(a) The coefficient of friction for all contacting surfaces in Figure 3(a) is 0·2. Does the 25 kg force move the block A up, hold…

(a) Equilibrium of block A and wedge B with friction

Data: θ = 15°, μ = 0.2 on all contact surfaces, W_A = 100 kgf, W_B = 50 kgf, and a 25 kgf force pushing B to the left. The wall is on the left of A, and B's top face rises to the right, so the reaction N of B on A is normal to the incline and points up and to the left. Let H be the wall reaction on A and R the floor reaction on B. The test is to find the force F needed to start A upward and the force needed to stop A sliding down, and to see where 25 kgf lies.

Case 1: impending motion of A up and B left. Friction on A from B acts down the incline, wall friction on A acts down, and floor friction on B acts to the right.

For A: x: H = N sinθ + μN cosθ = N(sinθ + μ cosθ) y: N cosθ = W_A + μN sinθ + μH Substituting H: N[(1 − μ²)cosθ − 2μ sinθ] = W_A, so N = 100/(0.96 × 0.9659 − 0.4 × 0.2588) = 121.4 kgf H = 121.4 × (0.2588 + 0.2 × 0.9659) = 54.9 kgf

For B (vertical): R = W_B + N(cosθ − μ sinθ) = 50 + 121.4 × 0.9136 = 161.0 kgf For B (horizontal): F_up = H + μR = 54.9 + 0.2 × 161.0 = 87.1 kgf

Since 25 kgf is less than 87.1 kgf, the force cannot push A up.

Case 2: impending motion of A down and B right. All friction directions reverse.

For A: N[(1 − μ²)cosθ + 2μ sinθ] = W_A, so N = 100/(0.96 × 0.9659 + 0.4 × 0.2588) = 97.0 kgf H = N(sinθ − μ cosθ) = 97.0 × 0.0656 = 6.37 kgf For B: R = 50 + 97.0 × (0.9659 + 0.0518) = 148.7 kgf, so the maximum floor friction is μR = 29.7 kgf.

The push of A on B towards the right is only H = 6.37 kgf, which is less than 29.7 kgf. Even with no applied force, B cannot slide out, and the system is self-locking. The balance F = H − μR = −23.4 kgf is negative, meaning no leftward force is required to stop A coming down.

Conclusion: the required force lies between 0 (strictly, a negative value) and 87.1 kgf, and 25 kgf is inside this range. The 25 kg force holds the block A in equilibrium; it neither moves A up nor is it too small.

(b) Stresses on planes rotated clockwise by 15°

Take x to the right and y upward. From the element: σ_x = −80 N/mm² (compressive), σ_y = +20 N/mm² (tensile). The shear on the top face acts to the left and on the right face downward, so τ_xy = −30 N/mm². A clockwise rotation of 15° gives θ = −15°, so 2θ = −30°, cos 2θ = 0.8660 and sin 2θ = −0.5.

Transformation equations: σ_x′ = (σ_x + σ_y)/2 + (σ_x − σ_y)/2 cos 2θ + τ_xy sin 2θ = −30 + (−50)(0.8660) + (−30)(−0.5) = −58.30 N/mm²

σ_y′ = (σ_x + σ_y)/2 − (σ_x − σ_y)/2 cos 2θ − τ_xy sin 2θ = −30 + 43.30 − 15 = −1.70 N/mm²

τ_x′y′ = −(σ_x − σ_y)/2 sin 2θ + τ_xy cos 2θ = −25 − 25.98 = −50.98 N/mm²

So the rotated element carries 58.30 N/mm² compression on the x′ face, 1.70 N/mm² compression on the y′ face, and a shear stress of 50.98 N/mm² in the negative sense.

Check: σ_x′ + σ_y′ = −60 N/mm² = σ_x + σ_y. The Mohr circle has centre −30 and radius √(50² + 30²) = 58.31 N/mm²; the point (−58.30, −50.98) lies at distance √(28.30² + 50.98²) = 58.31 N/mm² from the centre, so it is on the circle.

(c) Microstructure and properties of pearlite, bainite and martensite

Pearlite is a lamellar eutectoid of alternating ferrite and cementite (Fe₃C) plates, formed by diffusion-controlled decomposition of austenite at about 550–727°C. Fine pearlite, formed at lower temperature, has smaller interlamellar spacing and so higher strength and hardness. Pearlite has moderate strength (UTS about 600–900 MPa), hardness about 15–30 HRC, good ductility and good machinability.

Bainite is a non-lamellar aggregate of ferrite and fine carbides, formed isothermally at about 250–550°C. Upper bainite (about 400–550°C) has carbides between ferrite laths; lower bainite (about 250–400°C) has fine carbides inside the laths. Bainite gives a better strength–toughness combination than pearlite at equal hardness, and lower bainite is stronger and tougher than upper bainite. Hardness is about 40–55 HRC.

Martensite forms by diffusionless shear of austenite on rapid quenching below the Ms temperature. It is a body-centred tetragonal supersaturated solution of carbon in iron, lath-type in low-carbon steels and plate-type (twinned) in high-carbon steels. It is the hardest and strongest of the three (up to about 65 HRC) but brittle, with high residual stress and cracking tendency, so tempering is needed to restore toughness.