# Calculating Reactions (Equations and Moments)

• JMxBelfast
In summary, the equation for finding moments about A is M = (-30 x 0.5) + (-25 x 2.5) and the equation for finding moments about B is M = (-77.5 x 0.5) + (-25 x 2.5).
JMxBelfast

## Homework Statement

Need help at "Forming an equation based on the fact that all the vertical forces must be in equilibrium". I'm to calculate the reactions at the supports A and B by doing:

(i) Forming an equation based on the fact that all the vertical forces must be in equilibrium
(ii) Taking moments about A to find the reaction at B
(iii) Using your value for the reaction at B in the equation formed in part (i) to find the reaction at A

## Homework Equations

I thought that the equation that I need to form would have been an equation for finding moments,
moment = force x distance.

## The Attempt at a Solution

For part (i) I done:
M = (-30 x 0.5) + (-25 x 2.5)
= -77.5

It states that they are in equilibrium so does that mean I need to add '(30 x 0.5) + (25 x 2.5)' to the equation so that it become an equilibrium? Or is what I'm doing for part (i) actually what I should be doing for part (ii) as I need to take moments for that section of the question? If so then what equation must I form for part (i)?

Part (i), I believe, is simply summing all of the vertical forces and setting equal to 0. In other words, the sum of forces up is equal to the sum of forces down.

Part (ii), I believe, involves taking the sum of moments about point A (one of those moments being the upward force at point B) and setting them equal to 0. Or, as I prefer to think about it, the sum of CW torques is equal to the sum of CCW torques.

What I didn't understand is what that 1.25 kN/m is referring to. Also, I didn't see a weight for the tool that is laying on the two supports.

And yes, moment = force x distance.

Hmmm. It's hard not to just give the answer. Ok, you say you expected to be thinking about moments. That is one thing to think about, and you know the rotational form of Newton's law

Sum of torques = rotational Inertia x angular acceleration

Since the beam isn't spinning. Sum of torques = 0 about any axis you care to pick. That may come in handy at some point.

However, the beam also isn't moving, isn't accelerating. Do you know some similar facts for linear motion?

## 1. How do you balance chemical equations?

To balance a chemical equation, you need to ensure that the number of atoms of each element on the reactant side is equal to the number of atoms of the same element on the product side. This can be done by adjusting the coefficients of the reactants and products.

## 2. What is the difference between a reactant and a product?

A reactant is a substance that is present at the beginning of a chemical reaction and undergoes a change. A product is a substance that is formed as a result of the reaction. Reactants are written on the left side of the chemical equation, while products are written on the right side.

## 3. How do you calculate the moment of a reaction?

The moment of a reaction is calculated by multiplying the force applied to an object by the distance from the point of rotation to where the force is applied. This is represented by the equation M = F x d, where M is the moment, F is the applied force, and d is the distance.

## 4. What is the difference between a balanced and unbalanced equation?

A balanced equation has an equal number of atoms of each element on both the reactant and product side, while an unbalanced equation does not. In other words, a balanced equation follows the law of conservation of mass, which states that matter cannot be created or destroyed in a chemical reaction.

## 5. How do you handle coefficients when balancing equations with fractions?

If a balanced equation requires the use of fractions for coefficients, you can multiply the entire equation by the lowest common denominator to eliminate the fractions. Then, proceed to balance the equation as you would normally by adjusting the coefficients.

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