If m1 = 4.72kg, what mass m2 is required to balance the system?

(two masses are connected by a rope. m1 is sitting on an incline and m2 is hanging from a pulley)

To find the mass (m2) required to balance the system, we need to consider the equilibrium condition, which states that the sum of the forces acting on an object must be zero.

In this case, we have two forces involved: the force of gravity acting on m1 (F1) and the tension force in the rope (T). The force of gravity can be calculated using the formula F = m * g, where m is the mass and g is the acceleration due to gravity (approximately 9.8 m/s^2).

For m1 (4.72 kg), the force of gravity (F1) can be calculated as follows:
F1 = m1 * g = 4.72 kg * 9.8 m/s^2 = 46.256 N

Since the system is in equilibrium, the tension force in the rope (T) is equal to the force of gravity acting on m1:
T = F1 = 46.256 N

Now, let's consider m2, the mass required to balance the system. Since m2 is hanging from a pulley, it will experience the same tension force (T) as m1. So, we can set T equal to the force of gravity acting on m2.

By rearranging the formula for force (F = m * g), we can solve for m2:
F2 = m2 * g

Since F2 is equal to T, we can substitute T for F2:
T = m2 * g

Rearranging the equation, we find:
m2 = T / g

Plugging in the values we have:
m2 = 46.256 N / 9.8 m/s^2 ≈ 4.725 kg

Therefore, to balance the system, a mass (m2) of approximately 4.725 kg is required.