A steel girder AB, of mass 200 kg and length 12 m, rests horizontally in equilibrium on two smooth supports at C and D, where AC = 2 m and DB = 2 m - Edexcel - A-Level Maths Mechanics - Question 2 - 2013 - Paper 1

In the new configuration, since the reactions at C and X are equal, we denote the reaction at X as the same as R. The total load on the girder now includes the man and the girder's weight:

Therefore:

$2R = 80 ext{ kg} + 200 ext{ kg}$

This simplifies to:

$2R = 280 ext{ kg}$

Thus, the magnitude of the reaction R at X is:

$R = \frac{280}{2} = 140 ext{ N}$

Setting the sum of moments about point B results in the following:

The moments about B can be expressed as:

$M(B) = (80 ext{ kg} imes 8 ext{ m}) + (200 ext{ kg} imes 6 ext{ m})$

This gives:

$140 + (R imes 10) = (80 imes 8 ext{ m}) + (200 imes 6 ext{ m})$

Substituting the known values, we can write:

$140 + 1400 = 640 + 1200$

This simplifies down to determine x:

$140 + 1400 - 640 - 1200 = 0$

Thus, calculating for x yields:

$x = 2 ext{ m}$.