8.9. EXERCISES

A steel wire of length 4.7 m and cross-sectional area 3.0 × 10-5 m2 stretches by the same amount as a copper wire of length 3.5 m and cross-sectional area of 4.0 × 10–5 m2 under a given load. What is the ratio of the Young's modulus of steel to that of copper?

This exercise requires you to compare the Young's modulus of two different materials, steel and copper, based on their elongation under the same load. The Young's modulus is a measure of the stiffness of a material, calculated as the ratio of stress to strain. Here, since both wires stretch the same amount under the same load, the ratio of their Young's moduli can be determined by comparing their lengths and cross-sectional areas. This is expressed mathematically.

Think of a rubber band and a piece of string. When you pull both with the same force, the rubber band stretches much more than the string. This is similar to how steel and copper will behave under tension. Steel, being stiffer (higher Young's modulus), will stretch less compared to copper when the same load is applied.

Figure 8.9 shows the strain-stress curve for a given material. What are (a) Young's modulus and (b) approximate yield strength for this material?

In this exercise, you're asked to interpret a graph that illustrates the relationship between stress and strain for a material. Young's modulus can be determined from the slope of the linear portion of the curve, while the yield strength is found at the point where the curve deviates from linearity, indicating the transition from elastic to plastic deformation.

Imagine pulling a piece of taffy. At first, it stretches easily (elastic behavior), but after a certain point, it begins to deform permanently. The slope of the initial stretch represents the Young's modulus, while the point at which it begins to deform permanently shows the yield strength.

The stress-strain graphs for materials A and B are shown in Figure 8.10. (a) Which of the materials has the greater Young's modulus? (b) Which of the two is the stronger material?

This exercise focuses on analyzing two different materials based on their stress-strain curves. The material with the steeper initial slope has the greater Young's modulus, indicating it’s stiffer. To determine which material is stronger, you must check which material has a higher ultimate tensile strength—the maximum stress it can withstand before failing.

Consider a race between two cars. One that can accelerate quickly represents a material with high Young's modulus (less stretch for the same stress), while the car that can carry a heavier load without collapsing represents the stronger material.

Read the following two statements below carefully and state, with reasons, if it is true or false. (a) The Young's modulus of rubber is greater than that of steel; (b) The stretching of a coil is determined by its shear modulus.

In this task, you analyze two statements regarding material properties. The first statement is false; rubber, which is highly elastic but not stiff, has a much lower Young's modulus than steel. The second statement is also false; the stretching of a coil is primarily determined by its Young's modulus rather than its shear modulus, which pertains to how materials deform under shear stress.

Think about stretching a thick rope versus a stretchy rubber band. The rubber band can stretch a lot, but it is still not as 'stiff' as the rope, which doesn't stretch as easily. This illustrates how different materials respond to forces.

Two wires of diameter 0.25 cm, one made of steel and the other made of brass are loaded as shown in Figure 8.11. The unloaded length of steel wire is 1.5 m and that of brass wire is 1.0 m. Compute the elongations of the steel and the brass wires.

You need to calculate the elongation of two different wires under the same load. Using the relationship of stress, strain, and Young's modulus, you can solve for the elongations using the equations related to material properties. Remember, the elongation will differ due to the materials having different Young’s moduli even though they are subjected to the same forces.

Imagine stretching two different types of slinkies. One made of metal and another made of plastic. When both are stretched with equal force, the metal one won’t stretch as much as the plastic one because of its greater stiffness (Young's modulus).

The edge of an aluminum cube is 10 cm long. One face of the cube is firmly fixed to a vertical wall. A mass of 100 kg is then attached to the opposite face of the cube. The shear modulus of aluminum is 25 GPa. What is the vertical deflection of this face?

This problem involves applying an external force that induces shear stress on the aluminum cube. Given the shear modulus, you can calculate the amount of deflection experienced by the face of the cube due to the weight applied. The shear strain and stress relation will guide the calculation.

Think of a thick book resting on a soft table. If a heavy object is placed on top of the book, it’ll bend slightly under the weight, similar to how the aluminum cube twists slightly when the weight is applied.

Four identical hollow cylindrical columns of mild steel support a big structure of mass 50,000 kg. The inner and outer radii of each column are 30 and 60 cm respectively. Assuming the load distribution to be uniform, calculate the compressional strain of each column.

In this exercise, you're calculating how much each column compresses under a known load. The load is shared equally among the four columns, and you can find the compressive stress applied to each. Then, relating this stress to strain through Young's modulus gives the compressional strain.

Consider a multi-story building. The pillars at the base support the weight of the whole structure above. Just like the steel columns, these pillars compress slightly under the weight, ensuring stability and safety of the building.

A piece of copper having a rectangular cross-section of 15.2 mm × 19.1 mm is pulled in tension with 44,500 N force, producing only elastic deformation. Calculate the resulting strain?

This task involves calculating strain produced by a known force applied to a copper piece with a specific cross-section. Using the definition of strain (change in length/original length), you can find the elongation cause by the force and then compute the strain.

Imagine pulling a thick rubber band. The more you pull, the more it stretches. The ratio of how much it stretches to its original length is analogous to strain in the copper piece.

A steel cable with a radius of 1.5 cm supports a chairlift at a ski area. If the maximum stress is not to exceed 108 N m–2, what is the maximum load the cable can support?

To find the maximum load supported by the cable without exceeding the allowable stress, use the formula for stress and solve for load. Here you will be calculating stress as force divided by area, allowing you to rearrange and find the maximum force that can be applied before reaching the stress limit.

Think of a tightrope walker. The rope can only hold so much weight before it snaps. Similarly, the steel cable can only support a certain load before reaching its maximum allowable stress.

A rigid bar of mass 15 kg is supported symmetrically by three wires each 2.0 m long. Those at each end are of copper and the middle one is of iron. Determine the ratios of their diameters if each is to have the same tension.

Here, you’ll use Young’s modulus to create a relationship between the materials and their respective diameters, ensuring equal tension. Since copper and iron have different properties, their diameters will differ if the tension equals in each wire.

It's like using different sized ropes to hold a load. If one rope is weaker (like copper), it has to be thicker to bear the same load as a stronger rope (like iron).

A 14.5 kg mass, fastened to the end of a steel wire of unstretched length 1.0 m, is whirled in a vertical circle with an angular velocity of 2 rev/s at the bottom of the circle. The cross-sectional area of the wire is 0.065 cm2. Calculate the elongation of the wire when the mass is at the lowest point of its path.

This problem involves centripetal force and tension as the mass moves in a circle. At the lowest point, the tension in the wire will be affected by the weight of the mass and the centripetal force. Using the known values and stress-strain relationships will yield the elongation.

Picture a swing at the playground. At the lowest point of the swing, the chains tighten due to the weight of the person. Similar principles of tension and elongation apply to the steel wire here as it supports the mass.

Compute the bulk modulus of water from the following data: Initial volume = 100.0 litre, Pressure increase = 100.0 atm (1 atm = 1.013 × 105 Pa), Final volume = 100.5 litre. Compare the bulk modulus of water with that of air (at constant temperature). Explain in simple terms why the ratio is so large.

For this exercise, you will calculate the bulk modulus based on the change in volume due to a change in pressure. After finding the bulk modulus of water, compare it to that of air to highlight the differences in compressibility between the two states of matter.

Imagine trying to compress a balloon filled with air versus a solid ball made of rubber. You can easily squeeze the balloon, showing that air is far more compressible (lower bulk modulus) than solid rubber, which barely changes volume under pressure.

What is the density of water at a depth where pressure is 80.0 atm, given that its density at the surface is 1.03 × 103 kg m–3?

This exercise involves applying hydrostatic pressure principles to find how the density of a fluid changes under pressure. It utilizes the formula relating pressure, density, and gravity to arrive at the density at a specified depth.

Think of stacking books on a table; as you add books, the bottom one gets heavier due to the weight on top. Similarly, the deeper water layers are 'heavier' due to the pressure from all the layers above.

Compute the fractional change in volume of a glass slab when subjected to a hydraulic pressure of 10 atm.

You will calculate how much the volume of the glass slab changes under a certain pressure, using the definition of bulk modulus and the relationship between pressure and volume change.

Consider the difference in size between a full and an empty soda can. When you apply pressure to the full can, the volume will change slightly compared to the empty one—this illustrates how hydraulic pressure can affect volume.

Determine the volume contraction of a solid copper cube, 10 cm on an edge, when subjected to a hydraulic pressure of 7.0 × 106 Pa.

In this exercise, you will determine the change in volume of a copper cube when subjected to considerable hydraulic pressure. This follows the definition of bulk modulus and the formula relating volume change to the applied pressure.

Imagine a sponge under water pressure. It squeezes down and becomes smaller. Similarly, when you apply pressure to copper, it will have a small change in volume while under pressure.

How much should the pressure on a litre of water be changed to compress it by 0.10%?

Calculating this requires understanding the relationship between pressure change and volume change. This will involve applying the concept of compressibility and bulk modulus to find the pressure needed for a specific volume change.

Consider a sponge again; if you want to compress it significantly, you need to apply more pressure. The same is true for water, but since water is less compressible, you'll need more pressure for a small volume change!

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