A student investigates how the position of a load on a beam affects the forces on the supports. She uses a uniform metre rule of weight 1.0 N supported at t...

(a) Why the sum of the support forces is always 5.0 N

The total downward force is the weight of the rule (1.0 N) plus the hanging load (4.0 N) = 5.0 N. [1]

For the system to be in equilibrium, the total upward force from the two supports must equal the total downward force, so the support forces always sum to 5.0 N. [1]

(b) Expected force at the 0 cm support when the load is at 20 cm

Taking moments about the 100 cm mark eliminates the force at that support:

\( F(0) \times 1.00 = (4.0 \times 0.80) + (1.0 \times 0.50) \) [1]

The load at 20 cm is \( 100 - 20 = 80 \text{ cm} = 0.80 \text{ m} \) from the 100 cm support. The rule's weight acts at 50 cm, which is 0.50 m from the 100 cm support.

\( F(0) \times 1.00 = 3.2 + 0.5 = 3.7 \) [1]

\( F(0) = 3.7 \text{ N} \) [1]

This matches the measured value in the table.

(c) Relationship between load position and support forces

As the load moves towards a support, the force on that support increases. As the load moves away from a support, the force on that support decreases. [1]

(d) Forces if the load is placed at the 0 cm mark

Force at 0 cm support = 5.0 N (all the weight is directly on this support); force at 100 cm support = 0 N. [1]

(e) One source of error

Friction at the supports, or the metre rule may not be perfectly uniform, or difficulty reading the newton meters accurately. [1]

(f) Why forces are equal at the 50 cm mark

At the 50 cm mark, the load is at the exact centre, equidistant from both supports. The rule is uniform, so its own weight also acts at the centre (50 cm). Both loads act at equal distances from each support, producing equal moments, so the support forces are equal. [1]