if a 120-g mass is placed at the 25-cm mark and a 25-g mass at the 10-cm mark, where should a 500-g mass be placed to balance the system?
Where it the fulcrum? I assume this is a beam blance resting on some sort of a fulcrum support
The total moment about the fulcrum must be zero to achieve balance
To determine where the 500-g mass should be placed to balance the system, we need to consider the principle of the lever arm. The lever arm is the perpendicular distance from the pivot point (fulcrum) to the line of action of the force.
In this case, we have a system with multiple masses placed at different positions on a meter stick. The meter stick acts as a lever, and the pivot point or fulcrum is the center of the meter stick.
Let's break down the problem step by step:
1. Calculate the total torque (twisting force) acting on the meter stick:
Torque (T1) caused by the 120-g mass: T1 = (120 g) x (25 cm)
Torque (T2) caused by the 25-g mass: T2 = (25 g) x (10 cm)
2. The total torque exerted on one side of the meter stick is equal to the total torque on the other side for the system to be in equilibrium. Mathematically, T1 = T2.
3. Now, we need to find the torque (T3) caused by the unknown 500-g mass that needs to be placed on the meter stick. We can set up the equation: T3 = (500 g) x (x cm), where x is the distance of the 500-g mass from the fulcrum.
4. Equate T1 and T2 to T3:
(120 g) x (25 cm) = (25 g) x (10 cm) + (500 g) x (x cm)
5. Solve the equation for x:
(120 g) x (25 cm) - (25 g) x (10 cm) = (500 g) x (x cm)
Now, you can calculate the position where the 500-g mass should be placed by rearranging the equation and solving for x. Let's do that:
(120 g) x (25 cm) - (25 g) x (10 cm) = (500 g) x (x cm)
3000 g.cm - 250 g.cm = 500 g. x cm
2750 g.cm = 500 g. x cm
x cm = (2750 g.cm) / (500 g)
= 5.5 cm
Therefore, the 500-g mass should be placed at the 5.5-cm mark to balance the system.