SI Practice Test 3: Tools, Machines and Materials

Answer: D

Correct answer: A power shear. Match the job verb to the machine: cutting sheet along a line points to a power shear, while bending it into a flange would point to a press brake.

• A: A press brake bends sheet into angles and flanges; it does not cut the sheet apart.
• B: An arbor press pushes press-fit parts such as bushings into place.
• C: A surface grinder finishes flat surfaces with an abrasive wheel; it is not used to cut sheet to width.
• D (correct): A power shear cuts sheet along a straight line, which reduces its width.

Answer: B

Correct answer: Butt joint. Joint names describe geometry: edge to edge in one plane is a butt joint, overlapped pieces form a lap joint, and a right-angle meeting forms a T-joint.

• A: A lap joint overlaps one piece on top of the other.
• B (correct): A butt joint places pieces edge to edge without overlapping.
• C: A T-joint has one piece meeting another at roughly a right angle.
• D: A corner joint brings two pieces together at an angle to form a corner.

Answer: A

Correct answer: Hands may contact sharp chips or rotating tooling. Drilling chips are thin, hot and razor-edged, and long stringy chips often wrap around the bit, so reaching for them brings the hand into the rotating zone, where the chip itself can pull the hand in. The safe approach is to stop the spindle, let it come to rest, and then clear chips with a brush, hook or pliers. A glove is not a fix, since it can be snagged and drawn into the rotation.

• A (correct): Wait for safe machine condition and use an appropriate chip-removal tool.
• B: Color perception is not the relevant hazard.
• C: Ordinary machining chips are not inherently radioactive.
• D: Gloves may themselves increase entanglement risks around moving parts.

Answer: C

Correct answer: Connect a note or callout to a specified feature. A leader is a thin line, usually ending in an arrowhead on an edge or a dot on a surface, that ties a note or callout to the exact feature it describes. Hole callouts, chamfer notes and finish notes typically use leaders so the reader knows which feature gets the specification. The leader itself carries no requirement; the meaning comes from the note. The tempting misreadings give it extra roles, such as setting units or controlling concentricity, which come from general notes or geometric tolerances instead.

• A: Unit conventions generally come from notes or title block and cannot be inferred from the leader.
• B: That layout role is not the essential function of an annotation leader.
• C (correct): The leader directs an annotation to the part feature to which it applies.
• D: Concentricity requires appropriate geometry controls rather than any leader line.

Answer: C

Correct answer: It can distribute bearing load but does not guarantee anti-loosening. A flat washer spreads the clamping force of a bolt head or nut over a larger area, which protects soft or finished surfaces and can bridge an oversized hole. It has no feature that resists rotation, so it should not be counted on to prevent loosening under vibration. Locking requires a different method, such as correct preload, a prevailing-torque nut, a cotter pin or a threadlocker. A washer also cannot make up for too little thread engagement.

• A: Washer placement does not change the right- or left-hand helix.
• B: A plain flat washer has no locking feature, so it cannot be relied on to keep a nut from loosening under vibration.
• C (correct): A flat washer mainly serves a bearing or surface-protection role when properly specified.
• D: A washer only sits under the head or nut; it cannot replace the load-carrying engagement of the threads.

Answer: B

Correct answer: 15.18 mm. A vernier reading has two parts: the main scale gives whole millimeters up to the zero of the vernier, and the vernier line that aligns exactly with a main-scale line gives the fraction. Each vernier division is worth the least count, here 0.02 mm, so the ninth line adds

Treating the line number as tenths or hundredths without applying the least count gives readings such as 15.90 or 15.09 mm.

• A: The arithmetic adds more than the given nine vernier increments to the main reading.
• B (correct): The indicated fractional part is 9 multiplied by 0.02 mm = 0.18 mm; adding 15 gives 15.18 mm.
• C: Nine hundredths would require a 0.01-mm division rather than the specified 0.02 mm.
• D: 0.90 mm assumes each division equals a tenth of a millimeter.

Answer: C

Correct answer: Pipe wrench. A pipe wrench has an adjustable hook jaw with serrated teeth set at a slight angle. When you pull on the handle, the jaw cocks and wedges tighter against the round surface, so the harder you pull, the harder it bites. That makes it ideal for pipe and round stock, which offer no flats for an ordinary wrench. The same teeth chew into hex nuts, so use a box-end wrench where one fits rather than a pipe wrench.

• A: A box end fits polygonal fasteners.
• B: A mallet is for impact placement.
• C (correct): A pipe wrench grips suitable round stock by wedging toothed jaws.
• D: A caliper measures size and should not be used to grip under load.

Answer: A

Correct answer: A backsaw. A backsaw has a thin blade stiffened by a heavy strip of brass or steel along its top edge. The stiff spine keeps the blade from flexing, so it cuts straight and accurately, which suits joinery such as tenons, dovetails and miter-box work. The trade-off is that the spine limits cut depth. A coping saw and a keyhole saw use narrow blades for curves or openings, not stiff straight cuts.

• A (correct): A backsaw has a stiffening back that stabilizes many controlled straight cuts.
• B: A coping saw has a narrow blade for curves rather than a stiff backed blade.
• C: A bow drill rotates a bit and is not a stiff-backed saw.
• D: A keyhole saw is adapted for reaching openings and has no typical backsaw spine.

Answer: C

Correct answer: 200 N. Use

and solve for F.

• A: This treats the pull as a right angle; at 30 degrees only half the force turns the bolt, so more force is needed.
• B: This multiplies by sin 30° instead of dividing by it, which gives less force when an angled pull actually needs more.
• C (correct): Solving 40 = 0.40 × F × sin 30° gives F = 40 / 0.20 = 200 N.
• D: This uses cos 30° instead of sin 30°; the turning part of the force depends on the sine of the angle to the handle.

Answer: A

Correct answer: 324 kJ. Use

with the temperature change, not the final temperature:

• A (correct): The rise is 180 − 20 = 160 °C, and 4.5 × 450 × 160 = 324,000 J = 324 kJ.
• B: This uses the final temperature of 180 °C as the temperature change instead of the 160 °C rise.
• C: This enters the mass as 4,500 g while c is per kilogram, making the answer 1,000 times too large.
• D: This adds the two temperatures (200 °C) instead of subtracting them to find the change.