IGCSE Physics 0625: Support Forces on a Non-Uniform Girder Footbridge

A non-uniform steel girder, of weight $480\text{ N}$ and length $6.0\text{ m}$ , is used as a footbridge across a narrow stream. It rests horizontally on two concrete piers, P and Q, one at each end of the girder. Because the girder is not uniform, its centre of gravity is not at its midpoint. It lies $2.0\text{ m}$ from P. A worker of weight $720\text{ N}$ stands on the girder at a point $4.0\text{ m}$ from P. The girder is in equilibrium.

(a) State one reason why it is useful to take moments about P when finding the support force at Q. [1]

(b) By taking moments about P, calculate the support force at Q, $R_Q$ . [3]

(d) Show that your answers to (b) and (c) are also consistent with the resultant moment about Q being zero. [3]

Part (a): Why take moments about P

The support force at P, $R_P$ , acts at P itself, so its perpendicular distance from P is zero. This means $R_P$ produces no moment about P, so it drops out of the moments equation entirely, leaving only one unknown, $R_Q$ , to solve for.

Part (b): Taking moments about P

The girder’s weight acts at its centre of gravity, $2.0\text{ m}$ from P:

$\text{moment of girder's weight about P} = 480\text{ N} \times 2.0\text{ m} = 960\text{ N m}$

The worker’s weight acts $4.0\text{ m}$ from P:

$\text{moment of worker's weight about P} = 720\text{ N} \times 4.0\text{ m} = 2880\text{ N m}$

Both moments turn the girder the same way about P, so they add together. This is balanced by the moment of $R_Q$ , which acts $6.0\text{ m}$ from P:

$R_Q \times 6.0\text{ m} = 960\text{ N m} + 2880\text{ N m} = 3840\text{ N m}$

$R_Q = \frac{3840\text{ N m}}{6.0\text{ m}} = \boxed{640\text{ N}}$

Part (c): Using the resultant force condition

Since the girder is in equilibrium, the resultant force is zero, so the two support forces together must balance the total weight (girder plus worker):

$R_P + R_Q = 480\text{ N} + 720\text{ N} = 1200\text{ N}$

$R_P = 1200\text{ N} - R_Q = 1200\text{ N} - 640\text{ N}$

$R_P = \boxed{560\text{ N}}$

Part (d): Checking with moments about Q

As a check, the resultant moment about Q should also be zero. Measuring distances from Q instead of P:

distance from Q to the girder’s centre of gravity $= 6.0 - 2.0 = 4.0\text{ m}$ , so its moment about Q $= 480\text{ N} \times 4.0\text{ m} = 1920\text{ N m}$
distance from Q to the worker $= 6.0 - 4.0 = 2.0\text{ m}$ , so its moment about Q $= 720\text{ N} \times 2.0\text{ m} = 1440\text{ N m}$

These sum to $1920\text{ N m} + 1440\text{ N m} = 3360\text{ N m}$ .

This must be balanced by the moment of $R_P$ about Q, which acts $6.0\text{ m}$ from Q:

$R_P \times 6.0\text{ m} = 560\text{ N} \times 6.0\text{ m} = 3360\text{ N m}$

Since $3360\text{ N m} = 3360\text{ N m}$ , the two sides match exactly, confirming the resultant moment about Q is also zero, consistent with equilibrium.

Final answers

(a) $R_P$ produces zero moment about P, so it drops out of the equation
(b) $R_Q =$ $640\text{ N}$
(c) $R_P =$ $560\text{ N}$
(d) Moment of $R_P$ about Q $=$ moment of the two weights about Q $=$ $3360\text{ N m}$ , confirming equilibrium