Question

The diagram shows a horizontal girder AB of length l, from the ends of which a load of $10^3$ kg is suspended by two strings of length = l. The compression in the girder is (g = 10 $ms^{-2}$)

• A. $\frac{5000}{\sqrt{3}}N$
• B. $\frac{10^4}{\sqrt{3}}N$
• C. $5 \times 10^3 N$
• D. $5\sqrt{3}\times10^3 N$

Answer 1

Answer: (A)

$\frac{5000}{\sqrt{3}}N$

Answer 2

The weight of the load is $W = m \times g = 10^3 \text{ kg} \times 10 \text{ m/s}^2 = 10^4$ N. Since the lengths of the girder AB and the strings AP and BP are all equal to $l$, triangle ABP is equilateral. The angle each string makes with the horizontal girder is $60^\circ$. Let T be the tension in each string. For vertical equilibrium, the sum of the vertical components of the tensions balances the weight: $2 \times T \sin(60^\circ) = W$. Solving for T: $2 \times T \times \frac{\sqrt{3}}{2} = 10^4 \implies T\sqrt{3} = 10^4 \implies T = \frac{10^4}{\sqrt{3}}$ N. The horizontal component of tension at each end of the girder is $T_x = T \cos(60^\circ)$. This horizontal component, if acting inwards, would cause compression. Assuming the question implies a scenario leading to compression, the magnitude of the compression force is $T \cos(60^\circ) = \frac{10^4}{\sqrt{3}} \times \frac{1}{2} = \frac{5000}{\sqrt{3}}$ N.