Not calc this time is physics

[sonya]

ok...

u have built an exact replica of ur favorite sailboat on a scale of 1 to 20. u know that the boat weighs 3050 kg. to wat total weight, in grams, would u have 2 ballast the model so it floats in water like the real thing?

i thot it would just b 3050/x = 20/1
then x = 152.5 kg

but...thats not the rite answer...as per usual...
ne1 want to help me get started on this 1?

 

[HallsofIvy]

Although you didn't say, I would conclude that your "Scale of 1 to 20" is a LINEAR scale. That is, the model is 1/20 as long as the original sailboat. Since the width of the model would also be 1/20 the width of the actual sailboat, the AREA of the deck of the model, for example, would be (1/20)(1/20)= 1/400 of area of the deck of the actual sailboat.

Since volume is "length times width time height", that is, 3 lengths multiplied together (hence "cubic feet" or "cubic meters" to measure volume), the volume of the model sailboat would be (1/20)(1/20)(1/20)= 1/8000 of the volume of the actual sailboat.

Assuming that your model is made of the same material as the actual sailboat (and so has the same density) you would need
1/8000 of the mass: 3050/x= 8000/1 so x= 3050/8000= 0.38125 kg.

 

[M98Ranger]

Hi there, it seems like you are trying to calculate the weight of ballast needed for your model sailboat to float like the real thing. This is a physics problem, not a calculus problem. Calculus is a branch of mathematics that deals with rates of change and is not applicable in this situation. It might be more helpful to approach this problem using principles of buoyancy and density. The weight of the ballast needed will depend on the density of the material used and the volume of the model sailboat. You can start by calculating the volume of the model sailboat and then using the density of the ballast material to determine the weight needed for it to float. I hope this helps!