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### Course: Class 11 Physics (India)>Unit 9

Lesson 10: Tension

# Super hot tension

David shows how to solve a super hot tension problem where a can of peppers hangs from two diagonal strings. Created by David SantoPietro.

## Want to join the conversation?

• This works out beautifully and I understand all the steps involved in getting the problem that YOU have.. Although I'm stuck on the same sort of problem in which it isn't a perfect 30,60,90 triangle... meaning the numbers are in decimals and it gets messy and confusing. The two angles given in my problem are 35 and 48. the mass of the object pulling is 65kg. I got all the way down to the point you reached around 11 minutes in.
• Yes, good call everyone. That was bad form on my part by prematurely turning trig functions into decimals. Leaving things as symbolic as possible is almost always the way to go.
• T1 and T2 happen to be mg*Sin30 and mg*Sin60 respectively. Is this just a coincidence?
• It works out this way because the angle between the two ropes (and thus the two tensions) is exactly 90 degrees (T1 and T2 are at right angles to one another, you can see this by noting that the other two angles in the "inside" triangle are 30 and 60, and the angles in a triangle add up to 180, so the angle between T1 and T2 is 90).

Because of this, you could rotate your reference frame. Suppose you pick the new y axis to point along T2, and the new x axis to point along T1. Now, the x equation and the y equation are each simple and have one unknown. However, you need to break gravity into the two components you mentioned, which are the components along these "new" axes. Hope that helps.
• what if two ropes pull the can with angle zero i.e. along x-axis so hard that they balance gravitational pull.how to find force here. how do they really affect cans accelaration in y direction
• one possibility is to froze the ropes then hold the can between the two ropes while making them horizontal. then it is not in the relm of tension anymore as there's no pulling force by them to the can

in other words, if the rope is ex"tensi"ble, it must make an angle to balance out gravity in the situation like above. otherwise, there's no "tensi"on you need to concern
(1 vote)
• Earlier on, we learned that a vector can be moved and relocated within a vector diagram without changing its magnitude. Is it a shortcut in this problem to move the Gravitational Force vector head to tail with the Tension vector intersection? This would create 2 right triangles which would allow us to solve for T1 and T2 using trigonometry.

T1 = 30Ncos60 = 15N
T2 = 30Ncos30 = 15sqrt3N

This gives us the same answer, and eliminates an enormous amount of work. Is this a practical method? Or will the method shown in the video be of good use in more challenging problems thus the effort is worth it?
• It works well here because in this case the tensions form a right triangle as you mentioned. It would be tougher if they didn't
(1 vote)
• Will there be any sort of problems with three tensions? How will you solve that then?
(1 vote)
• Same way, break them into components and add them up.
• Very minor annoyance I'm having, is that everytime you work with newton's second law, you tend to negate a force variable to imply that it goes in the opposite direction, but shouldn't you instead negate the magnitude of the force variable, because if it is truly negative, and you negate a negative, that's a positive.
• Can't we just write everything in terms of Ty by using tan(theta)?
(1 vote)
• How? If you have an alternative solution, do share.
(1 vote)
• Why doesn’t T1(in Y direction)= T1*sin30 works for solving for T1 directly? After substituting it in Newton’s second law as follow: -30+T1sin30/3 = 0 solving this would bring T1 = 60N.
(1 vote)
• -30+T1sin30/3 = 0 is incorrect because the weight of the can is not only supported by T1(in Y direction), it is also supported by T2(in Y direction). So it is -30 + T1sin30 + T2sin60 = 0
(1 vote)
• When I tried to solve this problem by myself . I stopped when 0 = (-30N) + T1y + T2y . I noticed that T1y and T2y are equal because they are go out from the same point in the bottle to the roof so they have the same magnitude.
so I rewirte 0 = (-30N) + T1y + T2y to 30 = 2 T(1 or 2)y ... so T1y = 15 and T2y = 15 .... so sin30 = 15 / T1 ...so T1 = 15/sin30 = 30 , also i got T2 by the same way and both of them are wrong :D
I thought this way will work , why it dose not work ??
sorry for my bad E , I'm still learning .
(1 vote)
• It doesn't work because T1 and T2 have different directions and magnitudes. The magnitude of their y component would only be the same if they had the same angle. I think...
(1 vote)
• I managed to solve this on my own, I got magnitude of tension force 2 to be 27.7N though, I would assume inaccuracy & not any faults?
(1 vote)