Statics and Dynamics Quiz: 2d Rigid Body Equilibrium

What this quiz covers

This quiz focuses on 2d Rigid Body Equilibrium, giving you a quick way to practice the rules, question types, and explanations that matter most for Statics and Dynamics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. $R_A = R_B = 12\text{ kN}$, because the symmetric tent distribution has its resultant at the midpoint of the beam, and by symmetry both reactions must be equal and each carries half the total load of $24\text{ kN}$. (correct answer)
2. $R_A = 16\text{ kN}$ and $R_B = 8\text{ kN}$, because the resultant of the left triangle acts at $x = 8/3\text{ m}$ from A, which is closer to A, so A carries more load than B.
3. $R_A = R_B = 12\text{ kN}$, but only because the roller at B cannot exert a horizontal reaction, forcing A to carry all horizontal components while vertical components split evenly regardless of load shape.
4. $R_A = 8\text{ kN}$ and $R_B = 16\text{ kN}$, because the distributed load's peak is at the midpoint and the roller at B is a more rigid support than the pin at A, attracting more of the load.

Question 2

1. Link CD exerts only a horizontal force, but link AB can exert both a force along its axis and a perpendicular force because the rigid plate allows the pin at A or B (on the plate side) to develop a transverse reaction through bending of the link itself.
2. Each link is a two-force member, but since the plate is a rigid body (not a particle), the moment equation $\sum M = 0$ must also be satisfied; this introduces a third equilibrium equation that, combined with the two force equations, may over-constrain the system if the lines of action of the two link forces and F are concurrent.
3. The links can exert both force and moment reactions because they are pin-connected to a rigid plate, which behaves differently from a truss joint; therefore the reactions are statically indeterminate without additional information about the plate's stiffness.
4. Each link is a two-force member and can only exert a force along its own axis; link AB exerts a force along its $45°$ direction and link CD exerts a purely horizontal force. Together with the applied $F$, these three concurrent forces must satisfy $\sum F_x = 0$ and $\sum F_y = 0$ for equilibrium of the plate. (correct answer)

Question 3

1. $\mathbf{F}_3 = (0\hat{i} + 2\hat{j})\text{ kN}$, and it produces a moment of $0\text{ kN·m}$ about the origin because its purely vertical line of action passes through a point whose horizontal coordinate equals its vertical coordinate, creating a self-canceling moment.
2. $\mathbf{F}_3 = (0\hat{i} + 2\hat{j})\text{ kN}$, and it produces a moment of $+2\text{ kN·m}$ (counterclockwise) about the origin, computed as $r_x F_{3y} - r_y F_{3x} = 1(2) - 1(0) = +2\text{ kN·m}$. (correct answer)
3. $\mathbf{F}_3 = (0\hat{i} - 2\hat{j})\text{ kN}$, and it produces a moment of $-2\text{ kN·m}$ (clockwise) about the origin, computed as $r_x F_{3y} - r_y F_{3x} = 1(-2) - 1(0) = -2\text{ kN·m}$.
4. $\mathbf{F}_3 = (0\hat{i} + 2\hat{j})\text{ kN}$, and it produces a moment of $-2\text{ kN·m}$ (clockwise) about the origin, because even though $r_x F_{3y} - r_y F_{3x} = +2\text{ kN·m}$, equilibrium of the whole body requires the net moment to be zero, so $\mathbf{F}_3$'s contribution must be taken as negative to balance the other forces.

Question 4

1. The claim is incorrect: with three unknowns ($T_1, T_2, T_3$) and only two independent equilibrium equations ($\sum F_y = 0$ and $\sum M = 0$), the system is indeterminate, and a compatibility condition based on the bar's deformation geometry must supplement equilibrium to solve for all three tensions uniquely. (correct answer)
2. The claim is incorrect: the system has four unknowns because the bar can also translate horizontally, requiring an additional equilibrium equation $\sum F_x = 0$ that introduces a fourth unknown reaction, making the system indeterminate to the second degree.
3. The claim is correct for this specific loading case: because the load P is applied at x = a/2 (between wires 1 and 2), the bar rotates so that wire 3 goes slack, reducing the problem to two unknowns and two equations, which equilibrium alone can solve.
4. The claim is incorrect: equilibrium gives a unique solution only if the bar is rigid and all supports are pins; since wires are flexible tension-only members, a plastic analysis using limit loads is required instead of standard equilibrium equations.

Question 5

1. $A_x = 80\text{ N}$ directed to the right, because the pin at A must push in the same direction as the applied force $H$ to maintain moment equilibrium of the column about its base, balancing the overturning tendency.
2. $A_x = 40\text{ N}$ directed to the left, because the horizontal force $H$ is applied at the midpoint of BC and is shared equally between supports A and C by symmetry, so each takes half of the $80\text{ N}$.
3. $A_x = 80\text{ N}$ directed to the left, because the roller at C provides no horizontal resistance, so applying $\sum F_x = 0$ to the whole frame requires the pin at A to supply all $80\text{ N}$ of horizontal resistance. (correct answer)
4. $A_x = 0$ because $H$ is applied to beam BC and transmitted to the column only through the internal pin at B; since the beam is horizontal, pin B transfers only a vertical force to the column, leaving no horizontal reaction required at A.