A frame has member AB pinned to member BC at B, with an external moment M applied at joint B. When drawing separate FBDs for members AB and BC, how should the applied moment M be represented?
Answer: The moment M appears on both FBDs with the same magnitude and direction
When analyzing frames with applied moments at joints, you need to understand how moments are treated when you separate connected members into individual free body diagrams (FBDs).
An applied moment is a pure couple - it has magnitude and rotational direction, but no specific point of application within the joint. Unlike forces, which act along the line connecting two members, a moment affects the entire joint region. When you cut through a pin connection to create separate FBDs, this applied moment must appear on both members because the moment was applied to the joint system as a whole, not to one specific member.
Think of it this way: the moment M exists at joint B regardless of how you choose to analyze the structure. When you separate the members for analysis, you're not changing the physical reality - you're just choosing to look at each piece individually. The same moment that was applied to the joint must still be accounted for on both sides of your imaginary cut.
Option A is incorrect because moments at joints aren't inherently "left" or "right" - they're rotational effects that influence both connected members. Option B makes the same error from the opposite direction. Option D reflects a common misconception about equilibrium - while internal forces and moments at cuts do appear as equal and opposite pairs, an externally applied moment appears identically on both FBDs.
Remember: externally applied moments at joints appear on all connected members' FBDs with the same magnitude and direction. Only cut internal moments appear as equal and opposite pairs.
In a frame with members connected by internal pins, when should you include the weight of the members in your free body diagrams?
Answer: Only when the problem statement explicitly gives the weight values of each member
When analyzing frames with pinned connections, the decision to include member weights depends entirely on the information provided and the level of precision required for your analysis.
Answer A is correct because in statics problems, you should only include member weights when the problem explicitly provides these values. This indicates that the instructor considers member weights significant enough to affect your calculations and expects you to include them in your analysis. When weights aren't given, it's typically because they're either negligible compared to applied loads or the problem is designed to focus on other concepts.
Answer B is wrong because member weights don't always significantly affect pin forces. In many engineering problems, applied loads are much larger than member weights, making the latter negligible. Including estimated weights when not given would introduce unnecessary complexity and potential errors.
Answer C incorrectly suggests that member weights cancel out due to Newton's third law. While action-reaction pairs do exist at pins, member weights act downward due to gravity and don't automatically cancel. These forces still contribute to the equilibrium equations and can create moments about various points.
Answer D is incorrect because member orientation doesn't determine whether weights should be included. Both horizontal and vertical members can contribute moments about pin connections, depending on the location of the moment center and the member's center of gravity.
Study tip: In statics problems, let the given information guide your approach. If weights are provided, include them; if not, focus on the loads that are specified.
A frame consists of three members: AB, BC, and CD, where AB and BC are connected by a pin at B, and BC and CD are connected by a pin at C. The frame supports a 500 N downward load at point B. If you need to find the force in member BC, which free body diagram(s) must you draw first?
Answer: Free body diagram of the entire frame to find reaction forces, then free body diagram of member AB to find pin forces at B
To find the force in member BC, you need to know the pin forces at B and C. Since member BC has four unknowns (two pin force components at each end) but only three equilibrium equations, it's statically indeterminate by itself. You must first find the reaction forces from the entire frame FBD, then analyze member AB (which connects a known reaction to pin B) to determine the pin forces at B. Choice B fails because member BC alone is indeterminate. Choice C is unnecessarily complex. Choice D analyzes joint B before determining what forces act there from the connected members.
A frame consists of three members: AB, BC, and BD, with internal pins at B connecting all three members. When drawing the free body diagram for member AB, how many unknown force components act at pin B?
Answer: Two components, since a pin connection always provides two reaction components
When analyzing pin connections in statics, you need to understand how forces are transmitted through the connection itself, not how many members meet at that point. A pin connection is a mechanical joint that prevents translation but allows rotation, which means it can only provide reaction forces in two perpendicular directions (typically x and y components).
For member AB's free body diagram, you're isolating this single member and examining all forces acting on it. At pin B, the pin itself exerts forces on member AB to maintain equilibrium. Since a pin can only constrain motion in two directions, it provides exactly two force components regardless of how many other members are also connected to that same pin.
Answer A correctly identifies that pin connections always provide two reaction components - this is a fundamental property of how pins work mechanically. Answer B incorrectly assumes you need one force component per connected member, but the pin acts as a single constraint on member AB. Answer C suggests four components by somehow doubling the forces from the two other members (BC and BD), which misunderstands that you're drawing forces acting on AB, not forces between every possible member combination. Answer D takes this error even further by assuming each connected member contributes two components, but again, you're only concerned with forces the pin exerts on the specific member you're analyzing.
Remember: when drawing free body diagrams, focus on the constraint type (pin, roller, fixed support), not the number of members present. Each constraint type has a fixed number of reaction components it can provide.
A frame analysis involves members AB and BC connected by an internal pin at B. After drawing FBDs and writing equilibrium equations, you find that the horizontal pin force component is Bx = -180 N. What does the negative sign indicate?
Answer: The pin force acts in the opposite direction to your assumed positive direction
In statics problems involving pin-connected frames, you must choose a positive direction for each force component when drawing free body diagrams. The sign of your calculated result tells you whether the actual force direction matches your assumption.
When you assume a positive direction for the horizontal pin force component Bx and calculate Bx=−180 N, the negative sign simply means the actual force acts opposite to your assumed positive direction. If you assumed rightward was positive, the actual force acts leftward with magnitude 180 N. This is completely normal and expected in statics analysis.
Looking at the incorrect options: Option B is wrong because negative force values are perfectly valid in statics—they're just directional indicators, not errors. The mathematics of equilibrium naturally produces both positive and negative results. Option C incorrectly applies tension/compression concepts to pin forces. Pins transmit forces in any direction and don't experience tension or compression in the same way that axial members do. Option D confuses force direction with moment direction. The sign of a force component doesn't directly indicate whether it creates clockwise or counterclockwise moments—that depends on the force's line of action relative to the moment center.
Remember this key principle: in statics, always assume directions for unknown forces, solve the equilibrium equations, and let the signs tell you the actual directions. Negative results aren't mistakes—they're the solution telling you the force acts opposite to your assumption. This sign convention is fundamental to successful frame analysis.
A frame consists of two rigid members ABC and BCD connected by a pin at B. The frame is supported by a pin at A and a roller at D. A horizontal force P is applied at point C. When drawing the free body diagram for member ABC, which statement about the pin reaction at B is correct?
Answer: The pin reaction at B has both horizontal and vertical components, acting on member ABC in the opposite direction to how they act on member BCD
When analyzing multi-member frames connected by pins, you must understand how internal forces work between connected members. A pin connection allows rotation but prevents translation, creating reaction forces that maintain equilibrium at the joint.
For the pin reaction at B, you need to apply Newton's third law: forces between two bodies are equal in magnitude but opposite in direction. Since member ABC exerts forces on member BCD through the pin at B, member BCD must exert equal and opposite forces back on member ABC. The pin reaction will generally have both horizontal and vertical components because the applied horizontal force P creates moments and forces that must be balanced throughout the entire frame system.
Looking at the wrong answers: Choice A incorrectly assumes the pin reaction direction depends solely on the applied force direction, ignoring that pins typically develop reactions in multiple directions to maintain equilibrium. Choice B makes a critical error about Newton's third law - it suggests the pin forces act in the same direction on both members, which would violate the fundamental principle that action and reaction forces are opposite. Choice D wrongly assumes the pin reaction has only a vertical component and misunderstands how forces distribute through frame members.
Choice C correctly recognizes that pin reactions typically have both horizontal and vertical components, and crucially applies Newton's third law properly - the forces that member BCD exerts on member ABC through the pin are opposite in direction to the forces that member ABC exerts on member BCD.
Remember: at any pin connection between frame members, always apply Newton's third law to determine the correct direction of internal reaction forces on each member's free body diagram.
A frame has member ABC with an internal pin at B that connects to member BD. When drawing the FBD of member ABC, where should the pin forces at B be applied?
Answer: At point B, exactly where the physical pin is located
When analyzing frames with internal pins in statics, you need to understand that forces act at specific physical locations, not at abstract mathematical points. An internal pin creates a connection point where forces are transmitted between members, and these forces must be applied exactly where the physical connection exists.
The correct approach is option B - apply the pin forces at point B, exactly where the physical pin is located. This is fundamental to proper free body diagram construction because forces in statics problems represent real physical interactions that occur at specific locations. The pin at B creates reaction forces (typically horizontal and vertical components) that act precisely at that connection point. When you isolate member ABC, you must show these pin forces acting at B to properly represent how member BD influences member ABC through their connection.
Option A is incorrect because the centroid represents the center of mass for distributed loads or weights, not the location where concentrated forces act. Option C misapplies the concept of distributed loading - pin forces are concentrated forces, not distributed ones, so they don't spread along the member's length. Option D incorrectly suggests using moment balance to determine force location, but the physical location of the pin determines where forces act, not mathematical convenience.
Remember this key principle: in statics free body diagrams, always apply forces at their actual physical locations. Pin forces act at the pin location, applied loads act at their points of application, and support reactions act at the support points. This physical accuracy is essential for correct equilibrium analysis.
When analyzing a frame with internal pins, you draw separate FBDs for members AB and BC, which are connected by a pin at B. If the pin force components on member AB are Bx = 150 N (rightward) and By = 200 N (upward), what are the corresponding pin force components on member BC?
Answer: Bx = 150 N (leftward) and By = 200 N (downward)
When you encounter internal pins in frame analysis, you're dealing with Newton's Third Law in action. Internal pins create equal and opposite force pairs between connected members, which is crucial for solving these problems correctly.
To find the pin forces on member BC, you need to apply Newton's Third Law. Since the pin at B exerts 150 N rightward and 200 N upward on member AB, member AB must exert equal and opposite forces back on the pin. By Newton's Third Law, this means member BC experiences 150 N leftward and 200 N downward from the same pin.
Think of it this way: if you push on a wall with 10 N rightward, the wall pushes back on you with 10 N leftward. The same principle applies to pin connections - the forces are equal in magnitude but opposite in direction.
Looking at the wrong answers: Choice A gives the same direction for both force components, violating Newton's Third Law entirely. Choice C incorrectly swaps the force magnitudes (150 and 200 N) while getting the directions right. Choice D correctly reverses the x-component direction but fails to reverse the y-component direction.
The correct answer is B: the pin forces on member BC are 150 N leftward and 200 N downward - exactly opposite to those on member AB.
Study tip: Always remember that internal pin forces come in action-reaction pairs. When drawing separate FBDs for pin-connected members, the pin force components must be equal in magnitude but opposite in direction on each member. This is your consistency check for internal pin problems.
A frame consists of member ABC pinned to ground at A, connected to member CDE at C via an internal pin, with member CDE pinned to ground at E. A horizontal load acts at B. When you solve for the pin reactions at A and E using the entire frame free body diagram, you find Ax=150 N, Ay=200 N, Ex=−150 N, and Ey=300 N. What should be the next step to find the pin forces at C?
Answer: Draw the free body diagram of member ABC and solve its equilibrium equations using the known reactions at A
When analyzing multi-member frames with internal pins, you must methodically isolate individual members after finding the external reactions. This systematic approach prevents you from getting overwhelmed by too many unknowns at once.
Since you already know the reactions at A (Ax=150 N, Ay=200 N), the most efficient next step is to isolate member ABC and apply equilibrium. With the known pin reactions at A and the applied load at B, you have enough information to solve for the pin forces that member CDE exerts on member ABC at point C. This gives you three equations (∑Fx=0, ∑Fy=0, ∑M=0) with three unknowns (the two pin force components at C plus any unknown loads).
Option A is incorrect because joint equilibrium at C requires knowing the internal forces in both members, which you don't have yet. Option B creates an unnecessarily complex system of six equations with six unknowns when a simpler sequential approach works better. Option D, while theoretically possible, is less efficient because you'd be working with the more complex member CDE that spans from C to E, whereas member ABC is typically simpler to analyze first.
The key strategy for frame analysis is to work sequentially: solve the overall frame for external reactions, then isolate the simplest member first (usually the one with fewer loads or simpler geometry). This builds your solution step-by-step rather than trying to solve everything simultaneously.
In a frame with members AB, BC, and CD connected by pins at B and C, what is the primary advantage of drawing separate FBDs for each member rather than analyzing the entire frame as one unit?
Answer: Separate FBDs allow determination of internal pin forces, which cannot be found from the whole frame analysis
When analyzing structural frames in statics, you have two main approaches: treat the entire frame as a single rigid body, or "cut" the frame apart and analyze each member individually. The key difference lies in what internal forces you can discover.
The correct answer is B because separate free body diagrams (FBDs) are the only way to determine internal pin forces. When you analyze the whole frame as one unit, the pin forces at B and C are internal to the system and don't appear on your FBD - they're hidden inside the structure. But when you cut the frame at each pin and draw separate FBDs for members AB, BC, and CD, those previously internal pin forces become external forces on each member's diagram. You can then use equilibrium equations (∑Fx=0, ∑Fy=0, ∑M=0) for each member to solve for these pin reactions.
Option A is wrong because accuracy depends on correct application of principles, not the method chosen. Option C is incorrect - you still need external support reactions regardless of your approach; they're essential boundary conditions. Option D misunderstands equilibrium fundamentally - you always need to write and solve equilibrium equations whether using separate FBDs or whole-frame analysis.
Study tip: Remember this key principle: to find internal forces in any structure, you must "expose" them by cutting through the structure at that point. Internal forces only become solvable when they appear as external forces on your FBD.
A frame consists of members AB and BC connected by an internal pin at B. Member AB has length 3 m and member BC has length 4 m. When drawing the FBD for member AB, you need to apply a moment equilibrium equation. Which point would be most advantageous to use as the moment center?
Answer: Point B, because it's at the internal pin and eliminates pin force components from the moment equation
When analyzing frames with internal pins, the key principle for moment equilibrium is choosing a moment center that eliminates the maximum number of unknown forces from your equation. This strategic choice simplifies your calculations and reduces the number of variables you need to solve for simultaneously.
For member AB's free body diagram, taking moments about point B is most advantageous because the internal pin forces at B create no moment about point B itself. Since these pin forces are unknown and would appear in equations if you chose any other moment center, eliminating them immediately simplifies your analysis. This leaves you with a cleaner equation involving only the known applied loads and the reaction forces at A.
Option A is incorrect because while taking moments about point A does eliminate the reaction forces at A from the moment equation, the unknown pin forces at B would still appear with their full moment arms, creating a more complex equation. Option C is wrong because choosing the midpoint of AB offers no strategic advantage—all unknown forces (both at A and B) would remain in the moment equation with their respective moment arms. Option D is incorrect because point C is not even on member AB, making it an unnecessarily complicated choice that would include all unknown forces in the equation with large moment arms.
Study tip: Always choose your moment center to eliminate the maximum number of unknown forces. For any member in a frame, taking moments about a point where multiple unknown forces act (like pins or supports) will streamline your equilibrium equations significantly.
A frame member AB has an internal pin at A connecting it to member AC, and another internal pin at B connecting it to member BD. A vertical load P is applied at the midpoint of AB. How many equilibrium equations can be written for member AB?
Answer: Three equations: two force equilibrium equations and one moment equilibrium equation
When analyzing any structural member in statics, you must identify what type of body it is and what constraints act on it. Member AB is a two-force member with an applied load, making it a rigid body subject to three equilibrium conditions.
For any rigid body in 2D statics, you can write exactly three independent equilibrium equations: ∑Fx=0, ∑Fy=0, and ∑M=0 about any point. Member AB, despite having pins at both ends and a load at its midpoint, is still one rigid body, so these three equations apply.
Let's examine why the other options miss the mark:
Option A incorrectly limits you to only force equilibrium equations, ignoring that you can also write a moment equilibrium equation. This would leave the system underdetermined since you have unknown forces at both pin connections.
Option C suggests you can write two independent moment equations. While you could write moment equations about different points, they wouldn't be independent—they're related through the force equilibrium equations. In 2D, you only get three independent equilibrium equations total.
Option D applies 3D equilibrium conditions (six equations) to a 2D problem. This is incorrect since the problem clearly involves planar loading and constraints.
The key insight is that regardless of how many forces act on a rigid body or where they're applied, you're limited to three independent equilibrium equations in 2D statics. Remember this fundamental principle: one rigid body equals three equilibrium equations, no more, no less.
In a frame analysis, member ABC is connected to member DEF at point C via a pin. After solving the equilibrium equations, you find that the pin exerts forces Cx=−300 N and Cy=400 N on member ABC. What is the most likely error if a student reports the pin forces on member DEF as Cx=−300 N and Cy=400 N?
Answer: The student forgot to apply Newton's third law; the forces on DEF should be Cx=300 N and Cy=−400 N
Newton's third law requires that the forces between connected members be equal in magnitude but opposite in direction. If member DEF exerts (-300 N, 400 N) on member ABC through the pin, then member ABC must exert (300 N, -400 N) on member DEF. The student's answer shows the forces that DEF exerts on ABC, not the forces that ABC exerts on DEF (which is what should appear on DEF's FBD). Choice A incorrectly suggests the forces should be the same. Choice C misunderstands sign conventions. Choice D incorrectly states that pin forces must be perpendicular to members.
You are analyzing a frame where member AB is connected to member BCD at point B by an internal pin. Member BCD has a right-angle bend at point C. When drawing the FBD for member AB, what information do you need about the geometry of member BCD?
Answer: Only the location of point B relative to member AB
When analyzing frames with internal pins in statics, you're dealing with the fundamental principle that a pin connection allows free rotation between members while transmitting forces. The key insight is understanding what information you actually need to draw a proper free body diagram (FBD).
For member AB's FBD, you only need to know where point B is located relative to member AB itself. The pin at B will exert forces on member AB, and these forces can be represented as components (typically horizontal and vertical). The magnitude and direction of these pin forces will be determined when you solve the equilibrium equations, but to set up the FBD, you simply need to know where on member AB these forces act - which is the location of point B.
Answer A is incorrect because you don't need the complete geometry of member BCD. The pin forces depend on equilibrium requirements, not the detailed shape of the connected member. Answer B is wrong because while the angle between members affects the overall structural behavior, it doesn't determine what you need to draw member AB's FBD - the pin forces are still just applied at point B regardless of this angle. Answer D misses the mark because while pin forces are indeed independent of member shapes in terms of their nature, you do need to know where the pin is located geometrically on your member of interest.
Remember: when drawing FBDs for members connected by pins, focus on where the pin attaches to your specific member, not the geometry of what it's connected to.
In frame analysis, you have correctly drawn FBDs for members AB and BC connected by a pin at B. You notice that the pin force components you've shown satisfy Newton's third law. However, when you solve the equilibrium equations, you get inconsistent results. What is the most likely cause?
Answer: You made a sign error or computational mistake in the equilibrium equations
When analyzing frames with pinned connections, getting inconsistent results despite correctly applying Newton's third law typically points to algebraic errors rather than conceptual mistakes. Since you've already verified that your pin force components properly satisfy Newton's third law (equal magnitude, opposite directions), your free body diagrams are likely correct.
The most probable culprit is option B - a computational error in your equilibrium equations. Frame analysis involves multiple simultaneous equations with careful attention to sign conventions. Common mistakes include: mixing up positive/negative directions when writing moment equations, incorrectly applying the right-hand rule for moments, dropping negative signs during algebraic manipulation, or making arithmetic errors when solving the system of equations.
Let's examine why the other options are less likely. Option A suggests an error with Newton's third law, but you've already confirmed these forces are correctly shown. Option C about static indeterminacy would be evident from having more unknowns than equilibrium equations available - you'd know this before solving. Option D about member deformation is irrelevant for statics problems, where we assume rigid bodies and use only equilibrium principles.
The key insight is that if your free body diagrams are conceptually correct (Newton's third law satisfied), but your solution is inconsistent, the problem lies in the mathematical execution, not the physics setup.
Study tip: Always double-check your sign conventions and work through frame problems systematically. When you get inconsistent results with correct FBDs, retrace your algebra step-by-step before questioning your conceptual approach.
In a frame analysis, you draw the FBD for member ABC with internal pins at A and C. You represent the pin forces as Ax, Ay at point A and Cx, Cy at point C. What constraint should these force components satisfy before you even solve the equilibrium equations?
Answer: The forces must be consistent with Newton's third law on the connected members' FBDs
When analyzing frames with internal pins, you're working with a system where multiple members connect at pin joints. The fundamental principle governing these connections is that forces are interactions between bodies—and interactions must be mutual.
Before you even write equilibrium equations, you must ensure that the pin forces on your free body diagrams respect Newton's third law. If member ABC exerts forces Ax and Ay on the pin at point A, then the pin (and any other member connected there) must exert equal and opposite forces -Ax and -Ay back on member ABC. This means the pin forces you draw on different members' FBDs must be consistent with each other—they represent the same physical interaction viewed from different perspectives.
Choice A is incorrect because it describes equilibrium conditions that you'll solve for later, not constraints that must exist beforehand. The sum of forces on member ABC might not be zero if external loads are present.
Choice C is wrong because pin forces at different locations have no requirement to be equal in magnitude. The force magnitudes depend on the loading and geometry, not on any geometric constraint.
Choice D misses the point because moment equilibrium is another equation you'll solve later, not a pre-existing constraint on how you represent forces.
The key insight is that Newton's third law isn't something you verify after solving—it's a fundamental constraint on how you set up your force representations. Always check that your pin force directions are consistent across all connected members' FBDs before proceeding with equilibrium equations.
When drawing the FBD for a frame member that has two internal pin connections, what is the minimum number of unknown force components that will appear on the diagram?
Answer: Four components, with each pin connection contributing two force components
When analyzing frame members with internal pin connections, you need to understand how forces are transmitted through pins and what appears on a free body diagram (FBD).
A pin connection can transmit forces in any direction within the plane, but it cannot transmit moments. This means each pin connection contributes exactly two unknown force components to your FBD - typically represented as horizontal and vertical components (like Fx and Fy) or sometimes as a magnitude and direction.
For a frame member with two internal pin connections, you'll have two pins × two force components per pin = four total unknown force components. These represent the internal forces that adjacent members exert on your chosen member through the pin connections.
Option A is incorrect because you can't simply combine all pin forces into just two resultant components - each pin location creates its own set of force components that must be shown separately on the FBD. Option B incorrectly suggests that one pin somehow contributes fewer components than the other, but both pins behave identically in terms of force transmission. Option D makes the fundamental error of assuming pins can transmit moments - they cannot. Pins are specifically designed to allow rotation, so no moment can be transmitted through them.
Remember this key principle: each pin connection always contributes exactly two unknown force components to your FBD, regardless of how many members connect at that pin. Count your pins and multiply by two to find the minimum number of unknown force components from internal connections.
In drawing FBDs for a frame analysis, you have member AB connected to member BC by an internal pin at B. You decide to use the convention where Bx represents the horizontal component of the pin force. How should you consistently apply this convention?
Answer: Use opposite signs for Bx on the two members, with the choice of which is positive being arbitrary but consistent
When analyzing frames with internal pins, you're dealing with internal forces that must satisfy Newton's third law - forces between connected members are equal in magnitude but opposite in direction. The key insight is understanding how to represent these action-reaction pairs consistently in your free body diagrams.
The correct approach is to use the same sign convention for force components on both members, where positive indicates the same global direction. For example, if you define Bx as positive pointing right globally, then you show +Bx pointing right on member AB and −Bx pointing right (which visually appears as pointing left) on member BC. This automatically ensures Newton's third law is satisfied mathematically - the forces have opposite signs but the same magnitude.
Option A is incorrect because it suggests using the same sign for the same global direction but doesn't emphasize the critical point that one member gets the positive value while the other gets the negative value. Option B creates confusion by trying to manually enforce Newton's third law through sign manipulation rather than letting the math handle it naturally. Option C leads to inconsistent analysis since different sign choices for different members would violate Newton's third law. Option D is impractical because you don't know the resultant direction beforehand, and this approach would create inconsistencies between members.
Study tip: Always establish a global coordinate system first, then apply consistent sign conventions across all members. Let the equilibrium equations naturally produce positive or negative values - this tells you the actual direction of internal forces while maintaining mathematical consistency.
A frame contains member EFG connected to member HIJ at point F through an internal pin. Both members have additional loads and constraints. A student draws free body diagrams and finds that the pin forces at F are Fx=200 N (rightward) and Fy=150 N (upward) on member EFG. What should be shown on the free body diagram of member HIJ at point F?
Answer: Fx=200 N pointing leftward and Fy=150 N pointing downward
Newton's third law requires that forces between connected members be equal in magnitude but opposite in direction. If member HIJ exerts 200 N rightward and 150 N upward ON member EFG, then member EFG must exert 200 N leftward and 150 N downward ON member HIJ. This is what appears on HIJ's free body diagram. Choice B shows the same directions (violating Newton's third law). Choice C only reverses one component. Choice D incorrectly swaps the magnitudes.
When solving a frame problem, you determine that the pin force components at joint B are Bx = 120 N and By = 160 N. Later, you want to report the magnitude and direction of the resultant pin force. What is the magnitude of the resultant force at pin B?
Answer: 200 N
When you encounter pin force components in frame analysis, you need to find the resultant using vector addition. Pin forces at joints typically have both horizontal and vertical components that must be combined to determine the total force magnitude and direction.
To find the magnitude of the resultant force, you apply the Pythagorean theorem since the x and y components are perpendicular to each other. With Bx = 120 N and By = 160 N, the magnitude is:
∣FB∣=Bx2+By2=1202+1602=14400+25600=40000=200 N
Answer A (200 N) is correct using this standard vector magnitude calculation.
Answer B (280 N) represents the error of adding the components arithmetically: 120 + 160 = 280. This is incorrect because force components are vectors, not scalars, so you cannot simply add their magnitudes.
Answer C (140 N) might result from averaging the two components (120 + 160)/2 = 140, which has no physical basis in vector analysis.
Answer D (180 N) could come from incorrectly applying the Pythagorean theorem with calculation errors or from other mathematical mistakes in the vector addition process.
Remember that whenever you have perpendicular force components, always use the Pythagorean theorem to find the resultant magnitude. The components add as vectors, not as simple arithmetic sums. This pattern appears frequently in statics problems involving joint analysis, so master this fundamental relationship early.
