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Old Mar 5th 2018, 01:53 PM   #1
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Sphere at rest against wall

A sphere of weight 40N and radius 30 cm rests against a smooth vertical wall. The sphere is supported in this position by a string of length 20cm attached to a point on the sphere and a point on the wall. Find: a) the tension in the string
b) the reaction due to the wall
The answers are: a) 50N b) 30 N at a right angle to the wall

I began by stating that the vertical component of the tension (Tj) equals the weight of the sphere(40N) and the the horizontal component of the tension (Ti) equals the reaction from the wall.

I then related the vertical and horizontal distance components of the string with the different components of the tension. ie (O/H) = (Tj/T) ,
(A/H)=(Ti/T) , (O/A) = (Tj/T)

I also established using Pythagoras' theorem that Ti^2 + 40^2 = T^2
and O^2 + A^2 = H^2.

From here I tried to find the value of T by subbing in various values from my ratios for different variables. I kept getting trivial answers, however. I believe what I needed to do to solve the problem was to relate the 30 cm radius of the sphere to the other data somehow, but I wasn't sure how to go about doing that.

Help and solutions would be appreciated , thanks in advance.
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Old Mar 6th 2018, 11:08 AM   #2
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I think that a key point might be that sphere will naturally rotate, with respect to the string and the wall,
such that if you continue the line of the string from the attachment to the wall to the sphere and then into the sphere,
it will pass through the center of the sphere.
Try attaching a line to a cylindrical object (or even better a ball) and holding it against a wall and you will see how it behaves.

This allows you to define a triangle of forces from the center of the ball through the string to the wall.
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Last edited by Woody; Mar 6th 2018 at 11:10 AM.
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Old Mar 6th 2018, 04:55 PM   #3
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Originally Posted by Woody View Post
I think that a key point might be that sphere will naturally rotate, with respect to the string and the wall,
such that if you continue the line of the string from the attachment to the wall to the sphere and then into the sphere,
it will pass through the center of the sphere.
This can't be correct , woody .

Imagine if the string is only 2cm long , it will be almost horizontal , the projected line will pass above the center.

As we make the string longer , it will become more vertical , only at one specific length will it's line pass through the center ....I think...
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Old Mar 7th 2018, 05:48 AM   #4
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Hi Oz,
I may be wrong (rare but it does happen)
However having thought about it some more,
and played about with (empty) fizzy drink cans, string and sticky tape,
I still think I am correct.

A more important question however might be
does this help the original questioner?
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Old Mar 7th 2018, 06:56 PM   #5
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Originally Posted by Woody View Post
Hi Oz,
I may be wrong (rare but it does happen)
However having thought about it some more,
and played about with (empty) fizzy drink cans, string and sticky tape,
I still think I am correct.?
Ha ha ... I too have been giving this some more thought ,with experiment ...I think I've finally cracked it ...

Let's assume the line of the string does pass through the center , then we have a right angled triangle with the hypotenuse 0.5 ...one side 0.3 ... so that's a 345 triangle and the other side must be 0.4 ...and that gives the required answers of 50 and 30 Newtons ....

So you're right it does pass through the center ... but why should it . Does it always ?? for any length of string ??

It seems so , even with the string length 0 the sphere is pulled tight to the wall and inline with center , also for infinite string length it's inline.

If the string was not inline with the center then there would be a force acting tangentially at the point where the string joins the sphere and this force would roll the sphere up or down the wall till the string was inline with the center .

Last edited by oz93666; Mar 7th 2018 at 09:14 PM.
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