Brain Teasers ? or 50 Years On........... ?

Now I know some of the factors above for resolving Drift are approximations (0.7 and 0.9), but they are all easy to remember and easy to apply.

Of course the real factors would be √0/(2); √1/(2); √2/(2); √3/(2); √4/(2) 

[that is (root 1), divided by 2, NOT root(1/2) etc] but not so easy to apply when you're flying the plane as well !

Our Aeroplane

"Our aeroplane" is flying 270 deg at 120 kts at 30 degrees (270 – 240)  to a wind of 30 kts

Max Drift, using the 1 in 60 Rule is 15 deg

Resolving the Max Drift into the cross-wind component at 30 deg requires a factor of 0.5

Hence Cross-wind component (Drift) is 15 x 0.5 = 7.5 deg of Drift

Hence Required Heading is 270 – 7.5 = 263 deg (approx)

For the Headwind component use 0.9 x 30 = 27 kts

Hence Ground Speed = 120 – 27 = 93 kts  (approx)

ok Lionel,  you can come back now. Its all over !

The diagram shows three forces acting at a point on a horizontal plane. Find ;

• The magnitude of the resultant force
• The angle the resultant makes above the positive x-axis

I think this is a more typical A-Level question than the previous navigation application of vectors. But it’s still vectors

Probably completely wrong but here goes:

Magnitude: 17.5N

Angle: approx 36.5 degrees

I used pythag. to calculate the addition of the x and y vectors [(170^1/2)N, 39.5 degrees] then the Cos and sine rules (cos & sine tables courtesy of the web) to calculate the addition of the third vector.

Hi sj,

Good shot, but not quite close enough to hit the mark.

Best approach is to;

• Resolve the 4N force into x/y components, (you will need your sine and cosine tables for this)
• Use pythag to add the new x/y vectors. (you will need to find a square root for this)
• Then in the same triangle, use Tan = (side opp)/(side adj) to find the direction (you will need your tangent tables for this)

As I say, your figures are close, but not yet close enough. Good try !

I get the following:

17.01 N

30.72 degrees to the x-axis

I had followed the same method outlined by Don. Just hope that no "school-boy" errors have crept in ?

Yup I agree with sophiebear's figures now.

Length of vector = (289.41)^1/2 = 17.01

tan Θ = 8.69/14.625 = 0.594188

Θ = 30.72 degrees

Peter, sj,

Brilliant. No schoolboy errors. Top marks.

Takes you back, hey ?

Worth remembering that at work you need to get safety-critical things independently checked. One mistake and someone dies................well, not always, but sometimes !

A cyclist (hello winky !) starts from rest and accelerates at 1.5 m/s/s for 8s along a straight horizontal road and then continues at a constant speed.

A motorist (oh dear !) starts from alongside the cyclist at the same time and in the same direction and accelerates at a constant rate for 6s, reaching a maximum speed Vm/s. The car then decelerates at a constant rate until it comes to rest after a further 18s.

At the instant the car comes to rest it is overtaken by the cyclist.

Calculate :

• the greatest speed attained by the cyclist
• the distance travelled by the cyclist before overtaking the car
• the value of V

Just hope this isn't safety-critical.....

1) Greatest speed reached by cyclist = 12 m/s

2) Distance travelled before overtaking = 240 metres

3) V = 20 m/s

Bang-on with the answers Peter. Well done.

As for safety critical, this is where winky comes in....................(but you can all give it a go)

assume that the motorist opens his car door, just at the moment the cyclist is hurtling past at 12m/s and .......

..........................splat !

Assume that:-

the cyclist weighs 80kg

the cyclist leaves his bike (involuntary) at 12m/s and impacts the car door at 12m/s

the car door behaves as if it were a brick wall at rest

the cyclist is 300mm chest to back and impacts the door chest first (rather than head first)

the duration of the impact Δt can be adequately estimated by taking the length of the object (cyclist 300mm) and dividing by the relative velocity at impact (in this case 12m/s)

What is the average impact force on the cyclist ?

As a check, try a similar calculation between  an aeroplane at 180mph and a 500mm long duck at 20mph in a head-on collision (with the duck flying towards the plane). Again, assume it is the duck that distorts ! winky might find this calculation more palatable

Don

1) I think the cyclist experience 38,400 Newtons

2) I have a problem with the duck question - though arguably less of a problem than the duck himself. I believe we need the mass of the duck to work out the impact force. This could be difficult to determine post-impact.

For the purposes of calculation I have assumed a 2 kg duck (might feed a family of 4). I then get the impact force = 31,969 N

Regards,

Peter

Don

That 1 in 60 rule is quite cool having been explained so well.

Whoops !!!

Apologies, you are dead right. I had in mind a 1kg Mallard, but agree that a 2kg Aylesbury might be a more fulfilling  feast.

Favg = m*V^2÷L

V = (200x5280x0.3048) ÷ (60x60) = 89.4 m/s

Favg = 2x89.4x89.4 ÷ 0.5 = 31.98 kN

Which is about the same as being stationary and being hit by a three and a quarter tonne mass. Or in the case of winky, the slightly larger 38.4 kN force or a 3.9 tonne (say) 4 tonne mass

Probably a dead duck whichever way you look at it .....................

Thank you Lionel, that's very kind of you to say so.

Cheers

Don

Hi sj,

I have outlined my calcs above in response to sophiebear. and also the arithmetic for the aeroplane v duck

For the car v cyclist travelling at 12m/s i get Favg = 80 x 12 x 12 ÷ 0.3 = 38.4 kN

This is the same as sophiebears' answer.

Estimating impact loadings is very difficult. That's why I set out so many parameters in the initial question.

The deformation of the bodies involved is largely responsible for determining the impact loading and the consequences. Your estimate might well be far more realistic than mine. Only a practical test would help us determine who's estimate is the better one........................

................now where is winky when you most need him ?

Ok, a bit of light relief. This one doesn't need A-Level maths, but it does need a bit of care and the recollection of a really fundamental concept.

THE BARON'S TREASURE

Baron Von Brinkhoff kept his gold in a treasure house. In each room of the treasure house there were as many chests as there were rooms. In each chest there were as many gold coins, as there were chests in that room. All the gold coins were of the same size and equal value.

When he died, the Baron's will was that his faithful barber should receive one chest of gold coins. The remaining coins were to be divided equally between the Baron's three sons.

The three sons were proud and fierce men who would definitely resort to bloodshed if the coins could not be divided equally.

The question is simply: -

a) Was there blood shed?
b) Was there no bloodshed?
c) Is there no way of knowing whether there was blood shed or not.

PS. The proof is the real requirement, not simply a one in three guess.

There are a total of  x^3 - x  i.e. x(x^2 - 1) coins to be divided

This means there are x* (x+1) * (x -1) coins

since x-1, x and x+1 are 3 consecutive numbers then one of these factors must be divisible by 3.

No bloodshed (assuming there was more than 1 room/chest/coin)

hey guys, slow down ! I can't type the questions this fast !!

Well done sj. Nice, succinct answer.

BTW if there was only one room, one chest, one coin, then the barber would get it and the sons would have an equal share of nothing.....................er, yes, I see your point, the barrber had better run !

I think there should be no blood shed.

If the number of rooms is n

The number of chests is n^2

The number of coins is n^3

When the barber takes his 1 chest, then the number of remaining chests is (n^2-1)

So the remaining number of coins to be divided amongst the sons is n * (n^2 - 1)

Recognising n^2 - 1 = (n+1) (n-1)

Then the total number of coins to be divided amongst the sons is:

(n-1) * n * (n+1)

Since these are 3 consecutive numbers, then one of the numbers will always be exactly divisible by 3.

The special case of 1 room means the sons get nothing - but at least they all get the same !

SJ - looks like our responses crossed in the ether !

A crane moves a load

A crane moves a load of 3,000 kg A, suspended by two light cables AB and AC attached to a movable beam BC.

The load is moved in the direction of the line of the supporting beam BC during which time the cables maintain a constant angle of 40°to the horizontal. See diagram.

The load is initially moving with constant speed. What is the tension in each cable ?

The crane then moves the load with a constant acceleration of 0.4 m/s/s. What is the tension in each cable ?

Assume the cables are inextensible, the load acts as a point mass, there is no air resistance.

Well done Peter.

i was a bit lazy to substitute the numbers 

enjoy

ken