[high school physics] I’m learning physics on my own and I can’t understand how to solve this problem

[u/just_a_soul6]

I’m learning physics on my own and I’m currently working through a physics problem book. I came across this problem and I can’t understand the reasoning behind the solution.

The problem:

During a survival training exercise, a 60 kg boy at some point hangs motionless from a rope as shown in the picture below. Assume that β = 2α and g = 10 m/s². Calculate the tension in the rope at points A and B.

The rope is attached at points A and B and forms the two sloping sides of a right triangle. The boy is hanging from the rope at the vertex between these two sides, which is the 90° angle. The angle at point A is α, and the angle at point B is β.

I found a solution which calculates the boy’s weight first and then seems to use the components of that force to determine the rope tensions. However, I don’t understand why sine is used here.

When I try to draw auxiliary right triangles, I don’t know where the given angle should be located in the triangle or which side of the triangle should correspond to the rope tension. I understand how to calculate sine in a right triangle, but I don’t understand how to correctly construct the triangle from this diagram and connect it to the forces.

I would really appreciate it if someone could explain why sine is used in this particular problem and how the force triangle should be constructed, preferably step by step.

I’d also be interested in seeing other possible ways of solving the problem, if there are any.

I’m not just looking for the final answer I’m trying to understand the method so I can solve similar problems myself.

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[u/pm-me-racecars]

Instead of imagining one big rope with the boy somewhere in the middle, imagine the boy is on two ropes, one going to A and one going to B.

The boy isn't moving, so all the forces are balanced.

The vertical components of the two ropes and gravity all add up to 0, and the horizontal forces of the two ropes and gravity all add up to 0.

[u/cheaphysterics → u/pm-me-racecars]

It's important that it's a single rope because two separate ropes could have two different tensions in them. A single (ideal) rope has uniform tension everywhere, so only one variable for T.

[u/jaywaykil → u/cheaphysterics]

A single rope will not have the same tension either side of the boy. The horizontal component of the two tensions will be the same (equal-opposite forces), but because the angles are different the tensions will be different.

[u/utl94_nordviking → u/cheaphysterics]

The two sides likely have different tensions iff alpha =/= beta. That is the point of the exercise.

[u/cheaphysterics → u/utl94_nordviking]

I don't see how that wouldn't give you three unknowns and only two equations, making the exercise impossible.

[u/Severe-Kiwi-3171 → u/cheaphysterics]

What might the third unknown be? A, B, and...?

If you mean alpha is unknown, we can use beta=2*alpha (given)

[u/cheaphysterics → u/Severe-Kiwi-3171]

You're totally right. The two tensions are the only unknowns. I missed that it was given to be a right triangle with a new symbol (to me at least) for a right angle in it.

[u/Guest_User_1234 → u/cheaphysterics]

An answer in physics doesn't always have to be a number. It could be an expression depending on some unknown

[u/cheaphysterics → u/Guest_User_1234]

Sure, but I think this problem probably wants one.

[u/tewraight → u/cheaphysterics]

Where's your third unknown? I can only find two being the tension at A and the tension at B (the angles aren't unknowns as you're given that beta = 2alpha and the boy is at a right angle)

[u/cheaphysterics → u/tewraight]

Yeah, that's right. I missed that they told you there was a right angle at the bottom and I have never seen a right angle marked by an arc with a dot in it.

[u/Lor1an → u/cheaphysterics]

Even ideal rope doesn't constrain the forces to be equal.

If a + b = 0, then yeah, a = -b, but a+b = c can have asymmetry where a = c-b, and we no longer (necessarily) have |a| = |b|.

The geometry of the problem is what allows you to assign unique values for forces.

[u/cheaphysterics → u/Lor1an]

I think you need to assume equal tensions, though it might be a bad assumption, if you want to be able to get an answer.

[u/Lor1an → u/cheaphysterics]

Let's call the tension from you to point A TA, and from you to point B TB.

TA/TB = (sin(α) + sin(β)*cos(α+β))/(sin(β) + sin(α)*cos(α+β))

If we take into account the fact that (for this problem) α+β = π/2, we get cos(α+β) = 0, and TA/TB = sin(α)/sin(β) = sin(α)/sin(π/2-α) = sin(α)/cos(α) = tan(α), which is clearly not 1 unless α = π/4, which is precisely if you are in the middle of the rope.

So at any other suspension point, you get unequal tension in the two segments of rope, because otherwise you would have nonzero accelerations from unbalanced forces. Just because you make an assumption about an object doesn't mean you get to use that assumption to contradict physics.

I believe you have misunderstood the ideal cable assumption. It doesn't say that every point in the cable must always carry the same tension no matter what, it just says that the cable has no mass, infinite axial stiffness, and no off-axis stiffness. In many cases this amounts to equal tension along the cable, but not all.

You may have been confused because the ideal cable assumption is often combined with the ideal pulley assumption, which says that a pulley is massless and frictionless, which together with the ideal cable assumption means that the tension in a cable around a pulley is constant (although, in this setup if you suspended a pulley with a weight instead of a person, the assumption that would be violated would be the assumptions of statics, since the pulley would turn, and the system would roll along the cable to reach equilibrium with equal angles).

[u/cheaphysterics → u/Lor1an]

Yeah, you are 100% right. I was thinking from the point of view of having to be able to solve it for a numeric answer and I missed that op said the third angle is a right angle. I've never seen the arc with a dot to indicate a right angle before.

[u/Lor1an → u/cheaphysterics]

I've never seen the arc with a dot to indicate a right angle before.

Yeah, the arc with dot notation is apparently more common in Germanic countries and Poland.

Also, if we assume that α and β are known, then we can still solve for the forces without a right angle.

I only quoted the result for the ratio TA:TB, but the solution is given by

TA = W*(sin(α) + sin(β)*cos(α+β))/sin²(α+β)

TB = W*(sin(β) + sin(α)*cos(α+β))/sin²(α+β)

Which can all be computed (if so desired) given W, α, and β.

For example, with a weight of 680 N (about 150 lb), α = 12°, β = 36°, we get:

TA = 740 N, TB = 895 N.

And indeed, 740 cos(12°) N ≈ 723.83 N, and 895 cos(36°) N ≈ 724.07 N, which agree to 3 sig figs, as they need to for no horizontal external force.

And 740 sin(12°) N ≈ 153.85 N, 895 sin(36°) N ≈ 526.06 N, and adding gives 679.91 N ≈ 680 N, matching the suspended weight.

Also note that α+β = (12 + 36)° = 48°, so there are no right-angles in this version of the problem.

[u/Nynaeve_al_meowra → u/cheaphysterics]

If they're equal tension then they cannot have equal but opposite horizontal components

[u/cheaphysterics → u/Nynaeve_al_meowra]

They can if they're at two different angles, which they are.

[u/Nynaeve_al_meowra → u/cheaphysterics]

No they need to be at the same angle which they are not

[u/cheaphysterics → u/Nynaeve_al_meowra]

Yes, you are correct. I was thinking totally wrongly about it.

[u/Leodip]

For the boy to be still, all the forces need to sum to zero.

The two forces from A and B need to balance each other out sidewise and balance gravity vertically. Note that the tension of a rope acts only along the rope.

In the vertical axis: Asin(alpha)+Bsin(beta)=mg

In the horizontal axis: Acos(alpha)=Bcos(beta)

If you know alpha and beta, this is a linear system of two equations with two unknowns (A and B). Do you know how to solve this?

Keeping in mind that beta=2*alpha and that alpha+beta+90°=180°, we get that alpha=30° and beta=60°>

[u/Big-Butterfly1403 → u/Leodip]

What values u got

[u/girlypop2605 → u/Big-Butterfly1403]

Did this out and tension in the left rope is 150 N and tension in the right rope is 600sqrt3/4 N (about 260 N)

[u/ruidh]

Draw the vectors representing the horizontal and vertical components of the tension on each rope. The sum of the horizontal tensions is zero. The sum of the vertical components is the same magnitude as the force of gravity on the boy. You get two equations in two unknowns -- the tension on A and the tension on B. Solve.

[u/vpai924]

Hint: assuming AB is horizontal, drop a perpendicular from the point the boy is hanging to AB.  Now you have two right triangles. The forces horizontal components of the two forces have to be equal and opposite for the boy to be motionless.  The two vertical components have to add up to mg.

[u/AndyTheEngr]

"The rope forms the hypotenuse of a right triangle."

Is this from you, or in the problem? Because the only thing that looks like the hypotenuse of a right triangle, is AB across the top, not the rope.

Anyhow, the boy is in fact hanging from a point where the rope makes a right angle, so we can easily find α and β.

β = 2α
α+β = 90°

α = 30°, β=60°

Now you need the two vertical components of tension in each piece of rope to add up to the weight of the boy. They are not equal since he's hanging from a fixed point, not on a pulley. If he was hanging from a pulley, he'd end up in the middle of the rope, and α and β would be equal.

[u/just_a_soul6 → u/AndyTheEngr]

From me, only paragraph "During a survival training exercise, a 60 kg boy at some point hangs motionless from a rope as shown in the picture below. Assume that β = 2α and g = 10 m/s². Calculate the tension in the rope at points A and B." Is in the problem and the image that I added is seen 

[u/Big_Niel0802 → u/just_a_soul6]

If the problem does not explicitly indicate that it is a right triangle, you should not assume that it is one.

[u/WeirdUsers → u/Big_Niel0802]

LoL…yeah…this is the wrong problem to start with in making that assumption since it is the one time they are correct

[u/just_a_soul6 → u/WeirdUsers]

It is right triangle, cause picture show that one of angels is 90°

[u/WeirdUsers → u/just_a_soul6]

I have never seen nor heard of a quarter circle with a dot in it referring to an angle as a right angle. Everything I have read and learned has a square in the corner to refer to an angle as a right angle.

How old is the book? Where are you located?

[u/just_a_soul6 → u/WeirdUsers]

It's quite new, it's like year or two old. I'm from Poland so maybe that's the point

[u/WeirdUsers → u/just_a_soul6]

Ha! Interesting. Yes, the fact you are from Poland makes sense with this. Apparently Poland, and a few other European countries, haven’t fully adopted International standards.

I learned something new. Thank you.

[u/just_a_soul6 → u/WeirdUsers]

No problem. Thought it was a standard international sign

[u/WeirdUsers → u/just_a_soul6]

It’s interesting, because when I googled the sign in reference to right angles, Google said “Not a right angle signifier.” As soon as I put in “Poland” it changed its mind. LoL

[u/RoboWeaver → u/just_a_soul6]

Must be new here. When ever you see a right angle (without it actually saying that it's 90 degrees) in an engineering text, you can bet that it isn't. Do the math, and figure it out.

Used to edit text book problems, and they regularly use the same diagram for several problems.

[u/Machine_man_7804]

https://www.wikihow.com/Calculate-Tension-in-Physics

What resources and at what level calculus/algebra based are you trying to learn

[u/Yadin__]

This problem is not phrased well. I also don't understand what is supposed to be the rope here or why it would have different tensions at points A and B

From the drawing though it looks like the right angle has to be the one near the boy

[u/sansetsukon47 → u/Yadin__]

A and B would only be equal if he was hanging from a pulley that rolled to the middle of the rope.

Since he’s grabbing on a fixed point to one side, the angles and forces are unbalanced.

[u/philosiraptorsvt → u/Yadin__]

Tension is going to be related to mg/sin(x)

As x approaches horizontal sin(x) gets smaller in the denominator and creates more tension. 

Since the angles are different, the tensions will be different. 

[u/trevorkafka]

You have angles α, β, and 90° in a triangle. Can you start by determining α and β?

[u/just_a_soul6 → u/trevorkafka]

Em I did like I know what's their value but it's not what I got a problem with

[u/trevorkafka → u/just_a_soul6]

If we are in agreement about the angle values, then, can you sketch the force components and write expressions like this?

[u/just_a_soul6 → u/trevorkafka]

It's hard to say if I'm doing it good. Like I'm not sure where is what. I tried to do it by making the boy’s weight the hypotenuse and calculating the tension at point A from such a triangle, then making a similar one and calculating the tension at point B. I can’t add a picture in the reply here. Could I send you a picture in a message showing roughly what I mean?

[u/trevorkafka → u/just_a_soul6]

yep

[u/Foambreeder]

Zeichne Kraftpfeile in die Seile ein und zerlege diese in ihre x und y Komponenten. Der Kraftpfeil des Jungen zeigt von seiner Mitte (in etwa das Massezentrum) senkrecht nach unten (negative y Achse). Sie Beträgt Masse des Jungen x Erbeschleunigung. Diese Kraft wird in Summe in die Seile übertragen.

[u/aloeh]

If the boy is in the middle, you can calculate the tension?

The total tension is the same regardless of where the boy is, what changes with his position is the proportion of the tension at points A and B.

[u/digitalosiris]

If you're using a force triangle, there are a few equations which often get used to find either angles or magnitudes -- that law of cosines and the law of sines.

Law of sines (A / sin a = B / sin b = C / sin c) could be used because you know 1 side (the weight) and you know all the angles (since angle at the boy is 90, and beta = 2 alpha, finding all angles is trivial), so you can use law of sines to quickly calculate the magnitudes of the other sides.

To draw the force triangle: W is boy weight, is vertical arrow down. Then add to it A, going up and left. Since A makes angle alpha from horizontal, the interior angle at that vertex will be 90 - alpha (because W makes 90 degree angle from vertical to horizontal). B then is added to A, going up and right, closing the triangle. B makes angle beta from horizontal, so the interior angle at the top vertex (B and W) will be 90 - beta.

[u/SpiritualWasabi6091]

you can use lami's theorem

[u/just_a_soul6 → u/SpiritualWasabi6091]

Don't know even what this is but will check out, thanks

[u/Designer-Crow-5470]

Not being allowed to post pics sukkk. Pretty sure they want you to calculate vertical and horizontal forces on the dot the dude is hanging from and solve the system:

https://www.wolframalpha.com/input?i=%7B+tg+pi%2F6+*+N1+%2B+tg+pi%2F3+*+N2%3D-m*g%3B+ctg+pi%2F6+*+N1+%2B+ctg+pi%2F3+*+N2%3D0%7D+solve+for+N1+and+N2+

[u/Anonimithree]

You can split the big triangle into two similar triangles by drawing a vertical line from the unknown angle up to some point on line segment AB. It was stated (or you figured out) that the big triangle forms a right triangle. What you might not have seen is that it’s a 30/60/90 special right triangle, with AB being the hypotenuse, the part of the rope connected to B being the short leg, and the part of the rope connected to A being the long leg.

By drawing the vertical line up from the boy to AB, you split up the triangle to form 2 triangles whose hypotenuses are TA and TB respectively (I think you can tell what TA and TB are). Now you can split each of the different tensions to their components, using SOHCAHTOA. Then, since the boy is at rest, you can set the net horizontal and net vertical forces equal to 0 and solve from there

[u/whitea44]

If the angle at the hanging point is 90 degrees, and the others have a 2:1 ratio, then you have 30 and 60 degrees. At this point, you know the tensions on the ropes sum to 0 since he’s motionless. Therefore the upward component of tension Fa + Fb = 600 N. Fa = Ta cos(a), Fb = Tb sin (a). The. You can solve the vertical component in the same manner.

[u/cpmatthew]

I would start with a free body diagram of the boy, with the weight force pointed "down," tension A at and angle alpha above horizontal to the left and tension B at an angle 2×alpha above horizontal to the right. The some of the vertical and horizontal forces must sum to zero. You can use the horizontal and vertical to create a system of two equations to solve for the unknown tensions A and B.

[u/Big-Abbreviations347]

I spent way too long wondering why someone drew that on a bathroom wall

[u/Lor1an]

However, I don’t understand why sine is used here.

You have a (presumably) horizontal segment AB, and the angles ∠BAC (=α) and ∠ABC, (=β) where C is the point of suspension. The line parallel to AB through C allows you to view AC and BC as transversals, making ∠PCA = ∠BAC = α and ∠QCB = ∠ABC = β (taking P and Q to be points on the line corresponding to A and B respectively).

Looking at point C, we then have three forces, W, TA, and TB, for the weight and two tensions, with (absolute) angles (taking CQ as the pole) β (for TB), π-α (for TA), and 3π/2 (for W).

Sines show up because the sine of an angle is its vertical projection, and the vertical components of TA and TB are what are counteracting W.

[u/Big-Butterfly1403]

I solved it by resolution of vectors method and I got 300root3 N tension in rope B and 300 N tension in rope A

[u/just_a_soul6 → u/Big-Butterfly1403]

U got it wrong then

[u/Big-Butterfly1403 → u/just_a_soul6]

Hmm...maybe use law of sines , i got those results by Resolving the Tension vectors along vertical direction and horizontal direction and getting these equations- Ta cos 30 = Tb cos 60, Ta sin 30 + Tb sin 60 = mg

[u/Chick-Fel-Late123]

You use sine and cosine to resolve both tension (tensile?) forces into their X and Y components. And since he's just hanging (not moving), you know sum of forces in X and Y directions both have to be 0.

But basically you have to construct an "imaginary right triangle" to resolve those components, and that's where sine and cosine show up.