# Apparent Weight problem (falling beam)

• I
• Jackolantern
In summary, the conversation is about an "apparent weight" problem where a weightless beam with two 2 kg weights on each end is suspended by a rubber band and dropped. The task is to estimate the apparent weight of the weights at the moment when the beam has a velocity of 9.8 m/s and the rubber band exerts a force of 106 Newtons. Through calculations and the introduction of symbols to represent physical quantities, it is determined that the apparent weight of each weight is 53 N, and the total apparent weight of the beam with the weights is 106 N.

#### Jackolantern

Each weight: 2 Kg
Hello All, I'm trying to understand an "apparent weight" problem and check my answer. Please use the picture attached.

A weightless beam is at first resting over my palm, it has one 2 kg weight on each end of it. It is suspended to the ceiling by a rubber band. I drop it and it falls for 1 second. At this instant it is has a velocity of 9.8 m/s and the rubber band is exerting an elastic force on it of 106 Newtons. Now, the task is to estimate the "apparent" weight of the 2 kg weights on the beam in this instant as felt by the beam.

-First, I calculate the vertical acceleration induced by the force of the rubber band on the beam:

With a free body diagram of the beam, the only forces acting on it are the weights and the force from the rubber band

Each weight has a force of 19.62 N ( 2 kg * 9.81).
(F)sum = m * a

106 - 2*(19.62) = 4 kg * a
a = 16.69 m/s^2

The apparent weight of an object is its mass multiplied by the vector difference between the gravitational acceleration and the acceleration of the object.

F = m * a

F = 2 kg * (9.81 + 16.669) = 53 N

So, my question is, at the 2nd instant illustrated by my photo above, the apparent weight of each of the weights on the end of the beam is 53 N?
This is just half of the rubber band force...for every action there is an equal and opposite reaction?

Each mass has apparent weight 53 N so the total apparent weight of the beam with the masses is 106 N as expected.

Note that some things will often be clearer if you introduce symbols to represent physical quantities and only insert measured values at the very end. In your case, you could introduce the following:

F = 106 N
m = 2 kg
g = 9.1 m/s^2
a = acceleration of the masses
W = apparent weight of one mass

Force relation to acceleration:
$$F - 2mg = 2ma \quad \Longrightarrow \quad a = \frac{F}{2m} - g$$

Apparent weight of one mass:
$$W = m(a+g) = \frac{F}{2}.$$

It is then clear that the apparent weight of each mass is exactly half of the tension in the rubber band and that the masses and the acceleration and gravitational acceleration do not really matter for the result.

Lnewqban and Jackolantern
Orodruin said:
Each mass has apparent weight 53 N so the total apparent weight of the beam with the masses is 106 N as expected.

Note that some things will often be clearer if you introduce symbols to represent physical quantities and only insert measured values at the very end. In your case, you could introduce the following:

F = 106 N
m = 2 kg
g = 9.1 m/s^2
a = acceleration of the masses
W = apparent weight of one mass

Force relation to acceleration:
$$F - 2mg = 2ma \quad \Longrightarrow \quad a = \frac{F}{2m} - g$$

Apparent weight of one mass:
$$W = m(a+g) = \frac{F}{2}.$$

It is then clear that the apparent weight of each mass is exactly half of the tension in the rubber band and that the masses and the acceleration and gravitational acceleration do not really matter for the result.
That's much more clear to me now, thank you Orodruin.

Lnewqban and berkeman

## 1. What is the apparent weight problem when a beam is falling?

The apparent weight problem is a phenomenon that occurs when an object, in this case a beam, is falling and experiences a change in weight due to the acceleration of gravity acting on it. This can cause the beam to appear to weigh more or less than its actual weight.

## 2. Why does the apparent weight of a falling beam change?

The apparent weight of a falling beam changes because of the effects of gravity and acceleration. As the beam falls, it experiences an acceleration due to gravity, which can cause it to appear to weigh more or less than its actual weight. Additionally, the angle at which the beam falls can also affect its apparent weight.

## 3. How do you calculate the apparent weight of a falling beam?

The apparent weight of a falling beam can be calculated using the formula W = mg + ma, where W is the apparent weight, m is the mass of the beam, g is the acceleration due to gravity, and a is the acceleration of the beam as it falls. This formula takes into account the effects of gravity and acceleration on the beam's weight.

## 4. Can the apparent weight problem be observed in real-life situations?

Yes, the apparent weight problem can be observed in real-life situations, such as when a person is on a roller coaster or when a building is collapsing. In these scenarios, the acceleration and angle of the falling objects can cause their apparent weight to change.

## 5. How can the apparent weight problem be mitigated or avoided?

The apparent weight problem can be mitigated or avoided by ensuring that objects are securely fastened or supported when they are at risk of falling. Additionally, using proper safety precautions and following building codes can help prevent situations where the apparent weight problem may occur.