A) A 600 kg steel beam is supported by the two ropes shown in (Figure 1). Calculate the tension in the rope.

Question

Express your answer to two significant figures and include the appropriate units.
B) The rope can support a maximum tension of 3200 N . Is this rope strong enough to do the job? Choose the correct answer and explanation.
The rope can support a maximum tension of 3200 . Is this rope strong enough to do the job? Choose the correct answer and explanation.
a) Yes. The tension in the ropes does not exceed the maximum value, the ropes will not break.
b) No. The tension in the ropes does not exceed the maximum value, but since the ropes are not vertical the actual maximum tension they can support is lower and they will break.
c) Yes. The tension in the ropes exceeds the maximum value, but since we have two ropes they can support the beam without being broken.
d) No. The tension in the ropes exceeds the maximum value, the ropes will break.

Answer 1

What you need to do first is find the sum of the x/y force components (this is a lot easier if you make a free body diagram). In the diagram the tension force should be at an angle and has a Tx going along the x-axis and a Ty along the y-axis. With some simple trig. functions, we can say that Tx=sin(θ) and Ty=cos(θ). (check end note to see why these functions are used).

So now for the summations: the sum of all the forces in y= -weight + Ty. (check end note to see why weight is negative) (wondering why ∑Fy is used? check end note)
We can change this equation to be: ∑Fy= -W + Tcos(θ).
From Newton's second law, we know that F=ma, and in this problem the a=0 so F=0.
And now we can change the equation one more time to: W= Tcos(θ).
However, since we aren't given weight, you can make W=ma (a in this case would be 9.8m/s^2) --> ma= Tcos(θ). (don't understand why 'a' has taken on two different values? check end note).
Isolate T force. --> ma/cos(θ).
And then you have to divide by 2 because there are two ropes. Whatever you end up with is your answer!

Steps with numbers plugged in:
(1) ∑Fy= -W + Tcos(θ) = ma (but remember that a=0)
so ∑Fy= -W + Tcos(θ) is actually = 0
0= -W+ Tcos(θ)

(2) W = Tcos(θ) but we don't know W so we can make W =ma
ma = Tcos(θ)
600 • 9.8 = Tcos(θ)

(3) Isolate T
T= (600 • 9.8) / cos(θ)

(4) divide by 2 (because we have two ropes)
T = [(600 • 9.8) / cos(θ)]/2 => ANSWER!

I would help with the next part of the problem but there's no info for θ so I can't mathematically solve for you. But I can help with explanations: if you find that your answer above is less than 3200N, then the ropes cannot support the steel beam simply because it is too weak. The opposite is true: if the answer is greater than 3200N, then the ropes can support the beams because they are strong enough to hold.

End notes:
-On the free body diagram, depending on where θ is placed, this will determine how to assign Tx and Ty to sin or cos. For example, in this problem I have mentally placed θ above the tension force (closer to y-axis) and therefore Ty=cosθ, not sinθ. It's all trig!
-Weight is negative because if you draw the correct free body diagram then weight should be directed vertically down, along the negative portion of the y-axis. This is why it is negative in the equation.
-We use ∑Fy specifically because of where I placed θ on my free body diagram. Since a picture was not provided, I drew my own diagram; in it, θ was placed above the tension force and closer to the y-axis which is why I chose ∑Fy--> it's easier/makes more sense to calculate it from here.
-You might be wondering: wth, why does acceleration have two values in this problem? I'll try my best to explain: in the problem, we assume that the object is not moving and therefore has no acceleration and a=0. (which is value #1). But a couple steps later, we also define W as = m • a. 'a' in this case is actually equal to gravity, 9.8 m/s^2 so really, we haven't given acceleration two values at all; the variable is just being used twice.

I hope this helps!!! :D