Statics and work-energy questions as pdf and as doc.

Worked answers to statics questions

Worked answers to work-energy questions

**Ideas coming from this work**

Click here for summary as pdf or here as odt

**Plan of campaign**

**First**: draw your diagram (or use the diagram in the question paper). Mark in all the forces on the rod. Use a ruler. Draw long lines of action for the forces. Mark the angles (“α” and “not α”). Put all the information from the question into the diagram. Make the diagram big so you can put all the info in, and make corrections without making a mess.

**Second**: write your three equations

Resolve horizontally

Resolve vertically

Take moments about a chosen point

and simplify them as much as you can.

**Third**: do algebra to find your unknowns from your three equations.

**Choosing the point to take moments round**

In Gold 4 Question 7, for example, you could choose A or C. Both will give you the same information. A is better because moments round A gives you an equation with fewer unknowns in it.

100 (2 cos α) = 3 R_{2}

That’s good. Plug in cos α = (2√2)/3, which you can get from doing a little right-angled triangle with α and sin α = 1/3, and then:

R_{2} = (400 √2)/9

Straight away, just from that one equation.

Some students say: take moments around C instead, because then you get μ (which is what you want to find) in the moments equation. Sounds sensible, but isn’t sensible. You can’t get more information by taking moments round a different point. You can only get the same information in a simpler or more complicated form. Simpler is better.

You have three equations, resolving vertically, resolving horizontally, moments: μ will appear in at least one one of them. It’s easier for you if it appears in only one than if it appears in two or three.

The three simultaneous equations

μR_{1} = R_{2}

R_{1} = 10

R_{2} = 5

are simpler to solve than three equations every one of which contains all three of μ, R_{1}, and R_{2}. Aren’t they?

Take moments around the point which will give you fewest unknowns (usually: only one unknown) in your moments equation.

**What is an unknown?**

Sometimes in these equations you’re given the weight or mass of the rod, sometimes you’re just told weight = W or mass = m.

Sometimes you’re given the length of the rod, say 4 metres, sometimes you’re just told length = 4a or something like that.

Almost always, W or m or a are not really unknowns in your working. They are given quantities which just happen to be represented by letters rather than numbers. Your real unknowns will be things like the reactions and frictional forces.

**Which directions to resolve forces?**

In some mechanics questions with inclined planes, it is best to resolve along the plane and perpendicular to the plane.

In M2 statics? No. Best to resolve *vertically* and *horizontally*.

You’re pretty sure to have a number of vertical and horizontal forces on the rod (weight, normal reaction at ground, friction at ground, sometimes normal reaction and friction at a wall). There’ll be more forces that are vertical and horizontal than ones which are along the rod and perpendicular to the rod. So resolving *vertically* and *horizontally* is better.

**How to find moments when rod is resting on a drum or peg**

No surprise that when Edexcel collects together its hardest past-paper questions on statics, a number of them are about rods resting on pegs or drums. In those cases you have a force which is *not* vertical or horizontal: the normal reaction from the peg or drum on the rod. If the peg or drum is not smooth, i.e. there is friction from the peg or drum acting on the rod, you have two forces which are not vertical or horizontal.

Resolve those forces into horizontal and vertical components so that you can resolve all forces on the rod horizontally and vertically.

To calculate the moment of the force R_{2} around A in the diagram below, you can add the moment of the vertical component R_{2} cos α and the moment of the horizontal component R_{2} sin α

That gives you an anticlockwise moment of: 3 cos^{2}α + 3 R_{2} sin^{2}α

Since cos^{2}α + sin^{2}α = 1

moment = 3 R_{2}

You can get that more directly by forgetting for the time being about resolving vertically and horizontally. Just calculate directly the vertical distance from A to the line of action of R_{2}. That distance = 3 (along the rod). So moment = 3 R_{2}

**Using your calculator**

The general rule in pure maths is not to use your calculator until you have to, and to leave an expression like (400 √2)/9 as just (400 √2)/9 .

In mechanics, however, everything is approximate. You’re told that the rod is 4 metres long, but on a warm day it might be 4.001 metres. So there’s nothing lost by writing

R_{2} = 62.9 rather than R_{2} = (400 √2)/9

Nothing lost. Something gained if you find R_{2} = 62.9 easier to manage and to understand than R_{2} = (400 √2)/9

So in Gold 4 Q.7 you’d have

Moments around A: … R_{2} = 62.9

Resolving horizontally:… R_{2} = 3 μ R_{1}

Resolving vertically:… 3 R_{1} + 2.828 R _{2} = 300

Plug R_{2} = 62.9 into the third equation to get R_{1} , and then plug both R_{1} and R_{2} into the second equation to get μ.

Don’t give results in mechanics to more than 2 or 3 sig figs, unless specially asked. The exam papers tell you to take g = 9.8, which is correct only to two sig figs anyway. Writing μ = 0.5142594772 is worse, not better, than writing μ = 0.51, because the longer way of writing suggests that you have some accuracy to the tenth decimal place. Which you don’t.