How will you calculate tension now?

Classical Mechanics Level 5

Four rods having uniform mass distribution, each of weight $w$ , are used to make the sides of a square frame. In this frame, one of the diagonal is a light rod. This frame is hung from a vertex such that the frame lies in the vertical plane and the light rod is horizontal.

Now, at each of the other three vertices, identical blocks of weight $w$ are hung.

Find the tension in the light rod (in Newtons ( $N$ )) if the shape of the assembly doesn't vary.

Give your answer to 3 decimal places.

Take $w=1\, N$ .

Note: The tension along each of the rods having mass is a not a constant.

The answer is 4.000.

1 solution

Kishore S. Shenoy
May 6, 2016

Relevant wiki: Newton's Third Law

At point $P$ , balancing forces along vertical line gives $\dfrac{F_1}{\sqrt 2} = \omega + \dfrac{F_2}{\sqrt 2}$ Balancing forces along the horizontal gives $\dfrac{F_1}{\sqrt 2} = \omega + \dfrac{F_2}{\sqrt 2}$

Now consider a system.

Balancing net force on the vertical gives us $\dfrac{F_2}{\sqrt 2} = \dfrac32w$

Solving these three equations gives $\Large\boxed{T = 4w}=4N$

Small typo. It's 4N at the end and not 1N.

A Former Brilliant Member - 5 years, 1 month ago

Thank you!

Kishore S. Shenoy - 5 years, 1 month ago

Btw, don't the system as it is coz tention midway will be different. Or, put a comment saying 'for representative purposes'.

A Former Brilliant Member - 5 years, 1 month ago

@A Former Brilliant Member – What do you mean?

Kishore S. Shenoy - 5 years, 1 month ago

@Kishore S. Shenoy – As the rods have mass, the tension varies along the rod. The way you've presented it, it looks as if the rods are massless. So, add a comment saying 'Diagram only for repsentative purposes'.

A Former Brilliant Member - 5 years, 1 month ago

@A Former Brilliant Member – No. It's not only for representative purpose. The diagram is correct

Kishore S. Shenoy - 5 years, 1 month ago

@Kishore S. Shenoy – The tension along the rod changes exponentially with respect to length along the rod. So, the tension at the lower joint isn't $F_2$ !

A Former Brilliant Member - 5 years, 1 month ago

@A Former Brilliant Member – I did not say it's at the joint. It's at the centre of mass of that system!

Kishore S. Shenoy - 5 years, 1 month ago

virtual work method may also help

aryan goyat - 5 years ago

Can you elaborate...? :)

Kishore S. Shenoy - 5 years ago

Yeah, that's how I first did it!

A Former Brilliant Member - 5 years ago

https://brilliant.org/discussions/thread/virtual-work-method/

aryan goyat - 5 years ago

How will you calculate tension now?

The answer is 4.000.

1 solution

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