FE Other Disciplines (NCEES Fundamentals of Engineering) — All Questions

A pressure gauge reads 150 kPa. If atmospheric pressure is 101.3 kPa, what is the absolute pressure?

• a.150 kPa
• b.251.3 kPa✓
• c.48.7 kPa
• d.101.3 kPa

Absolute pressure = gauge pressure + atmospheric pressure = 150 + 101.3 = 251.3 kPa. Gauge pressure is measured relative to atmosphere; absolute is measured from a perfect vacuum.

In a hydraulic press, a 100 N force is applied to a small piston of area 0.01 m^2. What force is produced on the large piston of area 0.1 m^2?

• a.10,000 N
• b.10 N
• c.1,000 N✓
• d.100 N

By Pascal's principle, pressure is equal throughout: p = 100/0.01 = 10,000 Pa. Output force = p·A_large = 10,000 x 0.1 = 1,000 N. Force multiplies by the area ratio (10x).

Incompressible water flows at 2 m/s in a 0.1 m diameter pipe that narrows to 0.05 m diameter. What is the velocity in the narrow section?

• a.1 m/s
• b.8 m/s✓
• c.16 m/s
• d.4 m/s

Continuity A1·V1 = A2·V2 with area proportional to D^2 gives V2 = V1·(D1/D2)^2 = 2 x (0.1/0.05)^2 = 2 x 4 = 8 m/s. Velocity scales with the square of the diameter ratio, not linearly.

What is the velocity head of water flowing at 10 m/s (g = 9.81 m/s^2)?

• a.1.02 m
• b.51.0 m
• c.5.10 m✓
• d.10.2 m

Velocity head = V^2/(2g) = 10^2/(2 x 9.81) = 100/19.62 = 5.10 m. It is the kinetic-energy term of Bernoulli's equation expressed as an equivalent column height.

What is the pressure head corresponding to a water pressure of 98.1 kPa (rho = 1000 kg/m^3, g = 9.81 m/s^2)?

• a.10 m✓
• b.5 m
• c.100 m
• d.1 m

Pressure head = p/(rho·g) = 98,100/(1000 x 9.81) = 10 m. Pressure head expresses pressure as an equivalent height of the fluid column.

Water (surface tension 0.072 N/m, contact angle ~0) rises in a glass tube of radius 0.001 m. What is the approximate capillary rise (rho = 1000 kg/m^3, g = 9.81 m/s^2)?

• a.7.3 mm
• b.29.4 mm
• c.14.7 mm✓
• d.1.5 mm

Capillary rise h = 2·sigma·cos(theta)/(rho·g·r) = 2 x 0.072/(1000 x 9.81 x 0.001) = 0.144/9.81 = 0.0147 m = 14.7 mm. Rise is inversely proportional to tube radius.

What is the density of air modeled as an ideal gas at 101.3 kPa and 300 K (R = 287 J/kg·K)?

• a.1.00 kg/m^3
• b.2.35 kg/m^3
• c.1.18 kg/m^3✓
• d.0.85 kg/m^3

Ideal gas law rho = p/(R·T) = 101,300/(287 x 300) = 101,300/86,100 = 1.18 kg/m^3. Temperature must be absolute (kelvin).

For fully developed laminar pipe flow at a Reynolds number of 1600, what is the Darcy friction factor?

• a.0.016
• b.0.032
• c.0.080
• d.0.040✓

For laminar flow, f = 64/Re = 64/1600 = 0.040. This exact relation applies only below Re ~2100; turbulent flow requires the Moody chart or Colebrook equation.

A solid body of specific gravity 0.6 floats freely in water. What fraction of its volume is submerged?

• a.60%✓
• b.6%
• c.100%
• d.40%

For a floating body, the submerged fraction equals the ratio of densities: rho_body/rho_fluid = 0.6, so 60% is submerged. This follows from equating weight and buoyant force.

Open-channel flow moves at 3 m/s at a depth of 1 m. What is the Froude number (g = 9.81 m/s^2)?

• a.0.31
• b.0.96✓
• c.9.4
• d.3.0

Froude number Fr = V/sqrt(g·y) = 3/sqrt(9.81 x 1) = 3/3.13 = 0.96. Since Fr < 1 the flow is subcritical. Fr compares inertial to gravitational forces in free-surface flow.

Water (density 1000 kg/m^3) flows at 4 m/s with a static pressure of 100 kPa. What is the stagnation (total) pressure?

• a.100 kPa
• b.116 kPa
• c.108 kPa✓
• d.104 kPa

Stagnation pressure = static + dynamic = p + (1/2)rho·V^2 = 100,000 + 0.5 x 1000 x 4^2 = 100,000 + 8,000 = 108 kPa. The dynamic term is the pressure recovered when flow is brought to rest.

Pipe flow at a Reynolds number of about 3000 is best described as:

• a.Supersonic
• b.Laminar
• c.Turbulent
• d.Transitional✓

Between roughly Re 2100 and 4000, pipe flow is transitional, intermittently switching between laminar and turbulent behavior. Below ~2100 it is laminar; above ~4000 it is fully turbulent.

What height of a water column produces a pressure of 50 kPa (rho = 1000 kg/m^3, g = 9.81 m/s^2)?

• a.0.51 m
• b.5.1 m✓
• c.10.2 m
• d.2.5 m

From p = rho·g·h, h = p/(rho·g) = 50,000/(1000 x 9.81) = 5.1 m. This converts a pressure into an equivalent water-column height, as read on a manometer.

By how much does hydrostatic pressure in water increase per meter of depth (rho = 1000 kg/m^3, g = 9.81 m/s^2)?

• a.0.98 kPa/m
• b.1.0 kPa/m
• c.9.81 kPa/m✓
• d.98.1 kPa/m

The pressure gradient in a static fluid is dp/dz = rho·g = 1000 x 9.81 = 9810 Pa/m = 9.81 kPa/m. Pressure grows linearly with depth regardless of container shape.

Water flows from an orifice 2 m below the free surface of a large open tank. What is the ideal jet velocity (g = 9.81 m/s^2)?

• a.19.6 m/s
• b.39.2 m/s
• c.3.13 m/s
• d.6.26 m/s✓

Torricelli's theorem: v = sqrt(2·g·h) = sqrt(2 x 9.81 x 2) = sqrt(39.24) = 6.26 m/s. It comes from applying Bernoulli's equation between the surface and the orifice.

Pipe flow has a Reynolds number of 1500. The flow regime is:

• a.Choked
• b.Turbulent
• c.Laminar✓
• d.Transitional

Re = 1500 is below the ~2100 threshold, so the flow is laminar, dominated by viscous forces with smooth, orderly streamlines and no cross-mixing.

A Newtonian fluid with dynamic viscosity 1x10^-3 Pa·s experiences a velocity gradient of 100 s^-1 near a wall. What is the shear stress?

• a.0.01 Pa
• b.0.1 Pa✓
• c.10 Pa
• d.1 Pa

Newton's law of viscosity: tau = mu·(du/dy) = 1x10^-3 x 100 = 0.1 Pa. Shear stress is proportional to the velocity gradient, with viscosity as the proportionality constant.

A gauge pressure of -20 kPa indicates that the absolute pressure is:

• a.Above atmospheric
• b.Equal to absolute zero
• c.Below atmospheric✓
• d.Equal to atmospheric

A negative gauge pressure means the absolute pressure is below atmospheric, i.e., a partial vacuum. Absolute pressure = atmospheric + gauge = 101.3 + (-20) = 81.3 kPa, still positive.

For steady incompressible flow through a pipe with a single inlet and outlet, the continuity principle requires that:

• a.Mass in exceeds out
• b.Mass out exceeds in
• c.Mass in equals mass out✓
• d.Flow depends only on pressure

Conservation of mass for steady flow requires mass flow in to equal mass flow out. For incompressible flow this also means the volumetric flow rate Q is constant along the pipe.

A pump must deliver 8 kW of hydraulic power but operates at 80% efficiency. What shaft (brake) power is required?

• a.10 kW✓
• b.12.5 kW
• c.6.4 kW
• d.8 kW

Brake power = hydraulic power/efficiency = 8/0.8 = 10 kW. Efficiency below 100% means the input shaft power always exceeds the useful hydraulic power delivered to the fluid.

A pipe fitting has a minor loss coefficient K = 1.5. What head loss does it add at a flow velocity of 3 m/s (g = 9.81 m/s^2)?

• a.0.69 m✓
• b.1.38 m
• c.0.23 m
• d.0.46 m

Minor loss h_L = K·V^2/(2g) = 1.5 x 3^2/(2 x 9.81) = 1.5 x 0.459 = 0.69 m. Fittings, valves, and bends add losses proportional to the velocity head.

The hydrostatic paradox states that pressure at the bottom of a static liquid depends:

• a.On the liquid volume
• b.Only on depth and density✓
• c.On the container shape
• d.On the surface area

Hydrostatic pressure p = rho·g·h depends only on depth and fluid density (and g), not on the container's shape or the total volume it holds. Two differently shaped vessels filled to the same depth share the same bottom pressure.

Water flows at 4 m/s through a channel of cross-sectional area 0.05 m^2. What is the volumetric flow rate?

• a.0.2 m^3/s✓
• b.2.0 m^3/s
• c.0.4 m^3/s
• d.0.02 m^3/s

Volumetric flow rate Q = A·V = 0.05 x 4 = 0.2 m^3/s. Discharge is the product of cross-sectional area and average velocity.

Two forces of 50 N each act at a common point with a 60-degree angle between them. What is the magnitude of their resultant?

• a.100 N
• b.70.7 N
• c.50 N
• d.86.6 N✓

Using the parallelogram law, R = sqrt(F1^2 + F2^2 + 2·F1·F2·cos(theta)) = sqrt(2500 + 2500 + 2(2500)(0.5)) = sqrt(7500) = 86.6 N. Simply adding to 100 N is valid only when the forces are collinear.

A 100 N force acts at 30 degrees above the horizontal. What is its horizontal component?

• a.100 N
• b.70.7 N
• c.50 N
• d.86.6 N✓

The horizontal component is F·cos(theta) = 100·cos(30) = 100 x 0.866 = 86.6 N. The vertical component F·sin(30) = 50 N; resolving forces into components is the basis of equilibrium analysis.

A horizontal force of 100 N is applied at the top of a vertical post 2 m tall. What moment does it create about the base?

• a.100 N·m
• b.50 N·m
• c.200 N·m✓
• d.150 N·m

Moment = force x perpendicular distance = 100 N x 2 m = 200 N·m. The moment arm is the perpendicular distance from the base to the line of action of the force.

Two equal and opposite parallel forces of 50 N are separated by 0.4 m, forming a couple. What is the couple moment?

• a.12.5 N·m
• b.40 N·m
• c.100 N·m
• d.20 N·m✓

A couple moment = force x separation distance = 50 N x 0.4 m = 20 N·m. A couple produces pure rotation; its moment is the same about every point in the plane.

A particle is in equilibrium under a 30 N force pointing east and a 40 N force pointing north, plus one unknown force. What is the magnitude of that unknown (equilibrant) force?

• a.10 N
• b.70 N
• c.35 N
• d.50 N✓

For equilibrium the sum of forces is zero, so the equilibrant equals and opposes the resultant of the two known forces: sqrt(30^2 + 40^2) = 50 N (a 3-4-5 triangle), directed southwest.

At a truss joint, a 45-degree diagonal member supports a downward vertical load of 10 kN with no other vertical member present. What axial force must the diagonal carry?

• a.20 kN
• b.7.07 kN
• c.14.14 kN✓
• d.10 kN

The vertical component of the diagonal must balance the load: F·sin(45) = 10 kN, so F = 10/0.707 = 14.14 kN. Method of joints resolves forces at each pin, where all members are two-force members.

A simply supported beam carries a single 20 kN load at its exact midspan. What is the vertical reaction at each support?

• a.15 kN
• b.5 kN
• c.20 kN
• d.10 kN✓

By symmetry, each support carries half the load: 20/2 = 10 kN. Symmetry of loading and geometry lets you split the total load equally without writing moment equations.

A 6 m simply supported beam carries a 12 kN point load located 2 m from the left support. What is the left reaction?

• a.12 kN
• b.8 kN✓
• c.4 kN
• d.6 kN

Summing moments about the right support: R_left x 6 = 12 x 4, so R_left = 48/6 = 8 kN. The reaction nearer the load carries the larger share; the right reaction is 4 kN.

A uniformly distributed load of 5 kN/m acts over a 4 m length. What is the magnitude of the equivalent resultant force?

• a.10 kN
• b.40 kN
• c.5 kN
• d.20 kN✓

The resultant of a uniform distributed load equals intensity x length = 5 x 4 = 20 kN, acting at the centroid (midpoint) of the loaded region. This replaces the distributed load for reaction calculations.

A triangular distributed load varies from zero to a maximum of 6 kN/m over a 3 m length. What is the total resultant force?

• a.4.5 kN
• b.18 kN
• c.9 kN✓
• d.6 kN

The resultant equals the area of the load triangle = (1/2)·base·height = 0.5 x 3 x 6 = 9 kN. It acts at the centroid of the triangle, one-third of the base from the larger end.

A triangle has a base along the x-axis and a height of 9 m. How far above the base is its centroid?

• a.2.25 m
• b.4.5 m
• c.6.0 m
• d.3.0 m✓

The centroid of a triangle lies at one-third of its height measured from the base: 9/3 = 3.0 m. Centroid locations are essential for locating resultants of distributed loads and for area moments of inertia.

A block rests on an incline whose angle is slowly increased. If the coefficient of static friction is 0.3, at what angle does the block begin to slide?

• a.30 degrees
• b.45 degrees
• c.16.7 degrees✓
• d.3.0 degrees

At impending slip on an incline, tan(theta) = mu_s, so theta = arctan(0.3) = 16.7 degrees. This angle of repose is independent of the block's weight.

A 100 N block rests on a 30-degree incline. What is the component of its weight directed along (parallel to) the incline surface?

• a.25 N
• b.50 N✓
• c.86.6 N
• d.100 N

The weight component along the incline is W·sin(theta) = 100·sin(30) = 50 N. This is the force tending to slide the block; the perpendicular component W·cos(30) = 86.6 N presses it into the surface.

A 100 N block rests on a frictionless 30-degree incline. What is the normal force exerted by the surface?

• a.86.6 N✓
• b.70.7 N
• c.50 N
• d.100 N

The normal force balances the perpendicular weight component: N = W·cos(theta) = 100·cos(30) = 86.6 N. It is less than the full weight because part of the weight acts along the incline.

A 200 N weight hangs from two symmetric cables, each making a 30-degree angle with the horizontal. What is the tension in each cable?

• a.200 N✓
• b.100 N
• c.400 N
• d.115 N

Vertical equilibrium: 2·T·sin(30) = 200, so T = 200/(2 x 0.5) = 200 N. As the cables become more horizontal (smaller angle), the required tension rises sharply.

What is the area moment of inertia of a rectangle 0.1 m wide and 0.2 m tall about its horizontal centroidal axis?

• a.1.33x10^-4 m^4
• b.3.33x10^-5 m^4
• c.6.67x10^-5 m^4✓
• d.6.67x10^-4 m^4

For a rectangle about its centroid, I = b·h^3/12 = 0.1 x 0.2^3/12 = 0.1 x 0.008/12 = 6.67x10^-5 m^4. The height is cubed, so orientation strongly affects stiffness in bending.

A 100 N block sits on a surface with static friction coefficient 0.3. A horizontal push of 20 N is applied but the block does not move. What is the friction force acting on it?

• a.0 N
• b.20 N✓
• c.50 N
• d.30 N

Static friction is a reactive force that only rises to match the applied load, up to a maximum of mu_s·N = 0.3 x 100 = 30 N. Since 20 N is below that limit, friction equals the applied 20 N and the block stays still.

Two concurrent forces, 3 N east and 4 N north, act at a point. What is the magnitude of their resultant?

• a.7 N
• b.1 N
• c.5 N✓
• d.12 N

For perpendicular components, R = sqrt(Fx^2 + Fy^2) = sqrt(3^2 + 4^2) = sqrt(25) = 5 N. This is the standard 3-4-5 right triangle; adding directly to 7 N ignores the direction.

A structural member is loaded by forces applied at only two points (a two-force member). The internal force it carries is:

• a.Only axial force✓
• b.Bending only
• c.Shear only
• d.Torsion only

For equilibrium, a two-force member must have its two forces equal, opposite, and collinear along the line joining the points, so it carries only axial tension or compression. This is why truss members are treated as axial-only.

In truss analysis, a zero-force member is one that:

• a.Fails first
• b.Doubles the applied load
• c.Carries the maximum load
• d.Carries no load under the given loading✓

A zero-force member carries no axial force for a particular loading; it is often identified at unloaded joints connecting two collinear members plus one non-collinear member. It still provides stability and prevents buckling.

A simply supported beam of span 5 m carries a uniform load of 4 kN/m over its entire length. What is the vertical reaction at each support?

• a.8 kN
• b.5 kN
• c.20 kN
• d.10 kN✓

Total load = 4 x 5 = 20 kN; by symmetry each support carries half: 20/2 = 10 kN. A symmetric uniform load splits equally between the two supports.

For a body in equilibrium under exactly three non-parallel forces, the lines of action of those forces must be:

• a.Parallel
• b.Concurrent✓
• c.Perpendicular
• d.Zero

The three-force principle states that three non-parallel forces in equilibrium must be concurrent (their lines of action meet at a single point); otherwise an unbalanced moment would exist.

Varignon's theorem states that the moment of a force about a point equals:

• a.Zero for all forces
• b.The couple moment only
• c.The force divided by distance
• d.The sum of the moments of its components about that point✓

Varignon's theorem: the moment of a force about a point equals the sum of the moments of its rectangular components about that same point. It simplifies moment calculations by resolving forces into components.

In a two-dimensional statics problem, how many reaction components does a fixed pin (hinge) support provide?

• a.3
• b.1
• c.0
• d.2✓

A 2D pin support provides two reaction components (horizontal and vertical) but no moment, because it allows free rotation. A fixed support would add a third reaction: a resisting moment.

In a two-dimensional statics problem, how many reaction components does a roller support provide?

• a.1✓
• b.0
• c.2
• d.3

A roller provides a single reaction, perpendicular to the surface it rolls on. It permits both translation along the surface and rotation, restraining motion in only one direction.

A 100 N force acts with a perpendicular moment arm of 3 m about a pivot. What moment does it produce?

• a.300 N·m✓
• b.200 N·m
• c.150 N·m
• d.100 N·m

Moment = force x perpendicular distance = 100 x 3 = 300 N·m. The perpendicular (shortest) distance from the pivot to the force's line of action is the correct moment arm.

A block is on the verge of sliding down an incline when the incline angle equals the angle of repose. If the static friction coefficient is 0.5, what is that angle?

• a.45 degrees
• b.5 degrees
• c.30 degrees
• d.26.6 degrees✓

The angle of repose satisfies tan(theta) = mu_s, so theta = arctan(0.5) = 26.6 degrees. At this angle the friction force exactly balances the gravity component along the incline.