Question

The frame shown in the figure is loaded at S with a force of 2000 N. The reactions at T are denoted by \(T_x\) and \(T_y\), while the reaction at W is \(W_y\). Neglect the weight of the members.

Which one of the following options for the magnitudes of the forces (in N), \(T_x, T_y\) and \(W_y\), is CORRECT?

• A. \(T_x = 0, \, T_y = 1000, \, W_y = 1000\)
• B. \(T_x = 0, \, T_y = 1500, \, W_y = 500\)
• C. \(T_x = 0, \, T_y = 800, \, W_y = 1200\)
• D. \(T_x = 0, \, T_y = 500, \, W_y = 1500\)

Hint:

For frames, always begin with \(\Sigma M = 0\) about one support to find the distant reaction directly. Then use \(\Sigma F_y = 0\).

Answer 1

The Correct Option is D

Step 1: Equilibrium in horizontal direction.
Since no horizontal force acts externally, \[ \Sigma F_x = 0 \Rightarrow T_x = 0 \]

Step 2: Equilibrium in vertical direction.
\[ \Sigma F_y = 0 \Rightarrow T_y + W_y = 2000 \]

Step 3: Moment equilibrium about T.
- Vertical load of 2000 N acts at S, which is 3.6 m horizontally away from T. \[ M_{2000} = 2000 \times 3.6 = 7200 \, \text{Nm (clockwise)} \] - Reaction at W acts vertically upward at 4.8 m from T: \[ M_W = W_y \times 4.8 \, \text{(counter-clockwise)} \] Equilibrium: \[ W_y \times 4.8 = 7200 \Rightarrow W_y = 1500 \, N \]

Step 4: Find \(T_y\).
\[ T_y + 1500 = 2000 \Rightarrow T_y = 500 \, N \] Final Answer:
\[ \boxed{T_x = 0, \; T_y = 500, \; W_y = 1500} \]