Question

A rod is subjected to three forces as shown in the figure on the left. An equivalent force system with forces $F_1$, $F_2$ and moment $M$ is shown in the figure on the right. The value of $M$ (in N·m) is ................ (rounded off to one decimal place).

Hint:

Always resolve inclined forces into horizontal and vertical components. Only perpendicular components to lever arm create a moment.

Answer 1

Correct Answer: 117

Step 1: Break the 60 N inclined force into components.
The $60 \, N$ force at $45^\circ$: \[ F_x = 60 \cos 45^\circ = 60 \times 0.707 = 42.43 \, N \] \[ F_y = 60 \sin 45^\circ = 42.43 \, N \]
Step 2: Write down all forces acting.
- Downward force: $30 \, N$ at top. - Horizontal force: $40 \, N$ (to the right). - Inclined force components: $-42.43 \, N$ (horizontal, to the left), $+42.43 \, N$ (vertical upward).
Step 3: Compute net forces (for reference).
Horizontal resultant: \[ F_x = 40 - 42.43 = -2.43 \, N (\text{leftward}) \] Vertical resultant: \[ F_y = -30 + 42.43 = 12.43 \, N (\text{upward}) \] But the problem asks only for the {equivalent couple moment $M$}.
Step 4: Take moments about bottom of rod.
- 30 N downward force at $6 \, m$ above base: \[ M_{30} = 30 \times 6 = 180 \, N\!m \; (\text{clockwise}) \] - 40 N horizontal force at $2 \, m$ above base: \[ M_{40} = 40 \times 2 = 80 \, N\!m \; (\text{anticlockwise}) \] - Inclined 60 N force at $5 \, m$ above base: Vertical component $42.43 \, N$ at 5 m → clockwise: \[ M_{60v} = 42.43 \times 5 = 212.15 \, N\!m \; (\text{clockwise}) \] Horizontal component passes through line of action → no moment.

Step 5: Net moment.
\[ M = (180 + 212.15) - 80 = 312.15 \, N\!m \] Final Answer: \[ \boxed{312.2 \, N\!m} \]