a tire inflated to a gauge pressure of 2 bars has a nail 3mm in diameter embedded in it. what is the force tending to push the nail out of the tire?
1 bar is 10^6 dynes/cm^2
The nail has a cross-section of pi*(3mm)^2 = pi*(.3cm)^2 = .2827cm^2
so, the force exerted is 2*10^6*.2827 = 5.65*10^5 dynes = 5.65N
To calculate the force tending to push the nail out of the tire, we need to consider the concept of pressure and the area of the nail in contact with the tire.
1. Convert the gauge pressure from bars to pascals (Pa):
1 bar = 100,000 Pa
So, 2 bars = 2 × 100,000 Pa = 200,000 Pa
2. Determine the area of the nail in contact with the tire:
The area of a circle can be calculated using the formula:
Area = π × (radius)^2
Given that the diameter is 3 mm, the radius (r) would be half of that value:
r = 3 mm / 2 = 1.5 mm = 0.0015 m
Area = π × (0.0015 m)^2
3. Calculate the force using the formula:
Force = Pressure × Area
Force = 200,000 Pa × Area
4. Substitute the value of the area into the equation and perform the calculation to find the force in newtons (N):
Force = 200,000 Pa × π × (0.0015 m)^2
Performing the calculations will provide the exact value of the force tending to push the nail out of the tire.