A composite rod is made by attaching a glass-reinforced plastic rod and a nylon rod end to end, as shown.

Question 12

A composite rod is made by attaching a glass-reinforced plastic rod and a nylon rod end to end, as shown.

The rods have the same cross-sectional area and each rod is 1.00 m in length. The Young modulus Ep of the plastic is 40 GPa and the Young modulus En of the nylon is 2.0 GPa.

The composite rod will break when its total extension reaches 3.0 mm.

What is the greatest tensile stress that can be applied to the composite rod before it breaks?

A 7.1 × 10^-14 Pa

B 7.1 × 10^-2 Pa

C 5.7 × 10⁶ Pa

D 5.7 × 10⁹ Pa

Reference: Past Exam Paper – June 2014 Paper 12 Q21

Solution:

Answer: C. 

       

The composite rod will break when its total extension reaches 3.0 mm.

For a material,

Stress = Force / Area = F/A

Strain = extension, e / original length, L = e/L

Young modulus, E = stress / strain

E = Stress / (e/L) = Stress×L / e

Extension, e = Stress×L / E

The same tensile stress (the same force and since the cross-sectional area is the same for both, the same tensile stress is therefore applied) is applied to the different materials (glass-reinforced plastic and nylon). The different materials will extend by different amounts when the same stress is applied to each. 

The total extension cannot be equal to 3.0 mm (= 3.0×10^-3 m) as the composite rod would then break.

For the plastic rod, extension e_p = Stress × (1) / (40×10⁹) = Stress / (40×10⁹)

For the nylon rod, extension e_n = Stress × (1) / (2×10⁹) = Stress / (2×10⁹)

For the greatest tensile stress that can be applied to composite rod before it breaks, the sum of extensions of the plastic and nylon should be equal to 3.0 mm.

e_p + e_n = 3.0×10^-3 

Stress/(40×10⁹) + Stress/(2×10⁹) = 3.0×10^-3

Stress × [1/(40×10⁹) + 1/(2×10⁹)] = 3.0×10^-3

Stress = (3.0×10^-3) / [1/(40×10⁹) + 1/(2×10⁹)] = 5.71×10⁶ Pa

A uniform solid cuboid of concrete of dimensions 0.50 m × 1.20 m × 0.40 m and weight 4000 N rests on a flat surface with the 1.20 m edge vertical as shown in diagram 1.

Question 38

A uniform solid cuboid of concrete of dimensions 0.50 m × 1.20 m × 0.40 m and weight 4000 N rests on a flat surface with the 1.20 m edge vertical as shown in diagram 1.

What is the minimum energy required to roll the cuboid through 90° to the position shown in diagram 2 with the 0.50 m edge vertical?

A 200 J                        B 400 J                        C 1400 J                     D 2600 J

Reference: Past Exam Paper – June 2014 Paper 13 Q14

Solution:

Answer: A.

A force should be provided such that in the process, only one edge is in contact with the surface (in our case, this is the bottom right edge). This edge would act as a pivot about which the cuboid would roll. This is illustrated in the diagrams below.

Since the solid cuboid is uniform, its centre of mass may be considered to be at the centre of the cuboid, that is, initially at a height of 0.6 m.

Weight = mg = 4000N

Therefore before the force is applied,

Potential energy of cuboid = mgh = mg × h = (4000 × 0.6) J

As illustrated by the diagram (in the right), as the cuboid rolls about the edge (which is in contact with the surface), the centre of mass of the cuboid rises to a greater height. So, its potential energy increases.

The centre of mass rises to a maximum height when the top left edge is vertically above. So, the potential energy of the cuboid is maximum at that position. {Afterwards, the cuboid would fall under gravity. So, the height of its centre of mass decreases.}

Therefore, the minimum energy produced by the force should cause the potential energy of the cuboid to change from its original position to the position described above. That is the minimum energy is equal to the change in energy from these 2 positions.

Consider the position for maximum potential energy. The centre of mass would be at the middle of the (blue) dotted line. It can be seen that a right angle triangle is formed with the vertical (blue) dotted line as the hypotenuse and the other sides being 0.5 m and 1.2 m.

Using Pythagoras’ theorem.

Vertical (blue) dotted line = √(0.5²+1.2²) = 1.3 m

Height of centre of mass in new position = 1.3 / 2 = 0.65 m     

Potential energy of cuboid in new position = mgh = (4000 × 0.65) J

Minimum energy required to roll cuboid = Rise in potential energy

Minimum energy = (4000 × 0.65) – (4000 × 0.60) J

Minimum energy = (4000) × (0.65 – 0.6) = 200 J