A force of 12,000 n is exerted on a piston that has an area of 0. 020 m 2. What is the area of a second piston that exerts a for
1 answer:
The area of a second piston that exerts a force of 24,000 N is 0.04 .
Gravitational force is measured in newtons. Utilizing the physics formula pressure = force/area, you can compute pressure if you know the value of a force. Both pounds per square inch (psi) and newtons per square meter (N/m2) are acceptable units of measurement for this pressure.
Given force 1 = 12,000 N
The pressure which has produced a force of 12,000 N at the piston 1 = Force/ area
= 12,000N / 0.020
= 6,00,000 N/
Now force at piston 1 must be equal to force at piston 2. thus, force at piston 2 is given by:
Force at piston 2 = pressure × area
24,000 N = 6,00,000 × area
area = 0.04
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Answer:
The location of the shear center o is 0.033 or 33 m
Explanation:
Solution
Recall that,
The moment of inertia of the section is = I = 0.05 * 0.4 ^3 /12 + 0.005 * 0.2 ^3/12
= 30 * 10 ^ ⁻⁶ m⁴
Now,
The first moment of inertia is
Q =ῩA = [ (0.1 -x) + x/2] (0.005 * x)
= 0.5x * 10 ^⁻³ - 2.5 x * 10⁻³ x²
Thus,
The shear flow is,
q = VQ/I
so,
P = (0.5x * 10 ^⁻³ - 2.5 x * 10⁻³ x²)/ 30 * 10 ^⁻⁶
P = (16.67 x - 83. 33 x²)
The shear force resisted by the shorter web becomes
Vw,₂ = 2∫ = ₀.₁ and ₀ = P (16.67 x - 83. 33 x²) dx = 0.11x
Then,
We take the moment at a point A
∑Mₐ = 0
- ( p * e)- (Vw₂ * 0.3 ) = 0
e = 0.11 p * 0.3/p
which gives us 0.033 m
= 33 m
Therefore the location of the shear center o is 0.033 or 33 m
Note: Kindly find an attached diagram to the question given above as part of the explanation solved with it.
