# [Mechanics] Resolving Tensions

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Here is the question:

My current workings:

I have tried to resolve vertically and I found the vertical component of T1 but with The inextensible string since it doesn’t extend at all its tension I said is 0 therefore I found the upwards component and let it equal 2g because the system is at equilibrium and tried to solve for the extension x and I got 2m which isn’t the correct answer. The solution bank resolved diagonally however I don’t understand what mistake I’ve made resolving vertically.

Any help would be appreciated thanks

My current workings:

I have tried to resolve vertically and I found the vertical component of T1 but with The inextensible string since it doesn’t extend at all its tension I said is 0 therefore I found the upwards component and let it equal 2g because the system is at equilibrium and tried to solve for the extension x and I got 2m which isn’t the correct answer. The solution bank resolved diagonally however I don’t understand what mistake I’ve made resolving vertically.

Any help would be appreciated thanks

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#2

(Original post by

The inextensible string since it doesn’t extend at all its tension I said is 0

**BrandonS15**)The inextensible string since it doesn’t extend at all its tension I said is 0

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(Original post by

You sure about that ?

**RDKGames**)You sure about that ?

Last edited by BrandonS15; 11 months ago

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#4

(Original post by

It says inextensible: not extensible; incapable of being extended or stretched so extension (x) = 0 so using T = (lambda * x )/ l if x= 0 then T = 0 through that string? So it won’t have a vertical component?

**BrandonS15**)It says inextensible: not extensible; incapable of being extended or stretched so extension (x) = 0 so using T = (lambda * x )/ l if x= 0 then T = 0 through that string? So it won’t have a vertical component?

Inextensible simply means it wont change its length. Hookes law is not valid (spring), but it does transmit forces (tension).

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(Original post by

Im with rdkgames on this one.

Inextensible simply means it wont change its length. Hookes law is not valid (spring), but it does transmit forces (tension).

**mqb2766**)Im with rdkgames on this one.

Inextensible simply means it wont change its length. Hookes law is not valid (spring), but it does transmit forces (tension).

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#6

**BrandonS15**)

It says inextensible: not extensible; incapable of being extended or stretched so extension (x) = 0 so using T = (lambda * x )/ l if x= 0 then T = 0 through that string? So it won’t have a vertical component?

You resolve vertically and horiztonally. Two eqns in two tensions. Solve them.

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(Original post by

Hookes applies only to extensible objects.

You resolve vertically and horiztonally. Two eqns in two tensions. Solve them.

**RDKGames**)Hookes applies only to extensible objects.

You resolve vertically and horiztonally. Two eqns in two tensions. Solve them.

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#8

(Original post by

And when resolving horizontally how do I find the vertical component of T2 if T2 doesn’t equal 0?

**BrandonS15**)And when resolving horizontally how do I find the vertical component of T2 if T2 doesn’t equal 0?

T2 * cos(60) is horizontal

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Working:

Last edited by BrandonS15; 11 months ago

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#10

(Original post by

I resolved horizontally and found an equation solvable for x and I still get x = 2, I don’t see what I’m doing wrong:

Working:

**BrandonS15**)I resolved horizontally and found an equation solvable for x and I still get x = 2, I don’t see what I’m doing wrong:

Working:

t_1 cos(30) = t_2 cos(60)

t_1 sin(30) + t_2 sin(60) = 2g

If you resolve diagonally D must be less than 2g. Multiply by cos() rather than divide and you should get x=1/2 which agrees with the horizontal and vertical analysis.

The original force (2g) is always the hypotenuse and the perpendicular components in the directions of t_1 and t_2 are the other two sides, so must both be less than 2g.

Last edited by mqb2766; 11 months ago

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(Original post by

When you resolve horizontally and vertically you should get

t_1 cos(30) = t_2 cos(60)

t_1 sin(30) + t_2 sin(60) = 2g

If you resolve diagonally D must be less than 2g. Multiply by cos() rather than divide and you should get x=1/2 which agrees with the horizontal and vertical analysis.

The original force (2g) is always the hypotenuse and the perpendicular components in the directions of t_1 and t_2 are the other two sides, so must both be less than 2g.

**mqb2766**)When you resolve horizontally and vertically you should get

t_1 cos(30) = t_2 cos(60)

t_1 sin(30) + t_2 sin(60) = 2g

If you resolve diagonally D must be less than 2g. Multiply by cos() rather than divide and you should get x=1/2 which agrees with the horizontal and vertical analysis.

The original force (2g) is always the hypotenuse and the perpendicular components in the directions of t_1 and t_2 are the other two sides, so must both be less than 2g.

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#12

(Original post by

Also if the string PB is inextensible, how does it have tension when its extension is 0?

**BrandonS15**)Also if the string PB is inextensible, how does it have tension when its extension is 0?

https://physics.stackexchange.com/qu...ensible-string

Last edited by mqb2766; 11 months ago

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