FE Other Disciplines (NCEES Fundamentals of Engineering) — All Questions
33 questions
Two forces, 30 N and 40 N, act at the same point and are perpendicular to each other. What is the magnitude of their resultant?
- a.35 N
- b.10 N
- c.50 N✓
- d.70 N
For perpendicular forces the resultant is the vector sum magnitude sqrt(30^2 + 40^2) = sqrt(900 + 1600) = sqrt(2500) = 50 N. This is the 3-4-5 right-triangle relationship. Simply adding to 70 N would only be valid if the forces were collinear.
A 100 N force is applied perpendicular to a wrench handle at a distance of 0.5 m from the bolt. What moment does it produce about the bolt?
- a.200 N·m
- b.100 N·m
- c.150 N·m
- d.50 N·m✓
Moment M = force x perpendicular distance = 100 N x 0.5 m = 50 N·m. Because the force is already perpendicular to the moment arm, no angle factor is needed.
A block rests on a horizontal surface with a normal force of 200 N and a coefficient of static friction of 0.3. What is the maximum static friction force available?
- a.600 N
- b.60 N✓
- c.200 N
- d.0.3 N
Maximum static friction F = mu x N = 0.3 x 200 = 60 N. Friction is proportional to the normal force, not the contact area, and cannot exceed this value before sliding begins.
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.
A rigid body is in static equilibrium when:
- a.Both the sum of forces and the sum of moments are zero✓
- b.Neither condition is required
- c.The sum of forces is zero only
- d.The sum of moments is zero only
Complete static equilibrium requires both the net force and the net moment (about any point) to be zero. Force balance alone prevents translation; moment balance alone prevents rotation; both are needed.
Two parallel forces of 30 N and 40 N act in the same direction on a rigid body. What is the magnitude of their resultant?
- a.1200 N
- b.50 N
- c.70 N✓
- d.10 N
For parallel forces in the same direction, the resultant is the arithmetic sum: 30 + 40 = 70 N. (Only if they opposed each other would you subtract, giving 10 N.) Its line of action lies between them, closer to the larger force.
The center of gravity and the centroid of a body coincide when:
- a.They always differ
- b.They never coincide
- c.The body is painted
- d.The material has uniform density✓
The centroid is a purely geometric center, while the center of gravity is the weight-averaged center. They coincide when the body has uniform density (and a uniform gravitational field), so weight is proportional to volume.