Topic: Engineering Mechanics
A mass of 35 kg is suspended from a weightless bar AB which is supported by a cable CB and a pin at A as shown in the figure. The pin reactions at A on the bar AB are
- A. $R_{x}=343.4\text{ N}, R_{y}=755.4\text{ N}$
- B. $R_{x}=343.4\text{ N}, R_{y}=0$
- C. $R_{x}=755.4\text{ N}, R_{y}=343.4\text{ N}$
- D. $R_{x}=755.4\text{ N}, R_{y}=0$
Answer: D. $R_{x}=755.4\text{ N}, R_{y}=0$
Explanation: To find the pin reactions at A, we consider the equilibrium conditions and take moments about point C. The distance from the pin A to C is 125 mm, and the horizontal distance to the suspended mass is 275 mm. Summing the moments about C to zero gives: $R_{x} \times 125 = mg \times 275$. Substituting $m = 35\text{ kg}$ and $g = 9.81\text{ m/s}^{2}$, we get $R_{x} = \frac{35 \times 9.81 \times 275}{125} = 755.4\text{ N}$. The vertical reaction $R_{y}$ at the pin is 0 N because the vertical tension component in the cable perfectly balances the downward weight.