1.A 500.0 g mass is placed on a 20.0 cm cube of gelatin. The 500.0 g mass (a 5.00 cm diameter disk) depresses the gelatin 0.243 mm. What is Young's modulus for this gelatin? ____________N/m^2 Later, a 2.00 kg mass disk (5.00 cm in diameter) is placed on the gelatin and the gelatin depresses 0.753 mm. Do you think the elastic limit of the gelatin has been exceeded?______ Explain _________________________
2.A manufacturer wishes to produce crude benzene models for the chemistry department. The manufacturer plans to do this by punching regular hexagons out of steel. If the steel plate is 0.500 cm thick, What is the minimum shearing force needed to punch regular hexagons with sides of length 5.56 cm? ________N Note: Assume that if the shear stress in steel exceeds 4.0 x 10^8 N/M^2 the steel ruptures.
To find Young's modulus for the gelatin in the first question, we can use Hooke's law, which states that the deformation of a material is directly proportional to the applied force.
Young's modulus (E) is defined as the ratio of stress to strain in a material. Stress (σ) is the force applied per unit area, and strain (ε) is the ratio of the change in length to the original length.
In this case, the given mass (500.0 g) is acting as a force on the gelatin. The depression of the gelatin caused by this force can be considered as strain.
Given:
Mass of the disk = 500.0 g
Diameter of the disk = 5.00 cm
Radius of the disk (r) = diameter / 2 = 2.50 cm = 0.025 m
Depression of the gelatin (d) = 0.243 mm = 0.243 x 10^-3 m
To find the applied force, we can use the equation F = mg, where g is the acceleration due to gravity (9.8 m/s^2).
Applied force (F) = 500.0 g x 9.8 m/s^2
Next, to find the area of the gelatin that is being compressed, we can calculate the area of the circular disk. The formula for the area of a circle is A = πr^2.
Area of the circular disk (A) = π (0.025 m)^2
Now we can calculate the stress on the gelatin using the formula stress = force / area.
Stress (σ) = F / A
Finally, we can calculate the Young's modulus (E) using the formula E = stress / strain.
Young's modulus (E) = σ / ε
Substituting the values into the equations will give us the value of Young's modulus for the gelatin in N/m^2.
To answer the second question about whether the elastic limit of the gelatin has been exceeded when the 2.00 kg mass is placed on it, we can compare the strain caused by this mass to the strain caused by the 500.0 g mass. If the strain from the 2.00 kg mass is greater, then the elastic limit has likely been exceeded.
Now let's apply the same process to answer the second question.
Given:
Thickness of the steel plate (h) = 0.500 cm = 0.500 x 10^-2 m
Length of the sides of the hexagon (L) = 5.56 cm = 5.56 x 10^-2 m
Shear stress limit (τ) = 4.0 x 10^8 N/m^2
To find the minimum shearing force, we can use the formula:
Force (F) = shear stress x area
In this case, the shear stress will be limited to the maximum shear stress limit of 4.0 x 10^8 N/m^2.
To calculate the area of the hexagon, we need to determine the apothem (a). The apothem is the radius of the inscribed circle in a regular hexagon.
Given that the length of the side (L) is the same as the diameter of this inscribed circle:
Radius (r) = apothem (a) = L / 2 = 5.56 x 10^-2 / 2
Now, we can calculate the area (A) of the hexagon using the formula: A = (3√3 / 2) x (apothem)^2.
Substituting the values into the equation will give us the area of the hexagon.
Finally, we can calculate the minimum shearing force (F) using the formula F = τ x A.
If the calculated force is less than the shear stress limit, then the steel can withstand the punching process without rupturing. Otherwise, it will rupture.
1. To find Young's modulus for the gelatin, we can use Hooke's Law, which states that the amount that an object deforms (d) is directly proportional to the force applied (F) and inversely proportional to the cross-sectional area (A) and the Young's modulus (Y) of the material.
First, we need to calculate the cross-sectional area of the gelatin. Since it is a cube with a side length of 20.0 cm, the cross-sectional area is (20.0 cm)^2 = 400 cm^2 = 0.04 m^2.
Next, we can calculate the force applied to depress the gelatin using the equation F = mg, where m is the mass (500.0 g = 0.5 kg) and g is the acceleration due to gravity (9.8 m/s^2).
F = (0.5 kg)(9.8 m/s^2) = 4.9 N
Now, using Hooke's Law, we can write it as d = (F/A) * (1/Y).
Given that the gelatin is depressed by 0.243 mm = 0.243 * 10^(-3) m, we can rearrange the equation to solve for Young's modulus (Y).
Y = (F/A) * (1/d)
Y = (4.9 N) / (0.04 m^2 * 0.243 * 10^(-3) m)
Y ≈ 2.022 * 10^6 N/m^2
Therefore, the Young's modulus for the gelatin is approximately 2.022 * 10^6 N/m^2.
Later, when a 2.00 kg mass is placed on the gelatin and it depresses by 0.753 mm, we can follow the same steps to calculate the force applied and find Young's modulus for that scenario. If Young's modulus for this second scenario is higher than the previous one, it can indicate that the elastic limit of the gelatin has been exceeded.
2. To find the minimum shearing force needed to punch regular hexagons out of steel, we can use the equation for shear stress (τ) which is equal to the shearing force (F) divided by the area (A) over which it is applied.
Given that the steel plate is 0.500 cm = 0.005 m thick, we can calculate the area of the hexagon using the formula:
Area = (3√3/2) * (side length)^2
Area = (3√3/2) * (5.56 cm)^2 = (3√3/2) * (0.0556 m)^2
Now, we can calculate the minimum shearing force using the equation τ = F/A.
Given that the maximum shear stress before rupture is 4.0 x 10^8 N/m^2, we can rearrange the equation to solve for F.
F = τ * A
F = (4.0 x 10^8 N/m^2) * [(3√3/2) * (0.0556 m)^2]
F ≈ 1.681 x 10^5 N
Therefore, the minimum shearing force needed to punch regular hexagons with sides of length 5.56 cm from the steel plate is approximately 1.681 x 10^5 N.