# Solve Static Equilibrium Qs: 17.55kN & 17.45kN, 15.63kN & 16.37kN

• kiteboy
In summary: The net reaction force is then given by:\vec{F}=F_{\lambda} d\vec{x}+F_{\lambda} d\vec{y}\vec{F}=F_{\lambda}_0+F_{\lambda}_1\vec{F}=8kN+15kN\vec{F}=37.5kN
kiteboy
Hi

I have some questions to go through for static equilibrium...I have the answers but I have no idea how to do these really - I am guessing most of them without really understanding

I can't remember doing physics in shcool at all and since its been the best part of 20 years since school some help would be REALLY appreciated

Ok ill put two sample questions with the answers that were also given

A horizontal beam AB is simply supported at each end. The beam carries a point load of 12kN at a point 0.8 m from A, a point load of 8kN at a point 1.9m from A and a point load of 15kN at a point 3m from A. Calculate the reactions exerted by the supports on the beam.
The answers are : 17.55kB at A and 17.45kN at B

Another one

A horizontal beam AB is 6m long and simply supported at each end. The beam carries a uniformly distributed load of 3kN/m from each end A to the mid point of the beam, a point load of 9kN at a point 2.1m from A, and a point load of 14kN at a point 4.7m from A. Calculate the reactions exerted by the supports on the beam.
(answers are 15.63kN at A and 16.37kN at B)

If anybody can find the time to really explain how these are generally worked it out would be great

Ive already ordered books from amazon to hopefully help me out as this is beyond me at the moment - even google is failing me!

Thanks

For static equilibrium, two conditions must be met:
$$\Sigma \vec{F} = 0$$

$$\Sigma \vec{\tau} = 0$$

jhae2.718 said:
For static equilibrium, two conditions must be met:
$$\Sigma \vec{F} = 0$$

$$\Sigma \vec{\tau} = 0$$

I think that means the sum of the forces in the x and y direction must be 0
I think the sum of the moments must also be 0

I kinda get that but not really when it comes to the examples if you know what i mean

To show you the basic process, I'll walk you through a simpler example, given by this image:

Lets say we have two forces, $$F_1$$ and $$F_2$$ that are evenly spaced along a beam of length L. (For this, we assume that the mass of the beam is negligible. Sometimes the beam is given a mass; it acts at the center of mass.) There are simple reaction forces acting at A and B. That is, there is just a force in the vertical direction at each point.

The beam is in static equilibrium, so the following equations hold:
$$\Sigma \vec{F}=0$$

$$\Sigma \vec{M} = 0$$

We have two unknowns, A and B. Recall that the definition of the magnitude of a moment is $$M=Fd$$, where F is the magnitude of the force and d is the length of the moment arm, which is the distance between the line of action through the force and the line through the point which the moment is about.

(More generally, $$\vec{M}=\vec{r} \times \vec{F}$$, the cross product of the position vector and the force vector.)

If we take the net moment about one of the points, the reaction force at that point will produce no moment. This will give us an expression we can solve for the other reaction force.

I'll find $$\Sigma M_A$$, the net moment at A:
$$\Sigma M_A = F_1\frac{L}{3}+F_2\frac{2L}{3}-BL$$
This has to equal zero for it to be in static equilibrium, so:
$$0 = F_1\frac{L}{3}+F_2\frac{2L}{3}-BL$$
$$BL= F_1\frac{L}{3}+F_2\frac{2L}{3}$$
$$B= \frac{1}{3}\left( F_1 + 2F_2\right)$$

Now that we know B, we can use ∑F=0 to find A. There are no forces in the x direction, so we only need to look at the y direction. This gives:
$$\Sigma F_y = A+B-F_1-F_2$$
Letting ∑F=0 and solving for A gives:
$$A= F_1+F_2-B$$
$$A= F_1+F_2-\frac{1}{3}\left( F_1 + F_2\right)$$
$$A=\frac{1}{3}\left( 2F_1 + F_2\right)$$

Generally, you want to do the following:
1. Draw a diagram showing the forces.
2. Apply ∑M=0 through the point that eliminates the most unknown forces and solve for the remaining unknowns.
3. Apply ∑F=0 to solve for the remaining forces.

For your second problem, you have a part of the beam where a uniform load is applied. Calling the linear load density $$\lambda$$, the force it applies is given by $$F=\lambda d$$, where d is the length along which the load is applied. The load can then be treated as a point mass halfway between the points where the load is applied.

Last edited:

Hello,

I understand that you are struggling with some static equilibrium questions and you are looking for some help in understanding how to solve them. I can provide you with some guidance on how to approach these types of problems.

Firstly, it is important to understand the concept of static equilibrium. In simple terms, it means that the forces acting on an object are balanced and the object is not moving. This is achieved when the sum of all the forces acting on the object is equal to zero and the sum of all the moments (torques) acting on the object is also equal to zero.

To solve these types of problems, you will need to use the principles of Newton's laws of motion and the equations of equilibrium. In the first question, you are given the distances of the point loads from one end of the beam and the magnitudes of the point loads. You will need to use the equation of equilibrium for forces, which states that the sum of all the forces in the horizontal direction is equal to zero. This will allow you to solve for the reactions at the supports, which are the unknown forces in this problem.

In the second question, you are given the length of the beam, the uniformly distributed load, and two point loads at specific distances from one end of the beam. In this case, you will need to use the equations of equilibrium for both forces and moments. The equation for moments states that the sum of all the moments about any point is equal to zero. You can choose a convenient point to calculate the moments and use this equation to solve for the reactions at the supports.

I understand that it may seem overwhelming since you haven't studied physics in a while, but with practice and a good understanding of the principles involved, you will be able to solve these types of problems. I recommend reviewing the equations of equilibrium and Newton's laws of motion, and practicing with simpler problems before attempting more complex ones. Good luck!

## 1. What is static equilibrium?

Static equilibrium is a state in which all forces acting on a system are balanced, resulting in no movement or acceleration. This means that the sum of all forces in the horizontal and vertical directions must be equal to zero.

## 2. How do you solve static equilibrium problems?

To solve static equilibrium problems, you need to first identify all the forces acting on the system, including their magnitude and direction. Then, using the principles of equilibrium, you can set up equations to determine the unknown forces or angles. These equations typically involve the sum of forces in the horizontal and vertical directions, as well as the sum of moments about a point.

## 3. What is the unit for force in static equilibrium?

The unit for force in static equilibrium is typically Newtons (N). This is a derived unit in the International System of Units (SI) and represents the amount of force needed to give a mass of one kilogram an acceleration of one meter per second squared.

## 4. Can you use static equilibrium to solve problems with more than two forces?

Yes, static equilibrium can be used to solve problems with any number of forces. As long as the sum of forces and moments in all directions is equal to zero, the system will be in equilibrium. However, as the number of forces increases, the equations may become more complex and require more advanced techniques to solve.

## 5. How do you know if a system is in static equilibrium?

A system is in static equilibrium if all the forces acting on it are balanced and there is no net force or torque acting on the system. This means that the system will not move or rotate unless an external force or torque is applied. To verify if a system is in static equilibrium, you can use the equations of equilibrium to determine if the sum of forces and moments is equal to zero.

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