4.1.5 Using Moments - Harder

Moments: Nonuniform Rods

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Definition:

A nonuniform rod is one which does not (necessarily) have its centre of mass at the centre of the object. Reasons for this could include:

The rod has different thicknesses at different points.
The rod is made from different materials with different densities.

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Problem 1:

A nonuniform rod $AB$ , of length 4 m and weight 6 N, rests horizontally on two supports, $A$ and $B$ . Given that the centre of mass of the rod is 2.4 m from the end $A$ ,

Question : find the reactions at the two supports.

Solution:

Resolve the forces:

Let the reactions at supports $A$ and $B$ be $A$ and $B$ , respectively.
The weight of the rod (6 N) acts at the centre of mass, 2.4 m from $A$ .

Apply the equilibrium conditions: Vertical force balance:

A + B - 6 = 0 \quad \Rightarrow \quad A + B = 6

Taking moments about point $A$ :

6 \times 2.4 - 4B = 0

14.4 = 4B \quad \Rightarrow \quad B = \frac{14.4}{4} = 3.6 \text{ N}

Substitute $B$ back into the force balance equation:

A + 3.6 = 6 \quad \Rightarrow \quad A = 6 - 3.6 = 2.4 \text{ N}

Result:

Reaction at $A = :success[2.4 N]$ .
Reaction at $B = :success[3.6 N]$ .

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Problem 2:

A non-uniform bar $AB$ of length 5 m is supported horizontally on supports $A$ and $B$ . The reactions at these supports are 3 N and 7 N, respectively.

Questions:

(a) State the weight of the bar.
(b) Find the distance of the centre of mass of the bar from A.

Solution:

Part (a): Weight of the Bar

Vertical force balance:

Let the weight of the bar be $W$ .
The sum of the reactions at the supports equals the weight of the bar.

3 \text{ N} + 7 \text{ N} - W = 0

W = 10 \text{ N}

Result: The weight of the bar is 10 N.

Part (b): Distance of the Centre of Mass from A

Take moments about point $A$ :

Let $d$ be the distance of the centre of mass from $A$ .
The distance from $B$ to the centre of mass is $5 - d$ .

10g \times d = 7g \times 5

10gd = 35g

d = \frac{35g}{10g} = 3.5 \text{ m}

Result: The distance of the centre of mass of the bar from $A$ is 3.5 m.

Tips:

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Choose the pivot point wisely: Pick a point where multiple unknown forces act (e.g., a support or hinge). This simplifies the problem by eliminating those unknown forces from the moment equation, making it easier to solve for other quantities.
Resolve forces carefully: Ensure all forces, including angled or distributed loads, are resolved into horizontal and vertical components. For angled forces, use trigonometry: $F_x = F \cos(\theta), F_y = F \sin(\theta).$
Consider all equilibrium conditions: In addition to moments $( \sum \text{Moments} = 0)$ , ensure the system satisfies both horizontal and vertical force equilibrium: $\sum F_x = 0 \quad \text{and} \quad \sum F_y = 0.$

Using Moments - Harder Simplified Revision Notes for A-Level AQA Maths Mechanics

4.1.5 Using Moments - Harder

Moments: Nonuniform Rods

Definition:

Problem 1:

Solution:

Problem 2:

Solution:

Part (a): Weight of the Bar

Part (b): Distance of the Centre of Mass from A

Tips:

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