As part of Engineering Mechanics Series. Here is first problem we are going to solve. In case if you need some basics , Please visit this post as quick review

Here is an given problem

Problem 1

Ask

  • Find maginitude and direction of Force P
  • Find the contact force at bottom of block (say N)

Assumptions

  1. Block is in equilibrium - i.e. block is at rest or moving at constant velocity
  2. Frictionless surface - there is no frictional force involved.

Given data

  1. Mass of block (M) is 100 lb
  2. The force P is $10°$ inclined to horizontal
  3. The surface is inclined at $25°$ to horizontal base

In order to draw free body diagram follow these steps

  • Isolate the object - i.e. No external force or internal force. Imagine block is floating in space

  • Identify all forces acting ON the object

    • Applied External force (P) - $10°$ above horizontal
    • Weight - gravity pull straight down
    • Normal force (N)
    • Friction (f)

    Draw each force as an arrow FROM center of Mass

    Problem 1

  • Choose coordinate System

    1. Standard : x-axis horizontal and y-axis vertical. This is simple and straightforward, but need to solve N and f components

    2. Tilted : x-axis along incline surface and y-axis is perpendicular . This is good because

      a. Normal force N is already along y-axis b. f is already along x-axis (though it is zero) c. We need to resolve P and W

    3. Break Forces into components

    • Incline Angle: $10°$

    • Force $P$ Angle relative to the incline: The incline is $25°$ up from the horizontal. Force $P$ is $10°$ up from the horizontal. Therefore, the angle between Force $P$ and the incline is: $25°$ - $10^°$ = $15°$

      Force $P$ pushes “into” the slope at a slight angle of $15°$ relative to the $x$-axis.

    1. Resolving forces into components

      Weight ($W = 100 \text{ lb}$):

      • $W_x = -100 \sin(25^\circ)$ (Points down the slope)
      • $W_y = -100 \cos(25^\circ)$ (Points into the slope)

      Applied Force ($P$):

      • $P_x = +P \cos(15^\circ)$ (Points up the slope)
      • $P_y = -P \sin(15^\circ)$ (Points into the slope)

      Normal Force ($N$):

      • $N_x = 0$
      • $N_y = +N$ (Points away from the slope)

      Problem 1

    2. Equilibrium Equations

      Since the block is resting, the sum of forces in both directions must be zero.

      Equation 1: Sum of forces in the x-direction ($\Sigma F_x = 0$)

      $$P \cos(15^\circ) - W \sin(25^\circ) = 0$$

      $$P \cos(15^\circ) = 100 \sin(25^\circ)$$

      Equation 2: Sum of forces in the y-direction ($\Sigma F_y = 0$)

      $$N - W \cos(25^\circ) - P \sin(15^\circ) = 0$$

      $$N = 100 \cos(25^\circ) + P \sin(15^\circ)$$

  1. Calculations

    Step A: Solve for Force $P$

    Using Equation 1:

    $$P = \frac{100 \sin(25^\circ)}{\cos(15^\circ)}$$

    $$\sin(25^\circ) \approx 0.4226$$

    $$\cos(15^\circ) \approx 0.9659$$

    $$P = \frac{100 \times 0.4226}{0.9659}$$

    $$P \approx 43.75 \text{ lb}$$

    Step B: Solve for Contact Force $N$

    Using Equation 2 and the value of $P$ we just found:

    $$N = 100 \cos(25^\circ) + 43.75 \sin(15^\circ)$$

    $$\cos(25^\circ) \approx 0.9063$$

    $$\sin(15^\circ) \approx 0.2588$$

    $$N = (100 \times 0.9063) + (43.75 \times 0.2588)$$

    $$N = 90.63 + 11.32$$

    $$N \approx 101.95 \text{ lb}$$

Answer

  1. Magnitude and Direction of Force $P$:

    • Magnitude: $43.8 \text{ lb}$ (rounded to 3 significant figures)
    • Direction: $10^\circ$ above the horizontal, pointing up and to the right (as shown in the diagram)

  2. Contact Force on the bottom of the block:

    • Magnitude: $102 \text{ lb}$ (rounded to 3 significant figures)
    • Direction: Perpendicular to the incline, pointing upwards and to the left (away from the surface)