos(60^\circ)} = \frac{4\left(\frac{\sqrt{3}}{2}\right)}{6 + 4(0.5)} = \frac{2\sqrt{3}}{6 + 2} = \frac{2\sqrt{3}}{8} = \frac{\sqrt{3}}{4} \approx 0.4330\n    \alpha = \arctan(0.4330) \approx 23.41^\circ \text{ relative to } \mathbf{\vec{OA}}\n - Part (c): Division Ratio Using \lambda, \mu Theorem\n - With \lambda = 6$and$\mu = 4:\n      AC : CB = \mu : \lambda = 4 : 6 = 2 : 3\n - Part (d): Explanation of Why Point C Lies on the Line of Action\n - By definition of vector addition and section formula:\n      \mathbf{\vec{OC}} = \frac{\lambda \mathbf{\vec{OA}} + \mu \mathbf{\vec{OB}}}{\lambda + \mu}\n      (\lambda + \mu)\mathbf{\vec{OC}} = \lambda \mathbf{\vec{OA}} + \mu \mathbf{\vec{OB}} = \mathbf{\vec{R}}\n - Since the resultant vector \mathbf{\vec{R}}$is directly proportional to vector$\mathbf{\vec{OC}}$(differing only by scalar factor$\lambda + \mu$), the vectors$\mathbf{\vec{R}}$and$\mathbf{\vec{OC}} are collinear.\n - Thus, point C$must lie directly on the line of action of the resultant force passing through point$O$.

Recommended Literature and Textbooks

• Book 1:

  • Author: N.A. Shah
  • Title: Vector Analysis

• Book 2:

  • Authors: S. Timoshenko, D.H. Young, J.V. Rao, and Sukumar Pati
  • Title: Engineering Mechanics

• Book 3:

  • Author: Q.K. Ghori
  • Title: Mechanics

• Book 4:

  • Authors: Muhammad Iqbal Bhatti and Muhammad Nasir Ch.
  • Title: Mathematics for Engineers and Scientists
  • Publisher / Location: Allied Book Centre, Urdu Bazar, Lahore