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## Homework Statement

Just in case my interpretation of the diagram does not give a clear representation, here is a link to all of the questions. http://imgur.com/a/jsv9e

The question begins by asking me to state the necessary conditions for a body to be in a state of equilibrium.

It is a 1.5m wall on the left hand side to which a cable is connected from. This cable extends 2.5m to the right and is run over or connected to a beam (the diagram is very unclear to me but could you kindly help with both possibilities?), which is diagonally connected to a hinge at the bottom of the 1.85m wall, to where it is connected to a mass. The angle between the beam and cable, in the area where the cable runs over or is connected to the beam, is unknown and represented by θ. The beam has a weight of 900N in its centre and the mass has a weight of 4500N. The mass of the cable and beam are negligible.

I am asked to:

Draw a free body diagram showing the forces acting on the system. (3 marks)

Calculate the tension T in the cable. (2 marks)

Calculate the net force on the beam from the hinge. (3 marks)

The other question states that during the initial phase of take off, a BWIA jet has an acceleration of 4.0 ms^-2 lasting 5 seconds. The burner engines are then turned up to full power for an acceleration of 10 ms^-2. The speed needed for take off is 300 ms^-1.

Calculate the:

(i) length of the runway

(ii) total time of take off

Hence sketch a velocity-time graph for the motion of the jet

## Homework Equations

I am not sure if this is regarding equations that the question gave me to use or equations that I personally used. The question did not give me any equations to use as I had to use my own brain to figure which ones were needed. However, we do not use calculus at our level so I used:

Moments: F1 * d1 = F2 * d2

Linear Motion: v^2 = u^2 + 2as and s = ut + 1/2 at^2

## The Attempt at a Solution

For moments, I figured that the cable provides an anti-clockwise force to balance the clockwise forces of the beam and the mass (I included them both as I thought they were both connected to each other and the cable pulled them up). Thus T * 2.5 = (900cos35.6 * 2.5) + (4500 * 2.5)

I got 35.6° from using tan θ = opposite/adjacent > tan θ = 1.85/2.5 > θ = tan inverse 1.8/2.5 = 35.6°

I used cos as the vertical component is adjacent to the angle.

So T = 5231.79

T = 5230N (3 s.f.)

Then I figured that another force would be needed to balance the system as the cable would be pulling the entire system down to the left.

So to find the upward force:

total upward force - total downward force = 0

upward force - (5231.79 + 900cos35.6 + 4500) = 0

upward force = 10463.58

upward force = 10500N (3 s.f.)

Then for linear motion:

(i) u = 4 * 5 = 20ms^-1, v = 300ms^-1, a = 10ms^-2

v^2 = u^2 + 2as

300^2 = 20^2 + 2(10)s

(300^2 - 20^2)/20 = s

4480m = s

(ii) s = ut + 1/2 at^2

4480 = 20t + 1/2 (10)t^2

4480 = 20t + 5t^2

5t^2 + 20t - 4480 = 0

Using quadratic formula [-b +/- sqrt (b^2 - 4ac)]/2a

a = 5

b = 20

c = -4480

I worked out t = 28s and t = -32s, discarded the negative answer and used 28s as the time for the second part of the motion. The first part took 5s as it accelerated from 0 ms^-1 to 20ms^-1 in 5s. So 28s + 5s = 33s (the total time of take off).