What Are the Equations for Equilibrium and Motion in Physics Problems?

[Gundam101]

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).

 

[SteamKing]

For Problem 1, you should draw a free body diagram of the boom with all loads and reactions indicated.

Apply the equations of equilibrium to the boom in order to find the tensions and reaction at the attachment point of the boom.

BTW, your moment equations are incorrect. If you take moments about the attachment point of the boom with the wall, you will see a couple of mistakes in your moment equation. Why did you multiply the weight of the boom by the cosine of the angle between the boom and the cable? Why didn't you do the same for the load?

 

[SteamKing]

For Problem 2, you are not applying the equations of motion correctly.

For the first part of the takeoff, the jet accelerates at a constant velocity for a fixed amount of time.

It's not stated explicitly, but you should assume that when t = 0, the position of the jet s = 0 and the velocity of the jet v = 0. Applying the constant acceleration of 4 m/s^2 over the time interval of 5 sec. should give you the velocity directly at the end of this interval. You can also determine how far the jet has rolled in this time. This distance is part of the total distance required for takeoff.

Once the jet is rolling and the engines are throttled to max. thrust, then you have a situation where there is an initial velocity and a known acceleration and you have to find the time required to reach takeoff v = 300 m/s and find the distance it takes to reach this velocity.

 

[Gundam101]

Ahh, so for problem 2 I have to factor in the distance traveled from rest to 20 ms^-1 which is 50m. So the total length would be 50 + 4480 = 4530m.

However for problem 1 I am still lost. I also do not understand what you mean by "boom." Is that the beam? Are you saying that the cosine of the angle is irrelevant? This particular question is causing me serious headache.

 

[Nickie]

As for the velocity-time graph, it would have a straight line with a gradient of 10 and a starting point at (0,20), then a horizontal line at (33, 300) to represent the constant speed needed for take off.



Your attempt at solving the problem is commendable. However, there are a few errors in your solution. Let's start with the free body diagram. The forces acting on the system are the weight of the beam (900N), the weight of the mass (4500N), and the tension in the cable (T). Since the beam and the mass are connected, they can be considered as one object with a combined weight of 5400N. The tension in the cable is the only unknown force in this system.

Next, for moments, you have correctly set up the equation T * 2.5 = (900cos35.6 * 2.5) + (4500 * 2.5). However, you have made a mistake in solving for T. The correct solution is T = (900cos35.6 * 2.5) + (4500 * 2.5) / 2.5 = 6400N.

For the net force on the beam from the hinge, you have correctly set up the equation as upward force - total downward force = 0. However, you have made a mistake in calculating the upward force. The correct solution is upward force - (6400 + 5400) = 0, which gives an upward force of 11800N.

Moving on to linear motion, for the length of the runway, you have correctly used the equation v^2 = u^2 + 2as. However, you have made a mistake in substituting the values. The correct solution is 300^2 = 20^2 + 2(10)s, which gives a distance of 4480m.

For the total time of take off, you have correctly used the equation s = ut + 1/2 at^2. However, you have made a mistake in setting up the equation. The correct equation is 4480 = 20t + 1/2 (10)t^2, which gives a total time of 32s.

Finally, for the velocity-time graph, you have correctly identified the starting point at (0,20) and the horizontal line at