vector . .

QUESTION 1

 The diagram shows a stunt person of mass 82kg holding on to a rope. The rope on either side of the person makes an angle of 5 degrees to the horizontal

a) Determine the tension T in the rope.

b) What would be the consequence of making the angle between the rope and the horizontal equal to zero? 

Answers :

Let O be the point where the man or the weight force act towards earth.

Let the tension in  the ropes be T1  and T2.
Then the Component of tenesion force along and perpendicular dirctions should balance each other or thenet force is zero for equilibrium.

Horizontal componets T1cos(185) + T2cos(-5) = 0. This requires -cos5T1 and T2cos5 . Or the magnitudes of T1 = T2 = T say.
The vertical components: T1sin(185)+T2sin(-5) + 82g = 0
82 g = = 2T sin5.
T = 82g/2sin5, where g= 9.91 meter/s^2.

b)It is not possible as T = 82g/2sin0 =  82g/0 = infinite. No rope can bear this.



QUESTION 2

 A Crow flies in 2 successive displacements to a point that is 80m to the west. Its first displacement is 80m in a direction θ = 35degrees west of north.


a) What is the magnitude of its second displacement?


b)What is its direction as measured by the angle θ measured west from north?


answers :
Start from the origin.

The first displacement goes UP ( North ) and leans over to the left ( 35˚ West ) and is 80 m.

From the end of that line, the second displacement is down to x-axis line and along that line to the –80 mark.... which is the 80m West where the crow finishes up.

You need the length of the line that come down and its angle. ( and direction )

The direction is South 27.5˚ West.

That is easy because the triangle made by the two displacements and the x-axis, is an isosceles triangle with a vertex of 55˚ and the paired angles 62.5˚. Then you should see that you just have to take 35˚ from the top 62.5 ˚ and get the 27.5˚

The length - there are lots of ways to work this out.... easiest is the sine rule

A / sin A = B / sin B

80 / sin 62.5 = L / sin 55

L = 73.9 m




QUESTION 3

1. Fishing boat:
6m/s due North = wind velocity
10m/s due 240 degrees is the motor velocity
and the current velocity happens to be 3m/s due east.

The one below is for subtraction:
A spacecraft of mass 800kg is maneuvered by 2 rockets. The first rockets fires from the south west with a thrust of 1000N. The second fires from the south east with a thrust of 1000N.
What is the:
Net force on the spaceship
and the net acceleration of the spaceship?

Answers :

The simplest method is to split the vectors into Cartesian co-ordinates (North/South and East/West).

Problem 1:
Vector 1 is 6m/s North and 0m/s East/West
Vector 2 is at 240 degrees, aka West 30 degrees South. It breaks down into 10m/s * sin 30 degrees in the South direction plus 10m/s * cos 30 degrees in the West direction. These are 5.0 South and 8.66 West respectively.
Vector 3 is 0m/s North/South and 3m/s East.

Adding up the North/South components of the vectors, we get 6m/s North + 5m/s South + 0, which comes out to a total of 1m/s North.

Adding up the East/West components, we get 0 + 8.66m/s West + 3m/s East, which comes out to 5.66m/s West.

So the components of the final vector will be 1m/s North and 5.66m/s West. These are the vertical and horizontal lines in a triangle, with the third line being the final vector.
The size of the final vector will be the square root of the sum of the squares of the two components. sqrt (5.66^2 + 1^2) is 5.74m/s.

We already know (from the components) that the direction of the final vector will be somewhere between North and West, and will be more West than North. This means that the compass bearing will be between 270 degrees (West) and 315 degrees (North-West).

The cosine of the angle from due West will be the West component (5.66) divided by the total vector size (5.74). 5.66/5.74 is 0.986, and the inverse cosine of this is 9.58 degrees. Adding this to the West direction (270 degrees), we get a final direction of 279.58 degrees.

Given that the original question components were only accurate to one significant figure, your answer of "5.74m/s due 279.58 degrees" would become "6m/s due 280 degrees".

Problem 2:
Part A:
1000N from the south east towards the north west, plus 1000N from the south west towards the north east.

Vector 1 is 1000N * sin 45 degrees (707N) North plus 1000N * cos 45 degrees (707N) West
Vector 2 is 1000N * sin 45 degrees (707N) North plus 1000N * cos 45 degrees (707N) East

Final vector is 707N + 707N North (total 1414N North) plus zero newtons East/West
= 1414N due North (0 degrees)

The acceleration of the spaceship can be calculated from the equation F=ma

F = 1414N due North
m = 800kg
a = 1414 due North / 800kg = 1.77ms^-2 due North

Given that the original question components were only accurate to one significant figure, your answer of "1.77ms^-2 due North" would become "2ms^-2 due North".