Photo AI
5.1 A force of 16 N is applied to a 3 m long metal wire - NSC Technical Sciences - Question 5 - 2021 - Paper 1
Question 5
5.1 A force of 16 N is applied to a 3 m long metal wire. The wire stretches by 0.5 mm. The diameter of the metal wire is 2.5 mm. Calculate the: 5.1.1 Stress in the... show full transcript
Worked Solution & Example Answer:5.1 A force of 16 N is applied to a 3 m long metal wire - NSC Technical Sciences - Question 5 - 2021 - Paper 1
Step 1
5.1.1 Stress in the wire
Answer
Stress () in the wire is defined as the force (F) applied per unit area (A) of the cross-section.
Given:
- Force, F = 16 N
- Diameter, d = 2.5 mm = 0.0025 m
- Area, A = rac{ au}{4}d^2 = rac{ au}{4}(0.0025)^2 = 4.91 imes 10^{-6} ext{ m}^2
egin{align*}
au & = rac{F}{A}
au & = rac{16}{4.91 imes 10^{-6}}
au & ext{(in Pa)} = 3.26 imes 10^6 ext{ Pa}
extbf{Final Result:} Stress in the wire is .
Step 2
5.1.2 Strain in the wire
Answer
Strain () is defined as the ratio of the change in length () to the original length (L).
Given:
- Change in length,
- Original length, L = 3 m
The formula for strain is:
ext{ε} = rac{ ext{ΔL}}{L} = rac{0.0005}{3} = 1.67 imes 10^{-4}
extbf{Final Result:} Strain in the wire is $1.67 imes 10^{-4}$.
Step 3
5.1.3 Young's modulus of the wire
Answer
Young's modulus (K) is the ratio of stress to strain.
Using the results from 5.1.1 and 5.1.2:
K = rac{ ext{Stress}}{ ext{Strain}} = rac{3.26 imes 10^6}{1.67 imes 10^{-4}} = 1.95 imes 10^{10} ext{ Pa}
extbf{Final Result:} Young's modulus of the wire is $1.95 imes 10^{10} ext{ Pa}$.
Step 4
5.2.1 Define pressure at a particular point.
Answer
Pressure (P) at a specific point is defined as the thrust or force (F) acting on the unit area (A) around that point. Mathematically, it can be expressed as:
P = rac{F}{A}
This definition highlights that pressure is a measure of how concentrated a force is over a specific area.
Step 5
5.2.2 Calculate the fluid pressure in the hydraulic system.
Answer
Given:
- Force applied, F = 26 N
- Area of input piston, A = 7.855 × 10⁻⁶ m²
Using the formula for pressure:
P = rac{F}{A} = rac{26}{7.855 imes 10^{-6}} = 3.309 imes 10^{5} ext{ Pa}
extbf{Final Result:} Fluid pressure in the hydraulic system is $3.309 imes 10^5 ext{ Pa}$.
Step 6
5.2.3 Calculate the area of the output piston.
Answer
Using the relationship between forces and areas in a hydraulic system:
rac{F_1}{A_1} = rac{F_2}{A_2}
Where:
- is the force on the input piston,
- is the area of the input piston,
- is the force exerted by the output piston (1,278 N),
- is the area of the output piston.
Rearranging gives:
A_2 = A_1 rac{F_2}{F_1}
Substituting in the values:
F_2 = 1278 ext{ N}, F_1 = 26 ext{ N}$$ Now calculate: $$A_2 = 7.855 imes 10^{-6} imes rac{1278}{26} ext{ m}^2 = 3.861 imes 10^{-3} ext{ m}^2$$ extbf{Final Result:} Area of the output piston is $3.861 imes 10^{-3} ext{ m}^2$.