A) A 600 kg stole beam is supported by the two ropes displayed in (Figure 1). Calculation the tension in the rope.**Express your answer come two far-reaching figures and also include the proper units.B) The rope can support a maximum stress and anxiety of 3200 N . Is this rope solid enough to carry out the job? choose the correct answer and explanation.The rope deserve to support a maximum tension of 3200 . Is this rope solid enough to execute the job? choose the exactly answer and also explanation.a) Yes. The tension in the ropes does no exceed the best value, the ropes will certainly not break.b) No. The stress in the ropes does no exceed the maximum value, but because the ropes space not upright the yes, really maximum stress and anxiety they have the right to support is lower and also they will certainly break.c) Yes. The tension in the ropes exceeds the best value, but since we have actually two ropes they can support the beam without gift broken.d) No. The stress and anxiety in the ropes exceeds the maximum value, the ropes will break. **

You are watching: A 1100 kg steel beam is supported by two ropes.

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Nov 11, 2015

What you have to do very first is find the amount of the x/y force materials (this is a lot easier if you do a cost-free body diagram). In the chart the tension force should be in ~ an angle and has a Tx going follow me the x-axis and also a Ty follow me the y-axis. V some straightforward trig. Functions, we can say the Tx=sin(θ) and also Ty=cos(θ). (check end note to view why these functions are used). **So now for the summations: the sum of all the forces in y= -weight + Ty. (check finish note to check out why load is negative) (wondering why ∑Fy is used? inspect end note) us can adjust this equation to be: ∑Fy= -W + Tcos(θ). Native Newton's 2nd law, we know that F=ma, and in this trouble the a=0 for this reason F=0. And also now us can readjust the equation one an ext time to: W= Tcos(θ). However, due to the fact that we aren't given weight, you have the right to make W=ma (a in this case would it is in 9.8m/s^2) --> ma= Tcos(θ). (don't know why 'a' has taken top top two various values? check end note). Isolation T force. --> ma/cos(θ). And then you need to divide by 2 because there are two ropes. Whatever you end up v is her answer! steps with numbers plugged in: (1) ∑Fy= -W + Tcos(θ) = ma (but remember the a=0) for this reason ∑Fy= -W + Tcos(θ) is actually = 0 0= -W+ Tcos(θ) (2) W = Tcos(θ) however we don't know W therefore we have the right to make W =ma ma = Tcos(θ) 600 • 9.8 = Tcos(θ) (3) isolation T T= (600 • 9.8) / cos(θ) (4) division by 2 (because we have actually two ropes) T = <(600 • 9.8) / cos(θ)>/2 => ANSWER! ns would help with the next component of the problem however there's no details for θ so i can't mathematically resolve for you. Yet I can aid with explanations: if you find that her answer over is much less than 3200N, then the ropes cannot support the stole beam simply because it is as well weak. The opposite is true: if the answer is higher than 3200N, then the ropes have the right to support the beams because they are strong enough to hold. End notes: -On the free body diagram, relying on where θ is placed, this will certainly determine exactly how to entrust Tx and also Ty to sin or cos. Because that example, in this trouble I have actually mentally placed θ above the tension force (closer come y-axis) and therefore Ty=cosθ, not sinθ. It's all trig! -Weight is an adverse because if you attract the correct cost-free body diagram climate weight must be directed vertically down, along the an adverse portion the the y-axis. This is why it is negative in the equation. -We use ∑Fy specifically since of wherein I inserted θ ~ above my cost-free body diagram. Because a snapshot was no provided, I attracted my very own diagram; in it, θ to be placed above the stress and anxiety force and also closer come the y-axis which is why I determined ∑Fy--> it's easier/makes much more sense to calculation it from here. -You could be wondering: wth, why go acceleration have two worths in this problem? I'll shot my best to explain: in the problem, us assume that the thing is no moving and therefore has actually no acceleration and a=0. (which is worth #1). However a couple steps later, we also define W as = m • a. 'a' in this situation is in reality equal to gravity, 9.8 m/s^2 for this reason really, us haven't provided acceleration two values in ~ all; the variable is simply being supplied twice. Ns hope this helps!!! :D**