A low-density block with a weight of 10.0 N is placed in a beaker of water and t

Question

A low-density block with a weight of 10.0 N is placed in a beaker of water and tied to the bottom of the beaker by a vertical string of fixed length. When the block is 25% submerged, the tension in the string is 17.0 N. The string will break if its tension exceeds 70.0 N. As water is steadily added to the beaker, the block becomes more and more submerged.

What percentage of the block is submerged at the instant the string breaks?

Explanation / Answer

Based on this, set up variables and equations for what you know. There's mg, weight of the block. Then there's buoyant force, which is pVg, p being density. And don't forget Tension! At the moment when Tension is 17 N, we can imagine that the lower quarter of the block is submerged, and that the pVg and mg are contesting for the net force. The tension is created from something pulling on the string. In this case, the pVg is so strong, it's pushing on the block causes Tension. --> Net Force = pVg - mg = T. However, you do realize that the V here is the amount of block submerged. Let's call the original amount Vo. In this case, 0.25*Vo = V in the previous statement. So, 0.25 pV*o*g = T + mg. Now, I'll plug and chug to find pVog. 0.25 pVog = 27 N pVog = 108 N. That's the buoyant force if the entire block was submerged. However, as with the 17 N, we need to find the breaking point, when T = 70 N. Let's call the coefficient on the left side C. We're solving for that - it's your percentage. CpVog = T + mg = 70 + 10N. Now, sub in your value for pVog. C * 108N = 80 N. You should get 80/108 as your percentage when the string breaks. I don't know what the number is - I spent my energy on doing this problem. Hope that helped. :)

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