Answer:
P = 4.745 kips
Explanation:
Given
ΔL = 0.01 in
E = 29000 KSI
D = 1/2 in
LAB = LAC = L = 12 in
We get the area as follows
A = π*D²/4 = π*(1/2 in)²/4 = (π/16) in²
Then we use the formula
ΔL = P*L/(A*E)
For AB:
ΔL(AB) = PAB*L/(A*E) = PAB*12 in/((π/16) in²*29*10⁶ PSI)
⇒ ΔL(AB) = (2.107*10⁻⁶ in/lbf)*PAB
For AC:
ΔL(AC) = PAC*L/(A*E) = PAC*12 in/((π/16) in²*29*10⁶ PSI)
⇒ ΔL(AC) = (2.107*10⁻⁶ in/lbf)*PAC
Now, we use the condition
ΔL = ΔL(AB)ₓ + ΔL(AC)ₓ = ΔL(AB)*Cos 30° + ΔL(AC)*Cos 30° = 0.01 in
⇒ ΔL = (2.107*10⁻⁶ in/lbf)*PAB*Cos 30°+(2.107*10⁻⁶ in/lbf)*PAC*Cos 30°= 0.01 in
Knowing that PAB*Cos 30°+PAC*Cos 30° = P
we have
(2.107*10⁻⁶ in/lbf)*P = 0.01 in
⇒ P = 4745.11 lb = 4.745 kips
The pic shown can help to understand the question.
This question is about Circle Geometry. it evaluates connected and broken lines with respect to circles.
<h3>What is Circle Geometry?</h3>
This refers to the body of knowledge in mathematics that has to do with the various problems associated with the Circle.
In real-world scenarios, circle geometry is used in technologies involving:
- Camera lenses
- Circular Architectural structures
- Steering Wheels
- Buttons etc.
Learn more about Circle Geometry at:
brainly.com/question/24375372
Answer:
Hydrostatic force = 41168 N
Explanation:
Complete question
A triangular plate with a base 5 ft and altitude 3 ft is submerged vertically in water so that the top is 4 ft below the surface. If the base is in the surface of water, find the force against onr side of the plate. Express the hydrostatic force against one side of the plate as an integral and evaluate it. (Recall that the weight density of water is 62.5 lb/ft3.)
Let "x" be the side length submerged in water.
Then
w(x)/base = (4+3-x)/altitude
w(x)/5 = (4+3-x)/3
w(x) = 5* (7-x)/3
Hydrostatic force = 62.5 integration of x * 4 * (10-x)/3 with limits from 4 to 7
HF = integration of 40x - 4x^2/3
HF = 20x^2 - 4x^3/9 with limit 4 to 7
HF = (20*7^2 - 4*7^(3/9))- (20*4^2 - 4*4^(3/9))
HF = 658.69 N *62.5 = 41168 N