You are Here: Home >< Physics

# Tension and Newton's Second Law watch

Announcements
1. Right. Yes. Of course! I understand that bit now.

I have two questions which are now unanswered. Could you please explain them to me?

The first question:
(Original post by PHKnows)
When calculating the resultant force acting on the girder, why do we not include the upward tension of the supporting ropes acting on the girder?

we do, the upward tension of the supporting ropes is equal to the tension in C.
You have not understood me! This is what I mean.

I agree the tension in rope C must be 8000 N and in the vertical components of the tension of the supporting ropes combined must be 8000 N. I also agree that the weight of the lift is 5000 N if we take g = 10 m/s2.

When I now apply F = ma, I would do this:

F = ma => 8000 (from rope C) + 8000 (from the supporting ropes) - 5000 (weight of the girder) = 500a => 11000 = 500a => a = 22 m/s2.

However, the answer in the markscheme (the right answer) has only added the upward force of the 8000 N once. What I want to know is why this is the case, if both rope C and the supporting ropes exert upward forces?
2. My second question [for part (c)]:

I agree wholeheartedly that if its moving at constant speed, the acceleration is zero and therefore the resultant force acting on the girder is zero.

I also agree therefore that the upward forces acting on the girder should equal the downward forces acting on the girder, where in this case, the downward force is its weight.

What I am struggling with is what the upward forces are. As far as I can see, this is both the tension in the rope C and the tension of the supporting ropes. This causes the consequent dilemma where in the mark scheme they have simply equated the weight only to the combined vertical tension of the two supporting ropes only, neglecting the tension of rope C.

Do you understand my question now?
3. ohhh. right. ok, no you dont add the tensions throughout the rope. Consider the rope in two parts. the first part(C) has 8000N of tension, right? The second part is a combination of two sidey bits. the tension is 8000N throughout the rope in both parts of the rope, in the sidey bits, the vertical components equals 8000N. You cant add the vertical components of the sidey ropes to C just as you cant add two lots of 8000N in C itself
4. isnt it a shame we cant post diagrams, they speak a thousand words and would be ideal to explain what going on here.
5. lol.yeah. i'm sure there is a way. you could attatch a paint file or something. or if you had a scanner...i cant do neither, how do you use symbols and maths notation and stuff?
6. Thank you so much!!!!!

I honestly cannot believe how stupid I've been. I was treated the supporting ropes and the cable as two separate ropes almost, instead of one rope consisting of two parts. It's all because of my force diagram: the way I drew it was such that the supporting cables and C both had tensions of 8000 N. But I misinterpreted this as though they were two separate bodies almost. It's hard to think how I did that now that I look back at it.

Thank you so much to everyone that's helped me, especially PHKnows.

EDIT: Although I'll be very interested to see how you would explain this via diagrams....
7. No problem I think diagrams are just more understandable than "shudders" english when trying to explain physics.
8. Dharma, to throw a spanner in the works, you can consider them as three separate ropes. It is a perfectly valid method.
9. Oh, you've got me intrigued.

How would you use this method then?
10. It will also help you understand why PHKnows's explanation was not really accurate. In our model, ignoring friction/elasticity, three similar ropes tied together should produce the same effect as one rope which splits into two, so I don't see how the explanation given is enough to answers your question.

If you want to be completely correct then you need to take into account all the forces acting.

I shall consider the entire system in equilibrium, and show that at this point.

Let us call the point where C, X and Y meet P. Let the tension in rope C be T. Let the tension in rope X be U. Let the tension in rope Y be V.

Consider the forces acting about P. The tensions in C, X and Y are acting through this point. This smaller system is in equilibrium as there is no relative acceleration within the system. Remember that for a system to be in equilibrium, the vector sum of the forces (and torque) on the system must equal nought. Therefore, resolving upwards and parallel to C gives:

Resolving horizontally gives:

(1)

Consider the girder. The smaller system of the girder, X and Y is in equilibrium.

(2)

Eliminating U from (1) and (2) gives:

Therefore, when the system is in equilibrium, , as required.
------------

Applying this revelation to part (b) :

We know that a tension of 500g holds the system in equilibrium. But we want to find the acceleration of the girder when We know what tension gives equilibrium, so we can reduce the question to: what acceleration would our "left-over" tension, or force, that we have to create the acceleration give? That is: what acceleration would , give?

as required.
----

These are some of my thoughts. I hope this helps a little, Dharma.
11. Thanks so much Lusus. That was extremely helpful.

The problem I had was identifying which bodies the forces acted upon because of the different connections of the ropes.

The way I have got around this is to think:

We know that the tension in the supporting ropes must be the same as in C. We can redraw the supporting ropes as a single rope of tension equal to that in C. We now have two ropes which connect both with the same tension. This is the same as having a single rope of that tension.

Would you by any chance know how to apply the internal/external forces argument to this case?
12. (Original post by Dharma)
Would you by any chance know how to apply the internal/external forces argument to this case?
I'm afraid I'm not familiar with that argument.

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: July 17, 2007
Today on TSR

And I hate it

### University open days

• Heriot-Watt University
School of Textiles and Design Undergraduate
Fri, 16 Nov '18
• University of Roehampton
Sat, 17 Nov '18
• Edge Hill University
Faculty of Health and Social Care Undergraduate
Sat, 17 Nov '18
Poll

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE