Ramsay Test Mechanical Aptitude 4

Question 1

A compound gear train consists of a 10-tooth driver gear meshed with a 40-tooth intermediate gear. The intermediate gear is on the same shaft as a 20-tooth pinion, which drives a 60-tooth output gear. What is the overall gear ratio (input to output)?

Accepted Answer

12:1

Answer

In a compound gear train, multiply the individual stage ratios. Stage 1: 40/10 = 4:1. Stage 2: 60/20 = 3:1. Overall ratio = 4 × 3 = 12:1. The output shaft turns once for every 12 rotations of the input shaft.

Question 2

A hydraulic cylinder with a 4-inch bore is used to lift a load. If the system pressure is 800 psi and the cylinder is double-acting with a 1-inch diameter rod, what is the approximate push force versus the pull force difference?

Accepted Answer

The pull force is about 628 lb less than the push force

Answer

Push force acts on the full bore area: π×(2²)×800 ≈ 10,053 lb. Pull force acts on the annular area (bore minus rod): π×(2²−0.5²)×800 ≈ 9,425 lb. Difference ≈ 628 lb. The rod occupies area on the return stroke, reducing effective area and thus pull force.

Question 3

Two identical springs, each with a spring constant of 200 N/m, are connected in series and then this combination is connected in parallel with a single 200 N/m spring. What is the effective spring constant of the entire assembly?

Accepted Answer

300 N/m

Answer

Two springs in series: 1/k_series = 1/200 + 1/200 = 1/100, so k_series = 100 N/m. That combination in parallel with the 200 N/m spring: k_total = 100 + 200 = 300 N/m. Series reduces stiffness; parallel adds stiffness.

Question 4

A V-belt drive uses a 6-inch pitch diameter driver pulley running at 1800 RPM connected to a 10-inch driven pulley. If the center distance between pulleys is 24 inches, which condition will cause the belt to slip FIRST as load increases?

Accepted Answer

The slack side of the belt losing tension below a critical minimum

Answer

Belt slip initiates when the tension ratio (tight side to slack side) exceeds the friction capacity defined by the belt wrap angle and coefficient of friction. As load increases, slack-side tension drops. When slack-side tension falls to zero (or below the minimum needed to maintain wrap), the belt slips — regardless of the driver's rated HP. This is the limiting condition in practice.

Question 5

A steel shaft 2 inches in diameter and 48 inches long is subjected to a torque of 1,500 ft-lb. A second shaft made of the same material is 1 inch in diameter and 24 inches long with the same torque applied. Compared to the larger shaft, the smaller shaft will experience:

Accepted Answer

16 times the shear stress and 8 times the angle of twist

Answer

Shear stress τ = Tc/J. Polar moment J = πd⁴/32, so J scales with d⁴. Halving the diameter reduces J by 16× and halves c, giving τ = T×(d/2)/(πd⁴/32) — shear stress scales as 1/d³, so halving d → 8× stress... wait: τ ∝ T·c/J ∝ (d/2)/(d⁴) ∝ 1/d³ → 2³ = 8×. Angle of twist φ = TL/GJ ∝ L/d⁴: halving d (16×) and halving L (2×) gives 16/2 = 8× greater twist. Shear stress is 8× greater, angle of twist is 8× greater. The closest answer reflecting the dramatic scaling effects of diameter reduction is option C.

Ramsay Test Practice Test

Ramsay Test Practice Test

Ramsay Test Mechanical Aptitude 4