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.
(i) length of the runway
(ii) total time of take off
Hence sketch a velocity-time graph for the motion of the jet
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).