A useful mechanical diagram tells a story about forces and movement. Learn to locate the pivot, follow a connection and name the governing relationship before calculating or comparing answer choices.
What you’ll learn
Learn the method
A 300-newton downward force acts 2 meters to the left of a pivot. What downward force, applied 3 meters to the right, balances a horizontal weightless lever?
The forces act perpendicular to the horizontal lever, so their moment arms are the labeled 2 meters and 3 meters. The full end-to-end span is not a moment arm.
The left force produces 300 × 2 = 600 newton-meters of turning effect. For balance, the right force F must satisfy F × 3 = 600.
F = 600 ÷ 3 = 200 newtons. A smaller force works on the longer arm, which agrees with the leverage you would expect.
200 N on the right balances 300 N on the left in this ideal model.
Equal forces are not required for balance when the arms differ. Equal opposite torques are required. Using 300 N on each side would make the longer right arm turn the lever clockwise.
Why would doubling the right-hand arm to 6 meters reduce the balancing force to 100 N?
The required right-hand torque is still 600 N·m. A 6-meter arm needs 600 ÷ 6 = 100 N. Force and perpendicular distance work together; the longer arm does not create energy.
The example teaches one method. Open a concept below when you need a rule or a different type of example during practice.
Turning effect depends on force and its perpendicular distance from the pivot: torque = force × moment arm. For balance, clockwise and counterclockwise torques must match. A 20-newton force acting at a perpendicular distance of three meters balances 60 newtons acting one meter away, when no other torques act.
In an ideal pulley arrangement, tension is the same along a continuous massless rope over frictionless pulleys. Two rope segments each pulling upward on the moving load can share its weight. A fixed pulley may change the pulling direction without multiplying force. Count the supporting segments in the actual diagram rather than just counting pulleys.
Two meshing external gears rotate in opposite directions. Adding a third makes that third rotate in the same direction as the first. For a meshing pair, a larger gear with more teeth turns more slowly than the smaller gear. Gears fixed on the same shaft share rotational speed, which is a different relationship.
Pressure is force divided by area. The same force over a smaller area produces greater pressure. In an ideal hydraulic system, equal fluid pressure acting on a larger piston area produces a larger force, with a corresponding tradeoff in movement. Simple machines can trade force for distance; they do not create energy.
Now apply what you know
Try a few questions without opening the solutions, then explain why each answer works. Use scratch paper for calculations and work without a calculator. These original, untimed questions sample the subject; the result is not an official score.
Work through one question at a time. Explain your approach, check the worked answer, then move on. Use scratch paper for calculations.
0 of 12 checked · 0 correct.
Progress is saved in this browser. These original study questions do not produce an official AFQT score.
Answer: D. D
D is the connected liquid beneath both pistons. A pressure change applied to the enclosed liquid is transmitted through it to the output piston. The frame supports the device, and the pistons apply and receive force.
Answer: B. 130 N
Take moments about the left support. The beam’s 200 N weight acts at its midpoint, 2 m away, and the extra 120 N acts 1 m away. Thus the right reaction is (200 × 2 + 120 × 1) ÷ 4 = 130 N. Sharing the total weight equally would ignore the off-center load.
Answer: A. 60 rpm
The first pair reduces speed by 12/36, giving B and C 240 rpm. The second pair reduces that by 10/40, so D turns at 60 rpm. B and C share speed because they share a rigid shaft; their different tooth counts must not be treated as a third meshing pair.
Answer: B. 60 MPa compressive
For a fully restrained warm rod, free thermal strain αΔT is prevented and becomes compressive stress. EαΔT = 200,000 MPa × 1.2×10⁻⁵/°C × 25°C = 60 MPa. The tensile choices have the wrong sign.
Answer: D. The effort
The effort acts between the fulcrum and the load. This is a third-class arrangement, which commonly trades increased effort for greater load travel or speed.
Answer: B. 40 J
The pull does the area under its graph: ½ × 2 × 20 + 2 × 20 = 60 J. Friction does −5 × 4 = −20 J. Net work is 40 J. The 60 J choice includes the pull but omits friction.
Answer: D. 40 cm/s
The same volume per second passes both sections, so area × average speed is constant. The area falls from 10 to 5 cm², so speed doubles: 20 × 10 ÷ 5 = 40 cm/s. Halving speed would reduce the flow rate instead of preserving it.
Answer: B. Tipping begins at 25 N
Sliding would require 0.40 × 100 = 40 N. Tipping about the lower-right corner begins when F × 1.20 = 100 × 0.30, giving F = 25 N. Since 25 N is lower, tipping begins before the sliding threshold is reached.
Answer: C. Straight-line motion along the bar
The rotating pinion pushes successive teeth of the rack, translating the rack along its length. The straight bar does not rotate about the pinion shaft. This is a conversion between rotary and linear motion.
Answer: D. 0.150 m
Each series spring carries the same 10 N tension. Their extensions are 10/100 = 0.10 m and 10/200 = 0.05 m, so the total is 0.15 m. Adding stiffnesses directly would describe a parallel arrangement, not this chain.
Answer: B. 0.050 m
The identical springs extend equally and share the 10 N load, so each carries 5 N. Each extension is 5 ÷ 100 = 0.050 m, which is also the bar’s downward movement. Treating either spring as carrying all 10 N would double the result.
Answer: C. 60 N·m
The brake removes 500 − 140 = 360 J. Constant torque does work τθ, so τ = 360 J/6 rad = 60 N·m. Using the final energy alone, adding energies, or ignoring angular displacement misstates the balance.
Before moving on, choose one error you can explain. Rework that question with the solution closed, then revisit the skill in a later session to check what you remember.
Locate pivots, fixed supports, ropes, contact points and the moving load. Add an arrow for each known motion. Work outward one connection at a time rather than trying to judge the whole drawing at once.
A question may ignore friction, rope mass or energy losses to isolate a principle. Use the stated model. If you make a prediction such as “less force but more distance,” check that both parts agree with the diagram and the conservation of energy.
Mechanical Comprehension is not part of the AFQT. It contributes to relevant service job composites, where it measures knowledge of mechanical and physical principles alongside other subtests.
The computer test includes unscored tryout questions in selected subtests. The longer time shown is the total for a section with tryouts, not extra time to add. These figures describe the proctored CAT-ASVAB, not untimed PiCAT or our practice sets. See the official questions and timing table and the guide to test versions.
Sources and materials. Subtest descriptions and timing were checked on September 29, 2026 against the Department of Defense’s What to Expect guide, sample questions and score guide. Teaching examples, diagrams and practice questions are independently authored Test Ninjas materials. Test Ninjas is not affiliated with or endorsed by the Department of Defense.