Unit Systems & Conversions - Statics and Dynamics

A stress is 40 ksi. What is this stress in MPa?
Answer: 276 40 ksi is 40,000 psi. Since 1 MPa equals 145 psi, divide 40,000 by 145 to get about 276 MPa. Equivalently, 1 ksi equals 6.895 MPa, so 40 x 6.895 = 276. The tempting wrong 27.6 comes from using 0.69 instead of 6.895, an order-of-magnitude slip.
In the US customary system, the slug is the derived unit of mass such that $1 \ \text{lbf} = 1 \ \text{slug} \cdot \text{ft/s}^2$ . A body weighs $W = 160 \ \text{lbf}$ on Earth ( $g = 32.2 \ \text{ft/s}^2$ ). The same body is transported to a location where the local gravitational acceleration is $g' = 5.40 \ \text{ft/s}^2$ . Which of the following correctly states the body's mass in slugs and weight in lbf at the new location?
Answer: Mass $= 4.97 \ \text{slugs}$ ; Weight $= 26.8 \ \text{lbf}$ , because mass is invariant and weight scales proportionally with local $g$ . Whenever you see a unit-system question in dynamics, anchor yourself to two fundamental truths: mass is invariant (it doesn't change with location), and weight depends on local gravity. In the US customary system, mass in slugs is found by dividing weight by gravitational acceleration: $m = W/g$ . At Earth's surface, $m = 160 \text{ lbf} / 32.2 \text{ ft/s}^2 = 4.97 \text{ slugs}$ . This mass is a fixed property of the body — it's the same whether the object is on Earth, the Moon, or deep space. At the new location, weight is simply $W' = m \cdot g' = 4.97 \times 5.40 = 26.8 \text{ lbf}$ . That's exactly what A states, making it correct. B is wrong because it claims weight stays at 160 lbf regardless of location. Weight is not conserved across gravitational environments — it's a force that scales directly with local $g$ . The lbf is defined through Newton's second law, not fixed to a standard gravity in a way that freezes the weight reading. C confuses the numerical coincidence that sometimes occurs when mixing lbf and lbm. Mass is never equal to weight in the slug system — 160 slugs would imply an enormous object, and "recomputing" mass using new gravity is physically wrong since mass doesn't change. D describes logic from the lbm–lbf system, where the conversion factor $g_c$ appears. That framework doesn't apply here; the question explicitly uses the slug system, where $g_c$ is unnecessary. Your key takeaway: always identify which unit system a problem uses before plugging in numbers. In the slug system, $m = W/g$ , mass is constant, and weight varies with local $g$ .
Which of the following statements is correct regarding the two approaches and their numerical results? Use $g = 9.81 \ \text{m/s}^2$ , $1 \ \text{slug} = 14.594 \ \text{kg}$ , $1 \ \text{ft} = 0.3048 \ \text{m}$ , and $1 \ \text{lbf} = 4.448 \ \text{N}$ .
Answer: Both approaches yield $\approx 490 \ \text{lbf/ft}^3$ ; Student B's intermediate density in slug/ft³ is $\approx 15.2 \ \text{slug/ft}^3$ , and multiplying by $g = 32.2 \ \text{ft/s}^2$ gives the same final result as Student A's method. When converting units across measurement systems, you must track every dimensional factor carefully — mass, length, and time all need attention simultaneously. Start with what Student A does: multiply $\rho = 7850 \ \text{kg/m}^3$ by $g = 9.81 \ \text{m/s}^2$ to get specific weight $\gamma = 76{,}994 \ \text{N/m}^3$ . Converting to lbf/ft³: multiply by $\frac{1 \ \text{lbf}}{4.448 \ \text{N}}$ and by $(0.3048)^3 \ \text{m}^3/\text{ft}^3$ , giving $\approx 490 \ \text{lbf/ft}^3$ . Now trace Student B. Converting $\rho$ from kg/m³ to slug/ft³ requires dividing by 14.594 kg/slug and multiplying by $(0.3048)^3 \ \text{m}^3/\text{ft}^3 \approx 0.02832$ : $\rho_B = 7850 \times \frac{0.02832}{14.594} \approx 15.2 \ \text{slug/ft}^3$ . Then $\gamma = 15.2 \times 32.2 \approx 490 \ \text{lbf/ft}^3$ . Both methods agree — confirming D is correct. Choice A is wrong because slug/ft³ × ft/s² does indeed give lbf/ft³ (since 1 slug·ft/s² = 1 lbf), so Student B's method is perfectly unit-consistent. Choice B fabricates an error; it claims Student B omits the cubic volume factor, but the correct conversion does include $(0.3048)^3$ , and both students get ≈490. Choice C is wrong because Student B's result is not ≈1600 lbf/ft³ — that figure would arise from misapplying a linear (not cubic) volume conversion, and both methods converge on the same answer. Study tip: Whenever you convert a volumetric quantity (density, specific weight), always cube your length conversion factor. A quick dimensional analysis check — writing out every unit explicitly — will catch this trap before it costs you points.
A 25-lb force acts at a perpendicular 6-in distance from a pivot. The moment in lb-ft is
Answer: 12.5 Convert 6 inches to 0.5 feet first. Moment = force times perpendicular distance = 25 times 0.5 = 12.5 lb-ft. The tempting value 150 comes from using 6 inches as if it were 6 feet; the distance must be in feet to get lb-ft.
Water density is 1000 kg/m³. Its mass density in slug/ft³ is
Answer: 1.94 Multiply 1000 kg/ $m^3$ by $0.3048^3$ to get 28.3 kg/ $ft^3$ , then divide by 14.59 kg per slug to get 1.94 slug/ $ft^3$ . The tempting 62.4 is water density in pounds per cubic foot, not slugs per cubic foot; slugs are a much larger mass unit.
A 2-kg mass accelerates at 3 m/s². The net force in lbf is
Answer: 1.35 Multiply mass and acceleration: 2 times 3 = 6 newtons. Since 1 lbf = 4.448 N, divide 6 by 4.448 to get about 1.35 lbf. The tempting 6.00 is the force in newtons, not lbf, because it skips the unit conversion.
At standard gravity, a 50-lb weight is acted on by a net 10-lb horizontal force. Its acceleration in ft/s² is
Answer: 6.44 A 50-lb weight means its weight force is 50 lb, so its mass is 50 divided by 32.2 ft/s², about 1.55 slugs. Dividing the net force 10 lb by that mass gives 6.44 ft/s². The tempting wrong move is dividing 50 by 10 to get 5.00, which confuses weight with mass and ignores gravity.
In SI, Newton's second law is written $\mathbf{F} = m\mathbf{a}$ with no conversion constant. In the gravitational metric system, mass is measured in technical mass units (TMU), where $1 \ \text{TMU} = 1 \ \text{kgf}\cdot\text{s}^2/\text{m}$ , and force is measured in kgf. A body has an SI mass of $60 \ \text{kg}$ and is subjected to a net force of $F = 200 \ \text{kgf}$ . What is the body's acceleration in m/s²? Use $g = 9.807 \ \text{m/s}^2$ .
Answer: $a = 32.7 \ \text{m/s}^2$ , found by converting 200 kgf to newtons ( $200 \times 9.807 = 1961 \ \text{N}$ ) and applying Newton's second law directly with the SI mass: $a = 1961 \ \text{N} / 60 \ \text{kg}$ . When a problem mixes unit systems, your safest strategy is to convert everything into one consistent system before applying Newton's second law. Here, the cleanest path is converting to SI entirely. Start by converting the force: $1 \ \text{kgf} = 9.807 \ \text{N}$ , so $200 \ \text{kgf} = 200 \times 9.807 = 1961 \ \text{N}$ . The mass is already given in SI as $60 \ \text{kg}$ . Applying $\mathbf{F} = m\mathbf{a}$ directly gives $a = 1961 \ \text{N} / 60 \ \text{kg} = 32.7 \ \text{m/s}^2$ , confirming C is correct. Choice A makes a critical double-conversion error: it correctly converts kgf to newtons (1961 N) but then divides by both the mass and $g$ , as if the gravitational factor needs to be applied a second time. Each unit system requires the conversion factor once — applying it twice compounds the error and artificially shrinks the acceleration. Choice B skips unit conversion entirely, plugging $F = 200$ and $m = 60$ directly into $F = ma$ as though kgf and kg are dimensionally interchangeable. They are not — kgf is a force unit, not a mass unit, and treating them as equivalent produces a physically meaningless result. Choice D inverts the mass conversion: instead of dividing 60 kg by $g$ to get TMU, it multiplies, inflating the mass and producing a wildly incorrect acceleration that is then further distorted by an extra factor of $g$ . Study tip: When you see mixed unit systems, always ask yourself: "Have I converted every quantity into the same system?" Apply each conversion factor exactly once — never twice.
Which of the following statements best characterizes the error and its source?
Answer: The first student is correct; 'kip' is a US customary unit equal to 1000 lbf (kilo-pound-force). The second student confuses 'kip' with 'kilo-kgf,' conflating SI and US customary prefixes, and over-estimates the force by a factor of approximately 2.2. When you encounter unit conversion problems in statics, your first move should always be to identify what system each unit belongs to before doing any arithmetic. "Kip" is a purely US customary unit — it comes from kilo pound-force — and equals exactly 1000 lbf. With that definition locked in, the first student's chain is clean: $P = 5 \ \text{kip} \times 1000 \ \frac{\text{lbf}}{\text{kip}} \times 4.448 \ \frac{\text{N}}{\text{lbf}} = 22{,}240 \ \text{N}$ This is correct, making B the right answer. The second student's error isn't a rounding issue — it's a conceptual one. They invented a hybrid interpretation where "kip" means 1000 kgf, mixing an SI-style "kilo-" prefix onto a metric mass-force unit. Since 1 kgf ≈ 2.205 lbf, treating 5 kip as 5000 kgf inflates the force by roughly that same factor of 2.2, yielding ~49,035 N instead of ~22,240 N. A is wrong because "kilogram-pound" is not a real unit in any engineering standard — this is pure fabrication, and kip has no international structural meaning tied to kgf. C misidentifies the source of error entirely; swapping $g = 9.807$ for $g = 9.81$ introduces less than 0.03% difference, nowhere near a 120% discrepancy — the real error is the wrong starting unit, not rounding. D is wrong because "kip" is unambiguously a unit of force, not moment; kip-inches or kip-feet are moment units, but the kip alone is force. Study tip: Always trace a unit back to its linguistic origin — "kip" = kilo + pound-force, firmly in the US customary world. When you see mixed-system problems, write out the unit family (SI vs. US customary) for every quantity before converting.
Which value correctly converts $250 \ \text{kPa}$ to psi, and which single-step error produces the most commonly seen wrong answer?
Answer: $36.3 \ \text{psi}$ ; the most common error is forgetting to square the inch-to-meter conversion factor, yielding $\approx 1420 \ \text{psi}$ instead. When converting pressure between unit systems, you must treat pressure as a ratio — force divided by area — and apply conversion factors to both the numerator and denominator correctly. The key trap here is the area term: because area has units of length², any length conversion factor must be squared. Starting with $250 \ \text{kPa} = 250{,}000 \ \text{N/m}^2$ , convert to lbf/in² like this: $250{,}000 \ \frac{\text{N}}{\text{m}^2} \times \frac{1 \ \text{lbf}}{4.448 \ \text{N}} \times \left(\frac{0.0254 \ \text{m}}{1 \ \text{in}}\right)^2$ $= 250{,}000 \times \frac{1}{4.448} \times (0.0254)^2 = 250{,}000 \times 0.2248 \times 0.000645 \approx 36.3 \ \text{psi}$ This confirms A is correct — $36.3 \ \text{psi}$ — and correctly identifies the most common error: forgetting to square $(0.0254)^2$ , which means multiplying by $0.0254$ rather than $0.000645$ . That inflates the result by a factor of about 39, giving the erroneous $\approx 1420 \ \text{psi}$ . B is wrong because dividing by $4.448$ twice has no physical basis — force appears only once in the numerator. C is wrong because $1420 \ \text{psi}$ is actually the error value, not the correct answer; C has the roles reversed. D is wrong because $1 \ \text{in} = 0.0254 \ \text{m}$ and $1 \ \text{in} = 2.54 \ \text{cm}$ are mathematically identical — using them correctly produces the same result. Study tip: Whenever you convert a unit that is raised to a power (area = m², volume = m³), always raise the entire conversion factor to that same power. Missing the exponent is the single most common unit-conversion mistake on engineering exams.
A rotating shaft transmits a torque of $T = 800 \ \text{N}\cdot\text{m}$ at an angular velocity of $\omega = 1200 \ \text{rpm}$ . A US customary specification sheet requires the power output in horsepower (hp). Using $1 \ \text{hp} = 550 \ \text{ft}\cdot\text{lbf/s}$ , $1 \ \text{ft} = 0.3048 \ \text{m}$ , and $1 \ \text{lbf} = 4.448 \ \text{N}$ , what is the shaft power in horsepower?
Answer: $\approx 135 \ \text{hp}$ , obtained by converting $\omega$ to rad/s, computing power in watts ( $P = T\omega$ ), converting watts to ft·lbf/s using both the force and length conversions, then dividing by 550. When a rotating shaft transmits power, the fundamental relationship is $P = T\omega$ , where torque must be in N·m and angular velocity in rad/s — not rpm. This question tests whether you can correctly apply that formula and then navigate a multi-step unit conversion chain without dropping a factor. Start by converting $\omega$ : $1200 \ \text{rpm} \times \frac{2\pi \ \text{rad}}{1 \ \text{rev}} \times \frac{1 \ \text{min}}{60 \ \text{s}} \approx 125.66 \ \text{rad/s}$ . Then $P = 800 \times 125.66 \approx 100{,}530 \ \text{W}$ . To convert watts (N·m/s) to ft·lbf/s, apply both conversions: $100{,}530 \ \frac{\text{N}\cdot\text{m}}{\text{s}} \times \frac{1 \ \text{lbf}}{4.448 \ \text{N}} \times \frac{1 \ \text{ft}}{0.3048 \ \text{m}} \approx 74{,}130 \ \text{ft}\cdot\text{lbf/s}$ . Dividing by 550 gives $\approx 135 \ \text{hp}$ , confirming D is correct. A is wrong because it divides $T$ by $\omega$ instead of multiplying — a fundamental formula error that produces a physically tiny result around 0.067 hp. B omits the length conversion (the 0.3048 m/ft factor), so the intermediate result is off by a factor of ~3.281, yielding an artificially low ~41 hp. C uses the torque correctly converted to ft·lbf but then plugs in the raw number 1200 (rpm) as if it were already in rad/s or rev/s, inflating the answer dramatically to ~1290 hp. Your strategy: whenever you see rpm in a power calculation, immediately convert to rad/s before doing anything else. Then treat unit conversions as a checklist — force units and length units both need to change when going from SI to US customary.
A spring has a stiffness of $k = 350 \ \text{lbf/in}$ . An engineer needs this value in SI units of N/m. Using $1 \ \text{lbf} = 4.448 \ \text{N}$ and $1 \ \text{in} = 0.0254 \ \text{m}$ , which conversion is correct, and what is the magnitude of the most common conversion error on this type of problem?
Answer: $k = 61{,}320 \ \text{N/m}$ ; the most common error is applying the inch-to-meter conversion factor to the numerator instead of the denominator, yielding $\approx 39.5 \ \text{N/m}$ , which is off by a factor of $1/0.0254 \approx 39.4$ . When converting a spring stiffness between unit systems, you must treat the units algebraically — every unit in the original expression must be converted individually. Spring stiffness has units of force/length, so you need one force conversion factor and one length conversion factor. Starting with $k = 350 \ \text{lbf/in}$ , multiply by the force conversion in the numerator and apply the length conversion in the denominator: $k = 350 \ \frac{\text{lbf}}{\text{in}} \times \frac{4.448 \ \text{N}}{1 \ \text{lbf}} \times \frac{1 \ \text{in}}{0.0254 \ \text{m}} = \frac{350 \times 4.448}{0.0254} \approx 61{,}320 \ \text{N/m}$ This confirms B is correct. The second part of B describes the classic error: accidentally placing the inch-to-meter factor in the numerator instead of the denominator. Doing so gives $350 \times 4.448 \times 0.0254 \approx 39.5 \ \text{N/m}$ , which is off by a factor of $1/0.0254^2$ relative to the correct answer divided through — equivalently, a factor of $\approx 39.4$ from misplacing that single conversion. A identifies the correct numerical answer but mischaracterizes the error. Squaring $(0.0254)^2$ in the denominator would actually make the result larger than correct, not smaller — and the error factor described doesn't match. C gets both the final answer and the error scenario wrong. $k = 1557 \ \text{N/m}$ doesn't follow from any standard partial conversion of this problem. D correctly identifies the answer but invents an implausible error. Engineers rarely invert a well-known force conversion, and $4.448^2$ as an error factor has no logical basis in this single-unit conversion. Study tip: Always write units explicitly in every step. If units don't cancel cleanly to your target unit, you've made an error before reaching the numbers.
An impact force is measured using a load cell calibrated in kN. The recorded peak force is $F = 18.6 \ \text{kN}$ . The engineer must report this force in both US tons-force (ton $_f$ ), where $1 \ \text{ton}_f = 2000 \ \text{lbf}$ and $1 \ \text{lbf} = 4.448 \ \text{N}$ , and metric tonnes-force (tonne $_f$ ), where $1 \ \text{tonne}_f = 1000 \ \text{kgf}$ and $1 \ \text{kgf} = 9.807 \ \text{N}$ . By how many percent is the force magnitude represented by $1 \ \text{tonne}_f$ greater than the force magnitude represented by $1 \ \text{US ton}_f$ ?
Answer: Approximately $10.2\%$ , because $1 \ \text{tonne}_f = 9807 \ \text{N}$ while $1 \ \text{US ton}_f = 8896 \ \text{N}$ , giving a ratio of $9807/8896 - 1 \approx 10.2\%$ . When comparing force units across measurement systems, your job is to convert each unit to a common baseline — almost always SI (Newtons) — and then compare numerically. Start by finding the Newton-equivalent of each unit. For the metric tonne-force: $1 \ \text{tonne}_f = 1000 \ \text{kgf} \times 9.807 \ \text{N/kgf} = 9807 \ \text{N}$ . For the US ton-force: $1 \ \text{ton}_f = 2000 \ \text{lbf} \times 4.448 \ \text{N/lbf} = 8896 \ \text{N}$ . The tonne $_f$ is clearly larger, and the percent difference is $\frac{9807 - 8896}{8896} \times 100 \approx 10.2\%$ . This confirms C is correct. Choice A is a fabricated procedure — there is no physical justification for squaring a ratio of gravitational accelerations when comparing force magnitudes. The 20.4% figure is simply wrong, and the method has no basis in unit analysis. Choice B is wrong because the two systems do not share the same gravitational reference. The US ton-force derives from the pound-force, which traces back to a slightly different effective gravitational value ( $\approx 9.807 \ \text{m/s}^2$ for lbf is the standard, but the pound-mass unit itself differs), and the chain of conversion constants yields a meaningfully different Newton equivalent — 8896 N vs. 9807 N. Identical they are not. Choice D reverses the inequality entirely — a classic trap. Students who conflate "US ton = 2000 lbf sounds large" with "therefore it must be bigger" skip the actual calculation. Always convert; never assume from names. Study tip: When comparing units across systems, always reduce everything to Newtons first. Unit names can be misleading — the math never is.

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