# What is the effect of adding weight to a balanced beam?

In summary, a beam with negligible mass is balanced by two external forces at R1 = 1.5 meters and R2 = 6.5 meters. If F1 = 10 N, F2 = 2.31 N. However, if the beam has a weight of 3 N, the equation becomes 10(1.5) - F2(6.5) - 3(4-1.5) = 0, resulting in F2 = 4.62 N. The weight of the beam at its center of mass must be taken into account when determining the value of F2.
A beam with negligible mass is balanced by two external forces at R1 = 1.5 meters and R2 = 6.5 meters. If F1 = 10 N, what is F2.

i got this:

(10)(1.5) + F2(6.5) = 0

so F2 = 2.31 N

but the second part i don't get:

If the beam actually has a weight of 3 N, what would F2 be now?

A beam with negligible mass is balanced by two external forces at R1 = 1.5 meters and R2 = 6.5 meters. If F1 = 10 N, what is F2.

i got this:

(10)(1.5) + F2(6.5) = 0

so F2 = 2.31 N

but the second part i don't get:

If the beam actually has a weight of 3 N, what would F2 be now?
10(1.5) - F2(6.5) = 0
F2 = 2.31N
You've got to watch your plus and minus signs, one moment is clockwise, the other is counterclockwise.
For Part 2, the 3N weight of the beam acts at its c.m. (or c.g.), that is, since the beam is 8m long, at 4m from one end. Try it again to solve for F2, watching your plus and minus signs. Remmber also the pivot point is at 1.5m.

Since the beam now has a weight of 3 N, it will contribute to the forces acting on the beam. The new equation would be:

(10)(1.5) + F2(6.5) + 3 = 0

Solving for F2, we get F2 = -0.69 N. This means that the force at R2 would have to be in the opposite direction of the original calculation, at 0.69 N, in order to maintain static equilibrium with the added weight of 3 N.

When the beam has a weight of 3 N, the equation for static equilibrium becomes:

(10)(1.5) + F2(6.5) = 3

Solving for F2, we get:

F2 = 0.46 N

This means that in order to balance the beam with a weight of 3 N, F2 would need to be 0.46 N. This also shows the importance of considering the weight of the beam in addition to the external forces in order to achieve static equilibrium.

## 1) What is static equilibrium?

Static equilibrium is a state in which all forces acting on an object are balanced, resulting in no net movement or acceleration of the object.

## 2) How is static equilibrium different from dynamic equilibrium?

Static equilibrium refers to a system at rest, while dynamic equilibrium refers to a system in motion with a constant velocity. In static equilibrium, all forces are balanced, while in dynamic equilibrium, the net force is zero.

## 3) What is the importance of static equilibrium in physics?

Static equilibrium is important because it is the basis for understanding the behavior of objects at rest. It allows us to analyze the forces acting on an object and predict its motion.

## 4) How do you determine if an object is in static equilibrium?

An object is in static equilibrium if the sum of all forces acting on it is equal to zero and the sum of all torques acting on it is also equal to zero. This means that the object is not accelerating and is either at rest or moving with a constant velocity.

## 5) Can an object be in static equilibrium if it is moving?

No, an object cannot be in static equilibrium if it is moving. In order for an object to be in static equilibrium, it must be at rest or moving with a constant velocity. Any movement or change in velocity would indicate that the forces acting on the object are not balanced.

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