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### Topic: Numerical Problem relating to TENSION  (Read 4810 times)

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#### uzair

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##### Numerical Problem relating to TENSION
« on: March 24, 2006, 01:40:03 AM »
A metalic block produces a tension of 10 N when suspended with a string without immersing it in to the water, and when immersed in the water tension was reduced to 8N find the density of metalic block?

#### Borek

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##### Re:Numerical Problem relating to TENSION
« Reply #1 on: March 24, 2006, 03:48:35 AM »
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#### uzair

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##### Re:Numerical Problem relating to TENSION
« Reply #2 on: March 24, 2006, 12:24:11 PM »
Plz Boorek don't try to fool me I'm sure, I'm not gona get its solution from the forum rules. :fishing:

#### green-goblin

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##### Re:Numerical Problem relating to TENSION
« Reply #3 on: March 24, 2006, 02:30:51 PM »
Plz Boorek don't try to fool me I'm sure, I'm not gona get its solution from the forum rules. :fishing:

LOL   are i'm not sure if your being series with that reply or not

#### Hunt

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##### Re:Numerical Problem relating to TENSION
« Reply #4 on: March 25, 2006, 08:40:43 PM »
This seems like Fluid statics to me.

From Netwon's 2nd law, sum of forces = 0 at Eq
This means Tension = Weight of the mb ( before inserting it in water )

This implies Wt of the mb = 10 N

When inserted in water, I'll assume that the metallic block reaches equilibrium without reaching the buttom surface.

At Eq : Tmb + Fb = Wtmb

The force exerted by water is the Bouyant Force Fb.

Fb = 10N - 8N = 2N

Fb = Vimmersed x g x dwater

2 N =  Vimmersed x 9.8N/Kg x 1Kg/L

Vimmersed = 0.2 L

Since the mettalic block is totally immersed, we can conclude Vimmersed = Vtotal , that is the total volume of the metallic block.

Wtmb = dmb x g x VT

10 N = dmb x 9.8 N/Kg x 0.2 L

dmb = 5.1 Kg/L
« Last Edit: March 27, 2006, 07:34:24 AM by Vant_Hoff »