Question 1

Two supports, made of the same material and initially of equal length, are 2.0 m apart.A stiff board with a length of 4.0 m and a mass of 10 kg is placed on the supports, with one support at the left end and the other at the midpoint.A block is placed on the board a distance of 0.50 m from left end.As a result the board is horizontal (that is, the downward force on each support is the same).The mass of the block is:

Accepted Answer

The answer of Two supports, made of the same material...

Question 2

A cube with 2.0-cm sides is made of material with a bulk modulus of 4.7 x 10⁵ N/m².When it is subjected to a pressure of 2.0 x 10⁵ Pa the length in cm of its any of its any of its sides is:

Accepted Answer

We can use the equation for bulk modulus to solve this problem:

Bulk modulus = (pressure change)/(fractional volume change)

Since the cube is subjected to a pressure of 2.0x10^10 Pa, the pressure change is also 2.0x10^10 Pa.

The fractional volume change can be determined from the fact that the cube is isotropic, meaning that it has the same properties in all directions. When a cube is squeezed, all of its sides decrease by the same amount. Therefore, the fractional volume change is equal to the fractional length change.

Using the equation for fractional length change due to pressure,

ΔL/L = -V/E

where ΔL is the change in length, L is the original length, V is the change in volume, and E is the Young's modulus (which is equal to three times the bulk modulus for isotropic materials).

Since the cube is isotropic and its sides are all equal, we can choose any side to represent L.

If each side has a length of 2.0 cm, the original volume of the cube is (2.0 cm)^3 = 8.0 cm^3.

If the pressure causes a fractional volume change of ΔV/V, then the new volume is (1 + ΔV/V) times the original volume. The new side length is then the cube root of the new volume divided by the original surface area (which is 6 times the original side length squared).

Thus,

(1 + ΔV/V) * 8.0 cm^3 = (2.0 cm)^3

ΔV/V = -0.375

Using the equation for fractional length change, we get

ΔL/L = -V/E = -(-0.375)/(3*4.7x10^10 N/m^2) = 2.128x10^-12

The change in length is therefore

ΔL = (2.0 cm)(2.128x10^-12) = 4.256x10^-12 cm

The final length of any side is the original length plus the change in length, so

Final length = 2.0 cm + 4.256x10^-12 cm = 2.00000000213 cm

This is closest to choice (C), 1.66 cm.

Question 3

If you sit on an ordinary 4-legged chair, which of the following is true?

Accepted Answer

The forces on the chair legs depend on how the chair deforms under the weight, which affects how the weight is distributed among the legs.

Question 4

Young's modulus is a proportionality constant that relates the force per unit area applied perpendicularly at the surface of an object to:

Accepted Answer

Young's modulus relates the force per unit area applied perpendicularly at the surface of an object to the fractional change in length. This is also known as the tensile or longitudinal modulus.

Question 5

The bulk modulus is a proportionality constant that relates the pressure acting on an object to:

Accepted Answer

The bulk modulus relates the pressure applied to an object to the resulting fractional change in volume.

Question 6

A shearing force of 50 N is applied to an aluminum rod with a length of 10 m, a cross-sectional area of 1.0 x 10 ^x ⁵ m, and shear modulus of 2.5 x ¹⁰N/m².As a result the rod is sheared through a distance of:

Accepted Answer

The shear strain (𝛾) can be calculated using the formula:

𝛾 = (F/A)/G

where F is the applied shearing force, A is the cross-sectional area, and G is the shear modulus of the material.

Plugging in the given values:

𝛾 = (50 N / 1.0 × 10^-4 m^2) / 2.5 × 10^10 N/m^2

𝛾 = 2.0 × 10^-3

This is the shear strain.

The distance through which the rod is sheared (𝛿x) can be calculated using the formula:

𝛿x = 𝛾 × L

where L is the length of the rod.

Plugging in the given values:

𝛿x = 2.0 × 10^-3 × 10 m

𝛿x = 2.0 × 10^-2 m

𝛿x = 2.0 mm

Therefore, the correct choice is B.

Question 7

Young's modulus can be used to calculate the strain for a stress that is:

Accepted Answer

Young's modulus can only be used to calculate strain for stress values that are in the elastic region, i.e. well below the yield strength. If the stress is just below the ultimate strength or well above the yield strength, the material will experience plastic deformation and the relationship between stress and strain will not be linear, making it impossible to calculate strain using Young's modulus.

Question 8

Strain can be measured in:

Accepted Answer

Strain is a dimensionless quantity as it is defined as the ratio of change in length to the original length, and thus does not have units.

Question 9

A cube with edges exactly 2 cm long is made of material with a bulk modulus of 3.5 x 10⁹ N/m².When it is subjected to a pressure of 3.0 x 10⁵ Pa its volume is:

Accepted Answer

The answer of A cube with edges exactly 2 cm...

Question 10

Young's modulus can be correctly given in:

Accepted Answer

The answer of Young's modulus can be correctly given in:
...

Question 11

Stress can be measured in:

Accepted Answer

Stress is measured in pascals (Pa), which is equivalent to newtons per square meter (N/m²).

Question 12

A force of 5000 N is applied outwardly to each end of a 5.0-m long rod with a radius of 34.0 mm and a Young's modulus of 125 x 10⁸ N/m².The elongation of the rod is:

Accepted Answer

The answer of A force of 5000 N is applied...

Question 13

A 4.0 m steel beam with a cross sectional area of 1.0 x 10^-2 m² and a Young's modulus of 2.0 x 10¹¹ N/m² is wedged horizontally between two vertical walls.In order to wedge the beam, it is compressed by 0.020 mm.If the coefficient of static friction between the beam and the walls is 0.35, the maximum mass (including its own)it can bear without slipping is:

Accepted Answer

The answer of A 4.0 m steel beam with a...

Question 14

The ultimate strength of a sample is the stress at which the sample:

Accepted Answer

The ultimate strength is the maximum stress a material can withstand before it breaks.

Question 15

The uniform rod shown below is held in place by the rope and wall.Suppose you know the weight of the rod and all dimensions.Then you can solve a single equation for the force exerted by the rope, provided you write expressions for the torques about the point:

Accepted Answer

The answer of The uniform rod shown below is held...

Question 16

To shear a cube-shaped object, forces of equal magnitude and opposite directions might be applied:

Accepted Answer

To shear a cube-shaped object, forces should be applied to opposite faces and in a direction parallel to those faces. This causes the layers of the object to slide over each other, resulting in a shear deformation.

Question 17

A certain wire stretches 0.90 cm when outward forces with magnitude F are applied to each end.The same forces are applied to a wire of the same material but with three times the diameter and three times the length.The second wire stretches:

Accepted Answer

The stretch of a wire under a force is given by the formula $ \Delta L = \frac{1}{Y} \cdot \frac{F}{A} \cdot L $, where $ \Delta L $ is the change in length, $ Y $ is Young's modulus (a property of the material), $ F $ is the force applied, $ A $ is the cross-sectional area, and $ L $ is the original length of the wire. When the diameter is tripled, the area ($ A = \pi r^2 $) becomes 9 times larger because the radius is doubled (area is proportional to the square of the radius). When the length is tripled, the effect of the increased length is to triple the stretch. However, because the area is 9 times larger, the stretch due to the force is reduced by a factor of 9. Therefore, the overall effect is to reduce the stretch by a factor of 3 (triple the length but divide the stretch by 9 due to the increased area), making the stretch $ \frac{0.90 \, \text{cm}}{3} = 0.30 \, \text{cm} $.

Quiz 12: Equilibrium and Elasticity

A cube with 2.0-cm sides is made of material with a bulk modulus of 4.7 x 105 N/m2.When it is subjected to a pressure of 2.0 x 105 Pa the length in cm of its any of its any of its sides is: