Can you help me with this physics problem?
Hello Hello,
Can you help me with the following physics problem? Unfortunately, I can't get any further.
How big does a floating ice float 30cm thick have to be able to just about support a car weighing 1 ton?
I also have the following information: 920 kg per cubic meter and that the area of the ice floe is 5 square meters. It would be very nice if you could help me 🙂
(1 rating, 1 votes, rated)You need to be a registered member to rate this.
Loading...
I interpret “just so wearable” as follows: if you put the car on the ice floes, it has to immerse it less than 30 cm deep, so just a piece of water has to remain.
Now we can easily calculate that 1 m3 of ice weighs 920 kg and, during full immersion, displaces a cubic meter of water, ie 1000 kg; Consequently, a cubic meter of ice can bear a maximum of 80 kg load. To carry the whole car, we need at least 1000/80=12.5 m3 of ice.
Our ice floe should be d=30 cm thick and should have a volume of at least V=12.5 m3, and it needs an area of at least A=V/d=41.7 m2. This corresponds to a circle with a diameter of more than 7.3 m.
If you want to carry out the experiment in practice, I recommend the word I often use at least: to be very serious and to establish a major security factor.
Sorry, your answer wasn't visible when I wrote mine.
Water weighs 1000kg per cubic meter. Ice less like you know. This difference between the two masses ensures that the ice cholle floats. But there is also a little more difference and that is used for the car. So the difference once surface until it reaches.
It is the area size of the ice.
The acceleration of the ground is abbreviated to g.
The density of ice is according to ro(ice)=920 kg /m^3
The density of water may be assumed with crude (water) = 1000 kg / m3. This is relatively certainly not true, but the actual density of water at 0°C is somewhat lower.
We have to assume that the ice is practically completely immersed in the water.
The volume is then V =A * 0.3 m and its weight (ice) = A * 0.3 m* ro(ice) * g.
The buoyancy = V * ro(water) * g. The buoyancy must be 1000 kg *g larger than the weight of the ice.
So: lift = A * 0.3 m * ro(water) * g = 1000 kg * g + A * 0.3 m * ro(ice) * g.
Hopefully you can resolve this equation yourself to A.