Question

A uniform meterstick balances when the support is placed at its mid-point. a 400 g mass is placed at the 80.0 cm mark. at what location on the meterstick should a 700 g mass be placed so that the stick will be balanced?

Answer 1

To balance a meterstick with a 400 g mass at the 80.0 cm mark, a 700 g mass should be positioned at the 32.9 cm mark from the left end of the meterstick using torque balance principles.

Step-by-step explanation:

To determine at what location on the meterstick a 700 g mass should be placed to balance it, we apply the principle of moments (torque balance). The meterstick balances at its midpoint when no additional masses are added, meaning the center of gravity is at 50.0 cm. Placing a 400 g mass at the 80.0 cm mark introduces a torque that must be counterbalanced by the 700 g mass.

Let's use the following formula to achieve balance:
(Torque due to 400 g) = (Torque due to 700 g)

So, (400 g × 30.0 cm) = (700 g × x cm), where x is the distance from the 50.0 cm mark for the 700 g mass to balance the meterstick. We can now solve for x:

x = (400 g × 30.0 cm) / (700 g) = 12000 g·cm / 700 g = 17.14 cm

Since the 700 g mass must be on the opposite side of the pivot to the 400 g mass, we subtract this distance from the pivot point, i.e., 50.0 cm - 17.14 cm = 32.86 cm (or round to 32.9 cm using significant digits). Therefore, to balance the meterstick, the 700 g mass should be placed at 32.9 cm mark from the left end of the meterstick.