The ice can withstand a pressure of 90 kPa. Will a tractor with a mass of 5.4 tons pass on this ice, having a track area of 1.5 m2
Rm = 90 kPa = 90 * 10 ^ 3 Pa.
m = 5.4 t = 5.4 * 10 ^ 3 kg.
g = 9.8 m / s2.
S = 1.5 m2.
The pressure of the force P is the ratio of the force F to the area S on which this force acts: P = F / S.
The tractor acts by the force of its weight F on the area of contact of the tracks with ice S.
The weight of the tractor F is found by the formula: F = m * g, where m is the mass of the tractor, g is the acceleration of gravity.
Let’s find the pressure of the tractor on the ice: P = m * g / S.
P = 5.4 * 10 ^ 3 kg * 9.8 m / s2 / 1.5 m2 = 35280 Pa = 35.28 * 10 ^ 3 Pa.
We see that the pressure of the tractor on the ice is less than the maximum allowable pressure that the ice can withstand: P = 35.28 * 10 ^ 3 Pa <Pm = 90 * 10 ^ 3 Pa. The tractor can drive over ice.
Answer: the tractor will drive over the ice.
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